Answer:
Ammonium hydroxide, NH₄OH
Magnesium hydroxide, Mg(OH)₂
Sodium hydroxide, NaOH
Lithium hydroxide, LiOH
Explanation:
A base is a substance which neutralizes acids to produce salt and water. Bases are hydroxide or oxides of metals. Bases may be soluble or insoluble in water. Bases generally have a bitter taste and turn red litmus paper or indicator red.
Alkalis are bases which are soluble in water. They form the hydroxide of the alkali metals or alkaline earth metals in solution and they ionize to produce hydroxide ions. They are slippery to touch and turn red litmus blue being bases.
Therefore, all alkalis are bases but not all bases are alkalis. Insoluble bases are not alkalis.
From the given chemical compounds the alkalis present in the list are:
Ammonium hydroxide, NH₄OH; since it is soluble in water and produces hydroxide ions
Magnesium hydroxide, Mg(OH)₂; since it is slightly soluble in water and produces hydroxide ions
Sodium hydroxide, NaOH; since it is soluble in water and produces hydroxide ions
Lithium hydroxide, LiOH; since it is soluble in water and produces hydroxide ions
CuO(copper oxide) is a base but not an alkali as it does not produce hydroxide ions.
Zn(OH)2 (zinc hydroxide) is amphoteric and is insoluble
MgO(magnesium oxide) is a base but not an alkali as it does not produce hydroxide ions.
Na2O(sodium oxide) is a base but not an alkali as it does not produce hydroxide ions.
CoO(cobalt oxide) is a base but not an alkali as it does not produce hydroxide ions.
Label each formula and name pair as correct or incorrect.
Formula Name Correct/Incorrect
Aluminum tribromide
Sulfur dioxide
Beryllium hydride
Magnesium(II) oxide
Copper(II) oxide
Calcium sulfate
Nitric acid
Answer:
Aluminum tribromide: AlBr₃, however, it should be just aluminum bromide.
Sulfur dioxide: SO₂.
Beryllium hydride: BeH₂
Magnesium(II) oxide: MgO; however the roman numeral is not used in Mg as it just has one oxidation number.
Copper(II) oxide: CuO.
Calcium sulfate: CaSO₄
Nitric acid: HNO₃.
Explanation:
Hello there!
In this case, it seems that the formulas were not given, however, we can write the correct one for each given compound according to the widely used nomenclature rules as shown below:
Aluminum tribromide: AlBr₃, however, it should be just aluminum bromide.
Sulfur dioxide: SO₂.
Beryllium hydride: BeH₂
Magnesium(II) oxide: MgO; however the roman numeral is not used in Mg as it just has one oxidation number.
Copper(II) oxide: CuO.
Calcium sulfate: CaSO₄
Nitric acid: HNO₃.
Regards!
A scientist collects a sample that has 2.00 × 1014 molecules of carbon dioxide gas.How many grams is this, given that the molar mass of CO2 is 44.01 g/mol?
Answer:
1.46 × 10⁻⁸ g
Explanation:
Step 1: Given data
Molecules of CO₂: 2.00 × 10¹⁴ molecules
Step 2: Convert molecules to moles
We need a conversion factor: Avogadro's number. There are 6.02 × 10²³ molecules in 1 mole of molecules.
2.00 × 10¹⁴ molecules × 1 mol/6.02 × 10²³ = 3.32 × 10⁻¹⁰ mol
Step 3: Convert moles to mass
We need a conversion factor: the molar mass. The molar mass of CO₂is 44.01 g/mol.
3.32 × 10⁻¹⁰ mol × 44.01 g/mol = 1.46 × 10⁻⁸ g
Hydrofluoric acid is used in the preparation of numerous pharmaceuticals (e.g., Prozac) and industrial materials (e.g., Teflon). It can be produced by the reaction of hydrogen and fluorine gases. Starting with initial concentrations of 1.69 M for H2 and 1.69 M for F2, what would be the equilibrium concentration of HF? The equilibrium constant for the reaction Kc = 115.
a. 2.85 M.
b. 4.00 M.
c. 0.85 M.
d. 3.37 M.
e. 1.69 M.
Answer:
the equilibrium concentration of HF is 2.85 M
Option a) 2.85 M is the correct answer.
Explanation:
Given the data in the question;
H₂ + F₂ ⇄ 2HF
I 1.69 M 1.69 M 0
C -x -x +2x
E 1.69-x 1.69-x +2x
given that Kc = 115
Kc = [ HF ]² / [H₂][F₂]
we substitute
115 = [ 2x ]² / [ 1.69-x ][ 1.69-x ]
lets find the square root of both sides
10.7238 = 2x / [ 1.69-x ]
10.7238[ 1.69-x ] = 2x
18.123222 - 10.7238x = 2x
2x + 10.7238x = 18.123222
12.7238x = 18.123222
x = 18.123222 / 12.7238
x = 1.424356
Hence, equilibrium concentration of HF = 2x
that is;
HF = 2 × 1.424356
HF = 2.8487 ≈ 2.85 M
Therefore, the equilibrium concentration of HF is 2.85 M
Option a) 2.85 M is the correct answer.
What is the concentration of s solution that contains 55 mL of alcohol per 145 mL solution?
Answer:
37.9% v/v
Explanation:
Since both the alcohol and solution are presumed to be liquid, this concentration can be expressed as a volume concentration (or % v/v):
volume concentration = volume of solute / volume of solution
[tex]\% v/v = 55/145= 0.379[/tex]
Describe why corrosion is a natural process
Answer :
Answer :because it happens due to moisture and oxygenWhat type of organic compound is a reactant in all substitution reactions?
1.alkyne
2.alkane
3.alkene
The organic compounds that are capable of being a reactant in all substitution reactions would belong to the alkane group.
Alkanes and substitution reactionsAlkanes are saturated compounds that ordinarily will not participate in addition reactions due to their saturated nature.
Thus, alkanes are only able to participate in substitution reactions involving the substitution of one or more of their component atoms for another atom.
This is unlike alkyne and alkenes which are naturally unsaturated. The unsaturation makes them a candidate for additional reactions.
More on saturation and substitution reactions can be found here: https://brainly.com/question/3735006
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Ammonium sulfate (NH4)2SO4 is made by reacting 25.0 L of 3.0 mol/L H2SO4 with 3.1× 103 L of NH3 at a pressure of 0.68 atm and a temperature of 298 K according to the following reaction .
NH3(g) + H2SO4(aq) → (NH4)2SO4 (aq)
How many grams of ammonium sulfate are produced?
Answer: The mass of [tex](NH_4)_2SO_4[/tex] produced is 9910.5 g
Explanation:
For [tex]H_2SO_4[/tex]:Molarity is calculated by using the equation:
[tex]\text{Molarity}=\frac{\text{Moles}}{\text{Volume}}[/tex] ......(1)
Molarity of [tex]H_2SO_4[/tex] = 3.0 M
Volume of solution = 25.0 L
Putting values in equation 1, we get:
[tex]\text{Moles of }H_2SO_4=(3.0mol/L\times 25.0L)=75mol[/tex]
For [tex]NH_3[/tex]:The ideal gas equation is given as:
[tex]PV=nRT[/tex] .......(2)
where,
P = pressure of the gas = 0.68 atm
V = volume of gas = [tex]3.1\times 10^3L[/tex]
n = number of moles of gas = ? moles
R = Gas constant = 0.0821 L.atm/mol.K
T = temperature of the gas = 298 K
Putting values in equation 2, we get:
[tex]0.68atm\times 3.1\times 10^3L=n\times 0.0821L.atm/mol.K\times 298K\\\\n=\frac{0.68\times 3.1\times 10^3}{0.0821\times 298}=86.16mol[/tex]
For the given chemical equation:
[tex]NH_3(g)+H_2SO_4(aq)\rightarrow (NH_4)_2SO_4(aq)[/tex]
By stoichiometry of the reaction:
If 1 mole of [tex]H_2SO_4[/tex] reacts with 1 mole of [tex]NH_3[/tex]
So, 75 moles of [tex]H_2SO_4[/tex] will react with = [tex]\frac{1}{1}\times 75=75mol[/tex] of [tex]NH_3[/tex]
As the given amount of [tex]NH_3[/tex] is more than the required amount. Thus, it is present in excess and is considered as an excess reagent
Thus, [tex]H_2SO_4[/tex] is considered a limiting reagent because it limits the formation of the product.
By the stoichiometry of the reaction:
If 1 mole of [tex]H_2SO_4[/tex] produces 1 mole of [tex](NH_4)_2SO_4[/tex]
So, 75 moles of [tex]H_2SO_4[/tex] will produce = [tex]\frac{1}{1}\times 75=75mol[/tex] of [tex](NH_4)_2SO_4[/tex]
The number of moles is defined as the ratio of the mass of a substance to its molar mass. The equation used is:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
We know, molar mass of [tex](NH_4)_2SO_4[/tex] = 132.14 g/mol
Putting values in above equation, we get:
[tex]\text{Mass of }(NH_4)_2SO_4=(75mol\times 132.14g/mol)=9910.5g[/tex]
Hence, the mass of [tex](NH_4)_2SO_4[/tex] produced is 9910.5 g
The rate of the reaction is 1.6*10-2 M/s when the concentration of A is 0.15 M. Calculate the rate constant if the reaction is (a) first order in A and (b) second order in A.
Answer:
[tex]k_1=0.107s^{-1} \\\\k_2=0.711M^{-1}s^{-1}[/tex]
Explanation:
Hello there!
In this case, according to the given information and the attached picture in which we can see the units of the rate constant, it turns out possible for us to realize the two called rate laws are:
[tex]r=k[A]\\\\r=k[A]^2[/tex]
The former is first-order and the latter second-order; in such a way, we solve for the rate constant in both cases to obtain the following:
[tex]k=\frac{r}{[A]}=\frac{1.6x10^{-2}M/s}{0.15M}=0.107s^{-1} \\\\k=\frac{r}{[A]^2}=\frac{1.6x10^{-2}M/s}{(0.15M)^2}=0.711M^{-1}s^{-1}[/tex]
Regards!
Spell out the full name of the compounds
Help plz
Answer:
propanal
Explanation:
hope this helps :)
All the properties listed below are characteristic of the transition elements except __. a) most are paramagnetic b) most are colored c) most have high electronegativities d) most have multiple oxidation states e) most form many different complexes
Answer:
c) most have high electronegativities
Explanation:
They tend to have high electric CONDUCTIVITY because of the free-flowing d-orbital electrons, but have low electron affinity, ionization energy, and electronegativities.
Fe có tác dụng với HCL không
Fe có tác dụng với Hcl
Fe + 2hcl -> fecl2 +h2
Answer:
có
Explanatio:
Solution A has a pH of 7, and solution B has a pH of 14. Which statement
best describes these solutions?
Answer:
Option A. Solution B is basic, and solution A is neutral.
Explanation:
The pH of a solution is simply defined as the measure of acidity or alkalinity of a solution.
The pH scale ranges from 0 to 14 with the following readings:
0 to 6 => Acidic solution
7 => Neutral solution
8 to 14 => Alkaline / basic solution.
From the above, we understood that solutions with pH ranging from 0 to 6 are acidic solutions. Those with pH of 7 are neutral solutions while those with pH ranging from 8 to 14 are basic solutions.
With the above information in mind, let us answer the question given above. This is illustrated below:
pH of solution A = 7
pH of solution B = 14
Solution A has a pH of 7. This implies that solution A is a neutral solution
Solution B has a pH of 14. This implies that solution B is a basic solution.
Thus, option A gives the correct answer to the question.
Balance each of the following equations. Then, drag and drop each equation to match the coefficient of H2O in the balanced chemical equation. A coefficient for water may be used once, more than once, or not at all. Drag and drop your selection from the following list to complete the answer:
C2H5OH + O2 + CO2 + H2O NH3 + O2 + NO2 + H20 C3H2 + O2 + CO2 + H2O H2SO4 + NaOH → Na2SO4 + H20 NO2 + H2O → HNO3 + NO
Answer:
C₂H₅OH + 3 O₂ → 2 CO₂ + 3 H₂O
2 NH₃ + 3.5 O₂ → 2 NO₂ + 3 H₂O
C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O
H₂SO₄ + 2 NaOH → Na₂SO₄ + 2 H₂O
3 NO₂ + H₂O → 2 HNO₃ + NO
Explanation:
We will balance the equation using the trial and error method.
C₂H₅OH + O₂ → CO₂ + H₂O
1) We balance C atoms by multiplying CO₂ by 2 and H atoms by multiplying H₂O by 3.
C₂H₅OH + O₂ → 2 CO₂ + 3 H₂O
2) We balance O atoms by multiplying O₂ by 3.
C₂H₅OH + 3 O₂ → 2 CO₂ + 3 H₂O
NH₃ + O₂ → NO₂ + H₂O
1) We balance H atoms by multiplying NH₃ by 2 and H₂O by 3.
2 NH₃ + O₂ → NO₂ + 3 H₂O
2) We balance N atoms by multiplying NO₂ by 2.
2 NH₃ + O₂ → 2 NO₂ + 3 H₂O
3) We balance O atoms by multiplying O₂ by 3.5
2 NH₃ + 3.5 O₂ → 2 NO₂ + 3 H₂O
C₃H₈ + O₂ → CO₂ + H₂O
1) We balance C atoms by multiplying CO₂ by 3 and H atoms by multiplying H₂O by 4.
C₃H₈ + O₂ → 3 CO₂ + 4 H₂O
2) We balance O atoms by multiplying O₂ by 5.
C₃H₈ + 5 O₂ → 3 CO₂ + 4 H₂O
H₂SO₄ + NaOH → Na₂SO₄ + H₂O
1) We balance Na atoms by multiplying NaOH by 2.
H₂SO₄ + 2 NaOH → Na₂SO₄ + H₂O
2) We balance H and O atoms by multiplying H₂O by 2.
H₂SO₄ + 2 NaOH → Na₂SO₄ + 2 H₂O
NO₂ + H₂O → HNO₃ + NO
1) We balance H atoms by multiplying HNO₃ by 2.
NO₂ + H₂O → 2 HNO₃ + NO
2) We balance N atoms by multiplying NO₂ by 3.
3 NO₂ + H₂O → 2 HNO₃ + NO
You have run an experiment studying the effects of the molecular weight of a compound on the rate of diffusion in agar. Compound X has a molecular weight of 25.3 g/mol and compound Y has a molecular mass of 156.2 g/mol. On two separate agar plates, 0.1 g of each substance were transferred and allowed to diffuse for 2 hours. The results below were obtained. Use this information about this situation to answer the following three questions.
a. What was the independent variable in this experiment?
b. What was the independent variable in this experiment?
c. List two controls that were held constant for this experiment or that you would hold constant for this experiment
Answer:
A) Molecular weight
B) Rate of diffusion
C) Agar plates
Time of diffusion
Explanation:
A) The independent variable is the molecular weight because it is the variable that the researcher changes at will to aid his research. It doesn't depend on other variables.
B) The dependent variable is "Rate of diffusion because it depends on the molecular weight of the compound.
C) A control variable is one that would be held constant throughout the research.
In this case, the agar plates and the time of 2 hours diffusion remain the same throughout.
Which of the following choices is a source of groundwater pollution?
O sewage
very warm water
O silt
O All of these choices are correct.
Answer:
very warm water consuving
A company manufacturing KMnO4 wants to obtain the highest yield possible. Two of their research scientists are working on a technique to increase the yield.
Both scientists started with 50.0 g of manganese oxide.
What is the theoretical yield of potassium permanganate when starting with 50.0 g MnO2?
The equation for the production of potassium permanganate is as follows:
2 MnO2 + 4 KOH + O2 → 2 KMnO4 + 2 KOH + H2
Answer:
The theoretical yield potassium permanganate, KMnO₄ when starting with 50.0 g MnO₂ is 90.8 g
Explanation:
Molar mass of MnO₂ = (55 + 2 × 16) = 87.0 g/mol
Molar mass of KMnO₄ = (39 + 55 + 4 × 16) = 158 g/mol
Moles of MnO₂ in 50 g = reacting mass / molar mass
where reacting mass = 50 g
Moles of MnO₂ in 50 g = 50 g /87 g/mol = 0.575 moles
The equation for the production of potassium permanganate is as follows:
2 MnO2 + 4 KOH + O2 → 2 KMnO4 + 2 KOH + H2
From the equation of the reaction above, 2 moles of MnO₂ produces 2 moles of KmNO₄. The mole ratio of MnO₂ to KMnO₄ is 1 : 1
Therefore, 0.575 moles of MnO₂ will produce theoretically 0.575 moles of KMnO₄
Mass of 0.575 moles of KMnO₄ = number of moles × molar mass
Mass of 0.575 moles of KMnO₄ = 0.575 moles × 158 g/mol = 90.8 g of KMnO₄
Therefore, the theoretical yield potassium permanganate when starting with 50.0 g MnO₂ is 90.8 g
Select the net ionic equation for the reaction that occurs when magnesium sulfate and nickel(II) nitrate are mixed.
a. Ni2+(aq) + SO4^2- → NISO2 (s) + O2 (g).
b. Mg2+(aq) + 2NO3 (aq) → Mg(NO3)2(s).
c. Mg2+(aq) + NO3- → MgNO3 (s).
d. Mg2+(aq) + SO4^2- (aq) + Ni2+ (aq) + 2NO3- → Mg2+ (aq) + 2NO3 (aq) + NISO4 (s).
e. Ni2+(aq) + SO4^2- (aq) → NISO4 (s).
f. No reaction occurs.
Answer:
No reaction occurs.
Explanation:
The molecular reaction is as follows;
MgSO4(aq) + Ni(NO3)2(aq) ----> Mg(NO3)2(aq) + NiSO4(aq)
We can see from the reaction above that the both products of the reaction are soluble. Recall that a double replacement reaction often yields one insoluble product which separates as a precipitate.
This reaction does not occur since the two products that ought to be obtained are soluble in water.
Suppose a piece of silver jewelry contains 7.49x10^22 atoms of silver (Ag). how many moles of silver are in the jewelry?
Answer:
0.124 mol Ag
General Formulas and Concepts:
Atomic Structure
MolesStoichiometry
Using Dimensional AnalysisExplanation:
Step 1: Define
[Given] 7.49 × 10²² atoms Ag
[Solve] mol Ag
Step 2: Identify Conversions
Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.
Step 3: Convert
[DA] Set up: [tex]\displaystyle 7.49 \cdot 10^{22} \ atoms \ Ag(\frac{1 \ mol \ Ag}{6.022 \cdot 10^{23} \ atoms \ Ag})[/tex][DA] Divide [Cancel out units]: [tex]\displaystyle 0.124377 \ mol \ Ag[/tex]Step 4: Check
Follow sig fig rules and round. We are given 3 sig figs.
0.124377 mol Ag ≈ 0.124 mol Ag
Topic: AP Chemistry
5-Analysis of a 12.04-g sample of a liquid compound composed of carbon, hydrogen, and nitrogen showed it to contain 7.34 g C, 1.85 g H, and 2.85 g N. What is the percent composition of this compound
Answer:
%C = 60.9%; %H = 15.4%; %N = 23.7%
Explanation:
Step 1: Given data
Mass of the sample (m): 12.04 gMass of Carbon (mC): 7.34 gMass of Hydrogen (mH): 1.85 gMass of Nitrogen (mN): 2.85 gStep 2: Calculate the percent composition of this compound
To calculate the percent by mass of any element (E), we will use the following expression.
%E = mE/m × 100%
%C = 7.34 g/12.04 g × 100% = 60.9%
%H = 1.85 g/12.04 g × 100% = 15.4%
%N = 2.85 g/12.04 g × 100% = 23.7%
When stirred in 30°C water, 5 g of powdered potassium bromide, KBr, dissolves faster than 5 g of large crystals of potassium bromide. Which of the following best explains why the powdered KBr dissolves faster?
A. Potassium ions and bromide ions in the powder are smaller than potassium ions and bromide ions in the large crystals.
B. Powdered potassium bromide exposes more surface area to water molecules than large crystals of potassium bromide.
C. Fewer potassium ions and bromide ions have been separated from each other in the powder than in the crystals.
D. Powdered potassium bromide is less dense than large crystals of potassium bromide.
Answer:
B
Explanation:
Do diện tích tiếp xúc ở dạng bột cao hơn dạng tinh thể
When zinc nitrate is heated, zinc oxide, nitrogen dioxide(NO2) and oxygen gas are
produced.
i.
Calculate the mass of Zinc oxide produced if 38.5 g of zinc nitrate is heated.
ii.
Determine the volume of Nitrogen dioxide gas evolved at rtp
Answer:
i) 16.5g of ZnO
ii) 9.8 dm³ of NO2
Explanation:
The working is shown in the photo so kindly refer to it
Compare the solubility of silver iodide in each of the following aqueous solutions:
a. 0.10 M AgCH3COO
b. 0.10 M NaI
c. 0.10 M KCH3COO
d. 0.10 M NH4NO3
1. More soluble than in pure water.
2. Similar solubility as in pure water.
3. Less soluble than in pure water.
Answer:
Compare the solubility of silver iodide in each of the following aqueous solutions:
a. 0.10 M AgCH3COO
b. 0.10 M NaI
c. 0.10 M KCH3COO
d. 0.10 M NH4NO3
1. More soluble than in pure water.
2. Similar solubility as in pure water.
3. Less soluble than in pure water.
Explanation:
This can be explained based on common ion effect.
According to common ion effect the solubility of a sparingly soluble salt decreases further in a solution which has a common ion to it.
The solubility of AgI(s) silver iodide in water is shown below:
[tex]AgI(s) <=> Ag^{+}(aq)+I^{-}(aq)\\[/tex]
a. a. 0.10 M AgCH3COO has a common ion Ag+ with AgI.
So, AgI is less soluble than in pure water in this solution.
b. 0.10 M NaI has a common ion I- with AgI.
So, AgI is less soluble than in pure water in this solution.
c. 0.10 M KCH3COO:
This solution has no common ion with AgI.
So, AgI has similar solubility as in pure water.
d. 0.10 M NH4NO3:
In this solution, AgI can be more soluble than in pure water.
A recipe calls for 1/4 cup of butter. One ounce (oz) of butter is equivalent to 2 tablespoons (T), and there are 8 oz of butter in 1 cup. Which of the following statements correctly interpret the information provided?
a. The information provided gives the conversion factor 8oz/2T.
b. The information provided gives the conversion factor 8oz/1 cup.
c. The recipe requires 64 T of butter.
d. The recipe requires 4T of butter.
Answer:
D
Explanation:
8oz=1cup
Xoz=1/4cup
cross multiply
Xoz=8×1/4=2
2oz=1/4cup
1oz=2T
2oz=x
cross multiply
×=2×2=4T
The cup of butter, number of tablespoons and ounces of butter are all in direct proportion with one another. The interpretation of the information provided is:
(b). The information provided gives the conversion factor 8oz/1 cup.
Given that:
[tex]1\ recipe = \frac{1}{4}\ cup[/tex]
[tex]1\ oz = 2T[/tex]
[tex]8\ oz = 1\ cup[/tex]
Multiply both sides of [tex]1\ oz = 2T[/tex] by 8
[tex]8 \times 1oz = 8 \times 2T[/tex]
[tex]8 oz = 16T[/tex]
Substitute 8oz for 1 cup in: [tex]1\ recipe = \frac{1}{4}\ cup[/tex]
[tex]1\ recipe = \frac{1}{4} \times 8oz[/tex]
Substitute [tex]8 oz = 16T[/tex]
[tex]1\ recipe = \frac{1}{4} \times 16T[/tex]
[tex]1\ recipe = 4T[/tex]
The interpretations are as follows:
[tex]1\ recipe = 4T[/tex] means that 1 recipe requires 4T of butter [tex]8\ oz = 1\ cup[/tex] means that the information provided gives a conversion of 8oz/1 cupHence, (b) is correct
Read more about direct proportions at:
https://brainly.com/question/1266676
Name the following compound: Cuzs
O sulfur copperide (ll)
O sulfur copperide (1)
O copper(I) sulfide
copper(ll) sulfide
Answer:
THE ANSWER IS: copper(I) sulfide.
hope this helped <3
Explanation:
Suppose you are trying to separate three Proteins using Gel-Filtration chromatography. The sizes of each are given below:
Protein A: 1200 kDa
Protein B: 2000 kDa
Protein C: 800 kDa
Which protein will be the first to emerge from the column?
Answer: The correct answer is Protein B.
Explanation:
Gel-filtration chromatography is a separation technique that is based on the size of the molecules in a compound. It is also known as size-exclusion chromatography in which the eluent (carrier) used is an aqueous solution.
The matrix that is used is a porous material. When the sample is inserted in the column, the smaller particles interact strongly with the matrix than the large ones. Thus, as the eluent is passed through the matrix, larger molecules come out first, and the smallest molecule comes out last.
Given sizes of the proteins:
Protein A: 1200 kDa
Protein B: 2000 kDa
Protein C: 800 kDa
As protein B has the largest size of all the given proteins, it will emerge out first from the column.
Hence, the correct answer is Protein B.
The formula for europium oxide is Eu203. On the basis of this information, the formula for the chlorate of europium would be expected to be
Answer:
Eu(ClO3)3
Explanation:
The chlorate ion is written as follows, ClO⁻ ₃. We can see from this that the ion is univalent.
From the formula, Eu203, it is easy to see that the europium ion is trivalent.
Hence, when a compound is formed between the europium ion and chlorate ion, the compound will be written as Eu(ClO3)3.
This is so because, when ionic compounds are formed, there is an exchange of valence between the ions in the compound. This gives the final formula of the ionic substance.
What are the characteristics of an acid-base neutralization reaction?
Boiling point-methanol (65.0) 66.8c.Boiling point-unknown (record from video)——-c
Identify of unknown:
Possibilities are:Mathanol65.0c;Ethanol 78.5c; Acetone 56.0C
Perform the following
mathematical operation, and
report the answer to the correct
number of significant figures.
7.125 x 8.00 = [?]
Answer ?
Soda contains phosphoric acid (H3PO4). To determine the concentration of phosphoric acid in 50.0 mL of soda, the available phosphate ions are precipitated with excess silver nitrate as silver phosphate (418.58 g/mol). The dry Ag3PO4 is found to have a mass of 0.0576 g. What is the concentration of phosphoric acid in the soda?
Answer:
0.0270w/v% H3PO4 in the soda
Explanation:
All phosphates reacts producing Ag3PO4. To solve this question we must convert the mass of Ag3PO4 to moles. These moles = moles of H3PO4. We can find, thus, the mass of H3PO4 and the w/v% as follows:
Moles Ag3PO4 -Molar mass: 418.58g/mol-
0.0576g * (1mol / 418.58g) = 1.376x10⁻⁴ moles Ag3PO4 = moles H3PO4
Mass H3PO4 -Molar mass: 97.994g/mol-
1.376x10⁻⁴ moles Ag3PO4 = moles H3PO4 * (97.994g/mol) = 0.0135g H3PO4
w/v%:
0.0135g H3PO4 / 50.0mL * 100 =
0.0270w/v% H3PO4 in the soda