i’m so lost please help

Answer:

Invasive species are an organism that causes ecological or economic harm in a new environment where it's not native.

Explanation:

An invasive species can harm both the natural resources in the ecosystem as well it threaten the human use of these resources and invasive species can be introduced to a new area via the ballast water of oceangoing ships, intentional and accidental releases of aquaculture species, aquarium specimens or bait, and etc.

Invasive species is capable of causing extinctions to native plants and animals, reducing biodiversity, competing with native organisms for limited resources, and altering habitats. This can also result a huge economic impacts and fundamental disruptions of coastal and the great lakes of the ecosystems.

I hope it helps you.

Answer:

d) this is an exothermic reaction.

Explanation:

The reaction between HCl and KOH results in an increase in temperature in the solution. Select the correct statement from the list below.

a) this is an endothermic reaction . NO. This would cause a decrease in the temperature of the solution.

b) this is a phase change reaction . NO. All the species remain in the aqueous phase.

c) this is a vaporization reaction . NO. All the species remain in the aqueous phase.

d) this is an exothermic reaction. YES. The reaction releases heat, so it is exothermic.

Answer: it is B

Explanation:

Answer:

α-iron (BCC) has faster diffusion rate because of lower values in activation energy and pre-exponential value.

Explanation:

Taking each parameters or data at a time, we can determine the values/a constant for each parameters in the diffusion coefficient equation.

For α-iron (BCC), the diffusion coefficient = pre-exponential value,Ao × e^( -Activation energy,AE)/gas constant,R × Temperature.

Converting the given Temperature, that is 900°C to Kelvin which is equals to 1173.15K.

For α-iron (BCC), the pre-exponential value, Ao = 1.1 × 10^-6, and the activation energy, AE = 87400.

Thus, we have that the diffusion coefficient = 1.1 × 10^-6 × e(-87400)/1173.15 × 8.31.

Diffusion coefficient for α-iron (BCC) = 1.41 × 10^-10 m^2/s.

Also, For the γ-iron (FCC), the pre-exponential value, Ao = 2.3 × 10^-5 and the activation energy, AE = 148,00.

From these values we can see that both the exponential value, Ao and the activation energy for γ-iron (FCC) are higher than that of α-iron (BCC).

Thus, the diffusion coefficient for the γ-iron (FCC) = 2.3 × 10^-5 × e ^-(14800)/8.31 × 1173.15.

Then, the diffusion coefficient for the γ-iron (FCC) = 5.87 × 10^-12 m2/s.

Therefore, there will be faster diffusion in α-iron (BCC) because of lower activation energy and vice versa.

Answer:

We are given:

mass of the block = 3500 grams

volume of the block = 1000 mL

Finding the density:

Density = mass of the object (in grams) / volume of the object (in mL)

Density = 3500 / 1000

Density = 3.5 g / mL

Answer:

a) common ion effect

b) solubility

c) saturated solution

d) solubility product constant

e) molar solubility

Explanation:

When a substance, say BA2 is dissolved in a solution and another substance CA2 is dissolved in the same solution. The solubility of BA2 is decreased due to the addition of CA2. This is known as common ion effect.

The mass of a substance that will dissolve in a given Volume of solvent is known as it's solubility.

The molar solubility is the amount of moles of solvent that dissolves in 1 dm^3 of solvent.

A solution that contains just as much solute as it can normally hold at a given temperature is known as a saturated solution.

Lastly, the product of molar solubilites raised to the power of the molar coefficient is know as the solubility product constant.

The correct matches and their explanation are:

a) A decrease in the solubility of an ionic compound as a result of the addition of a common ion: Option 4. common ion effect

b) The mass of salt in grams that will dissolve in 100 mL of water: Option 1. solubility

c) A solution that has dissolved the maximum amount of a compound at a given temperature. Any further addition of salt will remain undissolved: Option 5. saturated solution

d)  The product of the molarities of the dissolved ions, raised to a power equal to the ion's coefficient in the balanced chemical equation: Option 3. solubility product constant

e) The maximum number of moles of a salt that will dissolve in 1 L of solution: Option 2. molar solubility

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Answer:

768g

Explanation:

We can use to formula [tex]N(A) = N_0(\frac{1}{2})^\frac{t}{t_{1/2}}[/tex] . Here, N(A) is the final amount. N0 is the initial amount. t is the time elapsed, and [tex]t_{1/2}[/tex] is the half life. Plugging in, we get the answer above.

The half life of I -137 is 8.07 days. if 24 grams are left after 40.35 days, 800 gram  were in the original sample.

The half-life (symbol t12) is the amount of time it takes for a volume (of material) to be reduced to half of its original value. In nuclear physics, the phrase is typically used to indicate how rapidly unstable atoms experience radioactive decay or even how long stable nuclei survive.

The phrase is sometimes used more broadly to describe any form of exponential (or, in rare cases, non-exponential) decay. The biological ½ of medications and other compounds in the human body, for example, is referred to in the medical sciences. In exponential growth, the inverse of half-life is doubling time.

ln P = kt + C

P = amount of I-137 at time t

C = constant

k = 1/time

t = time

1st condition:

P = Po, t = 0 days

2nd condition: (half-life)

P = 0.5Po, t = 8.07 days

3rd condition:

P = 25 grams, t = 40.35 days

Po = 800 grams

mass of I-137 = 800 gram

Therefore, 800 gram were in the original sample.

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Answer:

1L of hot water just below the Boling point

Explanation:

asking questions is best to learn please ask more questions

Answer:

its 1kg of boiling water

Explanation:

Answer:

[tex]Y=75.6\%[/tex]

Explanation:

Hello.

In this case, since no information about the reacting hydrogen is given, we can assume that it completely react with the 28.0 g of acetylene to yield ethane. In such a way, via the 1:1 mole ratio between acetylene (molar mass = 26 g/mol) and ethane (molar mass = 30 g/mol), we compute the yielded grams, or the theoretical yield of ethane as shown below:

[tex]m_{C_2H_6}^{theoretical}=28.0gC_2H_2*\frac{1molC_2H_2}{26gC_2H_2}*\frac{1molC_2H_6}{1molC_2H_2} *\frac{30gC_2H_6}{1molC_2H_6}\\ \\m_{C_2H_6}^{theoretical}=32.3gC_2H_6[/tex]

Hence, by knowing that the percent yield is computed via the actual yield (24.5 g) over the theoretical yield, we obtain:

[tex]Y=\frac{24.5g}{32.3g}*100\%\\ \\Y=75.6\%[/tex]

Best regards.

Answer:

Is like for solving for the solution for both equations??

Answer:

C3Br6

Explanation:

C= (1 X 12.011) = 12.011

Br= (2 X 79.904)= 159.808

159.808+12.011 = 171.819

515.46 divided by 171.819 = 3.00

so you mulitpy CBr2 by 3 which gives you C3Br6

Answer:S²⁻

Explanation:

Sulphur is in group 16. The atomic number of sulphur is 16.

However, sulphur can accept two electrons to form the sulphide ion S²⁻ which contains 18 electrons, hence the answer provided above.

Answer:

The value is  [tex]V_2  =   0.246 \  m^3/h[/tex]  

Explanation:

From the question we are told that

  The temperature at which the gas enters the compressor is  

         [tex]T_i = 298 \  K[/tex]

  The pressure  at which the gas enters the compressor is  

       [tex]P_I =  1.0 \ atm[/tex]

  The volumetric rate at which the gas enters the compressor is

      [tex]V  =  127 m^3/h[/tex]

   The temperature to which the gas is compressed to is  

      [tex]T_f  =  358 \ K[/tex]

  The pressure  to which the gas is compressed to is  

     [tex]P_f=  1000 \  atm[/tex]

Generally the volumetric flow rate of compressed oxygen is evaluated from the compressibility-factor equation of state as

    [tex]V_2  =  V_1 *\frac{z_2}{z_1} * \frac{T_2}{T_1} * \frac{P_1}{P_2}[/tex]

Here [tex]z_1[/tex] is the inflow compressibility factor with value [tex]z_1 = 1[/tex]

Here [tex]z_1[/tex] is the outflow compressibility factor with value [tex]z_2 = 1.61[/tex]

So

    [tex]V_2  =  127*\frac{1.61}{1} * \frac{358}{298} * \frac{1}{1000}[/tex]  

    [tex]V_2  =   0.246 \  m^3/h[/tex]  

Answer:

Formation of covalent bond structures. The image is essentially a Lewis dot structure.

The image of HOH has depicted the covalent bond formation in water. Thus, option B is correct.

The image has been the Lewis structure of the water molecule. It has been consisted of two hydrogen atoms bonded to oxygen molecules.

The oxygen has a presence of 4 dots that have been the representation of the valence electrons.  Thus, oxygen has been consisted of 2 lone pairs.

The water molecule has the presence of shared electrons between hydrogen and oxygen. The bond formed by the sharing of electrons has been covalent.

Thus, the image of HOH has been the representation of the formation of covalent bonds in water. Thus, option B is correct.

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Answer:

Your answer is d

Explanation: silver carbonate Ag2CO3 is insoluble in water

Answer:

I would say it is true

Explanation:

Answer:

d

Explanation:

Answer:

70.9 grams 3 sig figs

Explanation:

24.1% of 294 grams = 0.241(294 gram) = 70.854 grams ≅ 70.9 grams 3 sig figs

Answer:

Approximately [tex]0.551\; \rm J\cdot kg^{-1} \cdot \left(^\circ\! C \right)^{-1}[/tex].

Explanation:

The specific heat of a material is the amount of energy required to increase unit mass (one gram) of this material by unit temperature (one degree Celsius.)

Calculate the increase in the temperature of this sample:

[tex]\Delta T = (116.9 - 24.3)\; \rm ^\circ\! C= 92.6\; \rm ^\circ\! C[/tex].

The energy that this sample absorbed should be proportional the increase in its temperature (assuming that no phase change is involved.)

It took [tex]2911\; \rm J[/tex] of energy to raise the temperature of this sample by [tex]\Delta T = 92.6\; \rm ^\circ\! C[/tex]. Therefore, raising the temperature of this sample by [tex]1\; \rm ^\circ\! C[/tex] (unit temperature) would take only [tex]\displaystyle \frac{1}{92.6}[/tex] as much energy. That corresponds to approximately [tex]31.436\; \rm J[/tex] of energy.

On the other hand, the energy required to raise the temperature of this material by [tex]1\; \rm ^\circ\! C[/tex] is proportional to the mass of the sample (also assuming no phase change.)

It took approximately [tex]31.436\; \rm J[/tex] of energy to raise the temperature of [tex]57.07\; \rm g[/tex] of this material by [tex]1\; \rm ^\circ C[/tex]. Therefore, it would take only [tex]\displaystyle \frac{1}{57.07}[/tex] as much energy to raise the temperature of [tex]1\; \rm g[/tex] (unit mass) of this material by [tex]1\; \rm ^\circ \! C\![/tex]. That corresponds to approximately [tex]0.551\; \rm J[/tex] of energy.

In other words, it takes approximately [tex]0.551\; \rm J[/tex] to raise [tex]1\; \rm g[/tex] (unit mass) of this material by [tex]1\; \rm ^\circ \! C[/tex]. Therefore, by definition, the specific heat of this material would be approximately [tex]0.551\; \rm J\cdot kg^{-1} \cdot \left(^\circ\! C \right)^{-1}[/tex].

Answer :

Part 1: 4.93 moles of [tex]H_2O_2[/tex] contains 9.86 moles of oxygen atoms.

Part 2: 2.01 moles of [tex]N_2O[/tex] contains 2.01 moles of oxygen atoms.

Explanation :

Part 1: 4.93 mol [tex]H_2O_2[/tex]

In 1 mole of [tex]H_2O_2[/tex], there are 2 atoms of hydrogen and 2 atoms of oxygen.

As, 1 mole of [tex]H_2O_2[/tex] contains 2 moles of oxygen atoms.

So, 4.93 moles of [tex]H_2O_2[/tex] contains [tex]4.93\times 2=9.86[/tex] moles of oxygen atoms.

Thus, 4.93 moles of [tex]H_2O_2[/tex] contains 9.86 moles of oxygen atoms.

Part 2: 2.01 mol [tex]N_2O[/tex]

In 1 mole of [tex]N_2O[/tex], there are 2 atoms of nitrogen and 1 atom of oxygen.

As, 1 mole of [tex]N_2O[/tex] contains 1 mole of oxygen atoms.

So, 2.01 moles of [tex]N_2O[/tex] contains [tex]2.01\times 1=2.01[/tex] moles of oxygen atoms.

Thus, 2.01 moles of [tex]N_2O[/tex] contains 2.01 moles of oxygen atoms.

Answer:

The equilibrium concentration of [CH₃NH₂] = 0.23965 M.

The equilibrium concentration of CH₃NH₃⁺ and  OH⁻ =  0.01035 M respectively.

Explanation:

The first step is to write out the dissociation reaction. Therefore, the equation showing the dissociation is given as below;

CH₃NH₂ + H₂0   <--------------------------------------------------------> CH₃NH₃⁺ + OH⁻.

Kindly note that ''<----------->'' arrow shows that the reaction is an equilibrium reaction.

Therefore, at the start of the reaction [that is time, t =0], we have that the concentration of CH₃NH₂ = 0.25M, thus, the concentration of CH₃NH₃⁺  and  OH⁻ is zero respectively at this time, t =0.

At equilibrium, the concentration of CH₃NH₂ = 0.25M - x, thus, the concentration of CH₃NH₃⁺  and  OH⁻ is x respectively.

Therefore, kb =  4.47 × 10-4 = [CH₃NH₃⁺ ][OH⁻]/[CH₃NH₂].  Hence, slotting in the values into this equilibrium equation showing the relationship between kb and concentration of the species involved, we have that;

kb = 4.47 × 10⁻⁴ = x² /0.25 - x.

x² +  4.47 × 10⁻⁴x - 1.1175  × 10⁻⁴ = 0.

Solving this quadratic equation gives us the value of x as 0.01035 M.

Thus, the concentration of [CH₃NH₂] = 0.25 M - 0.01035 M = 0.23965 M

The equilibrium concentration of [CH₃NH₂] = 0.23965 M.

The equilibrium concentration of CH₃NH₃⁺ and  OH⁻ =  0.01035 M respectively.

Answer: 600 mL

Explanation:

Given that;

M₁ = 5.85 m

M₂ = 1.95 m

V₁ = 200 mL

V₂ = ?

Now from the dilution law;

M₁V₁ = M₂V₂

so we substitute

5.85 × 200 = 1.95 × V₂

1170 = 1.95V₂

V₂ = 1170 / 1.95

V₂ = 600 mL

Therefore final volume is 600 mL

Answer:

21.51g/cm³

Explanation:

To answer this question we need to know that, in 1 unit FCC cell:

Edge length = √8 * R  

Volume = 8√8 * R³

And there are 4 atoms per unit cell

The mass of the 4 atoms of the cell, in grams, is:

4 atom * (1mol / 6.022x10²³atom) * (195.09g / mol) = 1.2958x10⁻²¹g

Volume in cm³:

0.1386nm * (1x10⁻⁷cm / 1nm) = 1.386x10⁻⁸cm

Volume = 8√8 * (1.386x10⁻⁸cm)³

Volume = 6.02455x10⁻²³cm³

And density, the ratio of mass and volume, is:

1.2958x10⁻²¹g  / 6.02455x10⁻²³cm³  =

Answer:

10.801 amu

Explanation:

From the question given above, the following data were obtained:

Isotope A (¹⁰B):

Mass of A = 10

Abundance (A%) = 19.9%

Isotope B (¹¹B):

Mass of B = 11

Abundance (B%) = 80.1%

Atomic mass of Boron =?

The atomic mass of boron can be obtained as illustrated below:

Atomic mass = [(Mass of A × A%)/100] + [(Mass of B × B%)/100]

= [(10 × 19.9)/100] + [(11 × 80.1)/100]

= 1.99 + 8.811

= 10.801 amu

Thus, the atomic mass of boron is 10.801 amu

The atomic mass of boron with natural abundance of 19.9% of 10 B and 80.1%  of 11 B is 10.801 amu

Boron has 2 isotopes.

First isotopes

mass = 10

% abundance = 19.9%

Second Isotopes

mass = 11

% abundance = 80.1%

Therefore,

Atomic mass = (19.9% of 10) + (80.1% of 11)

Atomic mass = (19.9 / 100 × 10) + (80.1 / 100 × 11)

Atomic mass = 199 / 100 + 881.1 / 100

Atomic mass = 1.99 + 8.811

Atomic mass = 10.801

Atomic mass = 10.801 amu

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