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3 years ago

What is the total number of moles represented by 20 grams of CACO3

Chemistry

1 answer:

Nutka1998 [239]3 years ago

8 0

Answer:

B. 0.2.

Explanation:

We can use the relation:

n = mass/molar mass

mass of CaCO₃ = 20 g, molar mass of CaCO₃ = 100.0869 g/mol.

∴ n = mass/molar mass = (20 g)/(100.0869 g/mol) = 0.1998 ≅ 0.2 mol.

So, the right choice is: B. 0.2.

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What is the molarity of 6 moles (mol) of NaCl dissolved in 2 L of water?

Mama L [17]

Answer:

C: $\frac{6 mol}{2 L}$

Explanation:

we can use the molarity equation

$M = \frac{mol}{L}$

so to find M we plug in what we know, which is 6 moles of NaCl and 2 L of water, which gives us:

$\frac{6 mol}{2 L}$

8 0

2 years ago

How many L of a 7.5 H2SO4 stock solution would you need to prepare a dilute solution 100 L of 0.25M H2SO4

blsea [12.9K]

Answer:

3.33 L

Explanation:

We can solve this problem by using the equation:

C₁V₁=C₂V₂

Where the subscript 1 refers to one solution and subscript 2 to the another solution, meaning that in this case:

C₁ = 0.25 M

V₁ = 100 L

C₂ = 7.5 M

V₂ = ?

We input the data:

0.25 M * 100 L = 7.5 M * V₂

V₂ = 3.33 L

Thus the answer is 3.33 liters.

3 0

2 years ago

The activation energy of a certain uncatalyzed biochemical reaction is 50.0 kJ/mol. In the presence of a catalyst at 37°C, the r

rodikova [14]

Answer:

E₁ ≅ 28.96 kJ/mol

Explanation:

Given that:

The activation energy of a certain uncatalyzed biochemical reaction is 50.0 kJ/mol,

Let the activation energy for a catalyzed biochemical reaction = E₁

E₁ = ??? (unknown)

Let the activation energy for an uncatalyzed biochemical reaction = E₂

E₂ = 50.0 kJ/mol

= 50,000 J/mol

Temperature (T) = 37°C

= (37+273.15)K

= 310.15K

Rate constant (R) = 8.314 J/mol/k

Also, let the constant rate for the catalyzed biochemical reaction = K₁

let the constant rate for the uncatalyzed biochemical reaction = K₂

If the rate constant for the reaction increases by a factor of 3.50 × 10³ as compared with the uncatalyzed reaction, That implies that:

K₁ = 3.50 × 10³

K₂ = 1

Now, to calculate the activation energy for the catalyzed reaction going by the following above parameter;

we can use the formula for Arrhenius equation;

$K=Ae^{\frac{-E}{RT}}$

If $K_1=Ae^{\frac{-E_1}{RT}} -------equation 1$ &

$K_2=Ae^{\frac{-E_2}{RT}} -------equation 2$

$\frac{K_1}{K_2} = e^{\frac{-E_1-E_2}{RT}$

$E_1= E_2-RT*In(\frac{K_1}{K_2})$

$E_1= 50,000-8.314*310.15*In(\frac{3.50*10^3}{1})$

$E_1 = 28957.39292$ $J/mol$

E₁ ≅ 28.96 kJ/mol

∴ the activation energy for a catalyzed biochemical reaction (E₁) = 28.96 kJ/mol

8 0

2 years ago

For every action, there is an equal and opposite reaction.

Svetradugi [14.3K]

Answer:

yeah its newton's third law

Explanation:

7 0

2 years ago

Which of the following is a factor in natural selection?

Vera_Pavlovna [14]

Answer:

Explanation:

Although all of these answers are similar, it seems that all three are the answer. I am sorry if the answer ends up being B but I'm pretty sure the answer is D. Charles Darwin's evolution theory "evolved", get it, from the natural selection theory. The theory is described as "the process whereby organisms better adapted to their environment tend to survive and produce more offspring." However, some factors that effect natural selection are competition, selection and variation. I believe all three fit the answer and therefore I believe answer D is correct.

5 0

3 years ago

Read 2 more answers