Which of the following best describes exothermic chemical reactions?

Graphs are a way to present data in picture form t or f

Irina-Kira [14]

I’m pretty sure it’s true

6 0

3 years ago

How many atoms are in 1.75 moles of CHCl?

jok3333 [9.3K]

1.75 moles ChCl3 x (6.02 x 10 ^-23) / 1 mole = 1.0535 x 10^-22 atoms.

hope this was helpful :)

8 0

4 years ago

How does the law of conservation of mass apply to this reaction: Mg + HCl > H2 + MgCl2 ?

Gelneren [198K]

Answer:

Explanation:

Law of conservation of mass:

According to the law of conservation mass, mass can neither be created nor destroyed in a chemical equation.

Explanation:

This law was given by french chemist Antoine Lavoisier in 1789. According to this law mass of reactant and mass of product must be equal, because masses are not created or destroyed in a chemical reaction.

Chemical equation:

Mg + HCl → H₂ + MgCl₂

24 g + 36.5 g = 2 g+ 95 g

60.5 g = 97 g

The reaction does not hold the law of conservation of mass, because it is not balanced.

Balanced chemical equation:

Mg + 2HCl → H₂ + MgCl₂

24 g + 73 g = 2 g+ 95 g

97 g = 97 g

this equation completely follow the law of conservation of mass.

7 0

4 years ago

A buffer solution is composed of 1.00 mol of acid and 2.25 mol of the conjugate base. If the p K a of the acid is 4.90 , what is

Gemiola [76]

Answer: The pH of the buffer is 5.25

Explanation:

Let the volume of buffer solution be V

We know that:

$\text{Molarity}=\frac{\text{Moles of solute}}{\text{Volume of solution}}$

To calculate the pH of acidic buffer, we use the equation given by Henderson Hasselbalch:

$pH=pK_a+\log(\frac{[\text{conjugate base}]}{[acid]})$

We are given:

$pK_a$ = negative logarithm of acid dissociation constant of weak acid = 4.90

$[\text{conjugate base}]=\frac{2.25}{V}$

$[acid]=\frac{1.00}{V}$

pH = ?

Putting values in above equation, we get:

$pH=4.90+\log(\frac{2.25/V}{1.00/V})\\\\pH=5.25$

Hence, the pH of the buffer is 5.25

4 0

3 years ago

Suppose of copper(II) acetate is dissolved in of a aqueous solution of sodium chromate. Calculate the final molarity of acetate

uranmaximum [27]

Answer:

0.0714 M for the given variables

Explanation:

The question is missing some data, but one of the original questions regarding this problem provides the following data:

Mass of copper(II) acetate: $m_{(AcO)_2Cu} = 0.972 g$

Volume of the sodium chromate solution: $V_{Na_2CrO_4} = 150.0 mL$

Molarity of the sodium chromate solution: $c_{Na_2CrO_4} = 0.0400 M$

Now, when copper(II) acetate reacts with sodium chromate, an insoluble copper(II) chromate is formed:

$(CH_3COO)_2Cu (aq) + Na_2CrO_4 (aq)\rightarrow 2 CH_3COONa (aq) + CuCrO_4 (s)$

Find moles of each reactant. or copper(II) acetate, divide its mass by the molar mass:

$n_{(AcO)_2Cu} = \frac{0.972 g}{181.63 g/mol} = 0.0053515 mol$

Moles of the sodium chromate solution would be found by multiplying its volume by molarity:

$n_{Na_2CrO_4} = 0.0400 M\cdot 0.1500 L = 0.00600 mol$

Find the limiting reactant. Notice that stoichiometry of this reaction is 1 : 1, so we can compare moles directly. Moles of copper(II) acetate are lower than moles of sodium chromate, so copper(II) acetate is our limiting reactant.

Write the net ionic equation for this reaction:

$Cu^{2+} (aq) + CrO_4^{2-} (aq)\rightarrow CuCrO_4 (s)$