What do you use to test ph in swimming pool

Identify each description as either a physical or chemical change.

Anettt [7]

Answer:

1 - physical

2 -chemical

3 - chemical

4 - physical

5 - physical

6 - chemical

7 - physical

8 - chemical

4 0

3 years ago

If liquid carbon disulfide reacts with 450 mL of oxygen to produce the gases carbon dioxide and sulfur dioxide, what volume of e

ivolga24 [154]

If excess carbon disulfide reacts with 450 mL of oxygen, 150 mL of carbon dioxide and 300 mL of sulfur dioxide gases would be produced respectively.

<h3>Stoichiometric calculation</h3>

The reaction between liquid carbon disulfide and oxygen is represented by the equations below:

$CS_2 + 3O_2 -- > CO_2 + 2SO_2$

The mole ratio of oxygen to carbon dioxide and sulfur dioxide produced is 3:1:2.

Thus, for 450 mL oxygen, 1/3 x 450 = 150 mL of carbon dioxide will be required.

Also for 450 mL of oxygen, 2/3 x 450 = 300 mL of sulfur dioxide will be required.

More on stoichiometric calculations can be found here: brainly.com/question/27287858

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5 0

2 years ago

2 HI(g) ⇄ H2(g) + I2(g) Kc = 0.0156 at 400ºC 0.550 moles of HI are placed in a 2.00 L container and the system is allowed to rea

Alex_Xolod [135]

<u>Answer:</u> The concentration of hydrogen gas at equilibrium is 0.0275 M

<u>Explanation:</u>

Molarity is calculated by using the equation:

$\text{Molarity}=\frac{\text{Number of moles}}{\text{Volume of solution (in L)}}$

Moles of HI = 0.550 moles

Volume of container = 2.00 L

$\text{Initial concentration of HI}=\frac{0.550}{2}=0.275M$

For the given chemical equation:

$2HI(g)\rightleftharpoons H_2(g)+I_2(g)$

<u>Initial:</u> 0.275

<u>At eqllm:</u> 0.275-2x x x

The expression of $K_c$ for above equation follows:

$K_c=\frac{[H_2][I_2]}{[HI]^2}$

We are given:

$K_c=0.0156$

Putting values in above expression, we get:

$0.0156=\frac{x\times x}{(0.275-2x)^2}\\\\x=-0.0458,0.0275$

Neglecting the negative value of 'x' because concentration cannot be negative

So, equilibrium concentration of hydrogen gas = x = 0.0275 M

Hence, the concentration of hydrogen gas at equilibrium is 0.0275 M

4 0

3 years ago

A compound is found to contain 42.88 % carbon and 57.12 % oxygen by weight. To answer the questions, enter the elements in the o

Ipatiy [6.2K]

Answer:

The empirical formule is CO

Explanation:

Step 1: Data given

Suppose the mass of a compound is 100 grams

Suppose the compound contains:

42.88 % C = 42.88 grams C

57.12 % O = 57.12 grams O

Molar mass C = 12.01 g/mol

Molar mass O = 16.0 g/mol

Step 2: Calculate moles

Moles = mass / molar mass

Moles C = mass C / molar mass C

Moles C = 42.88 grams / 12.01 g/mol

Moles C = 3.57 moles

Moles O = 57.12 grams / 16.0 g/mol

Moles O = 3.57 moles

Step 3: Calculate the mol ratio

We divide by the smallest amount of moles

C: 3.57 moles / 3.57 moles = 1

O: 3.57 moles / 3.57 moles = 1

The empirical formule is CO

3 0

3 years ago

What particle has the most energy

Katyanochek1 [597]

Answer:

electrons

Explanation:

By particles im assuming you mean subatomic. The particle with the most energy in this case would be electrons.

5 0

3 years ago

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