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3 years ago

Find the number of cations present in 7g of sodium prosphate Na=23 g/mol O=16g/mol P=31g/mol

Chemistry

1 answer:

Snowcat [4.5K]3 years ago

3 0

Answer: This contains magnesium, Mg2+, and hydroxide, OH–

, ions. Each magnesium ion is +2 and

each hydroxide ion is -1: two -1 ions are needed for one +2 ion, and the formula for magnesium

hydroxide is Mg(OH)2. The (OH)2 indicates there are two OH–

ions. In a formula unit of

Mg(OH)2, there are one magnesium ion and two hydroxide ions; or one magnesium, two

oxygen, and two hydrogen atoms. The subscript multiplies everything in ( )

hope that helped!!

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Problem PageQuestionSteam reforming of methane ( ) produces "synthesis gas," a mixture of carbon monoxide gas and hydrogen gas,

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The question is incomplete. Her eis the complete question.

Steam reforming methane (CH4) produces "synthesis gas", a mixture of carbon monoxide gas and hydrogen gas, which is the starting point for many important industrial chemical syntheses. An industrial chemist studying this reaction fills a 125L tank with 20 mol of methane gas and 10 mol of water vapor at 38°C. He then raises the temperature, and when the mixture has come to equilibrium measures the amount of gas hydrogen to be 18 mol. Calculate the concentration equilibrium constant for the steam reforming of methane at the final temperature of the mixture. Round your answer to significant digits.

Answer: $K_{c}$ = 2. $10^{-2}$

Explanation: The reaction for steam reforming methane is:

$CH_{4} + H_{2}O$ ⇒ $CO_{} + 3H_{2}$

To calculate the concentration equilibrium constant, first calculate the molarity ( $\frac{mol}{L}$ ) of each molecule of the reaction.

At 38°C: At the initial temperature, there no products yet

Molarity of CH4:

CH4 = $\frac{20}{125}$ = 0.16M

Molarity of H20:

H2O = $\frac{10}{125}$ = 0.08M

At final temperature:

Molarity of H2:

H2 = $\frac{18}{125}$ = 0.144M

According to the chemical reaction, the combination of 1 mol of each reagents produces 1 mol of CO and 3 mols of H2, so, for the products, the ratio is 1:3.

Molarity of CO:

CO = $\frac{0.144}{3}$ = 0.048M

For the reagents, the proportion is 1:1, but they had an initial concentration, so, when in equilibrium, the concentration will be:

Molarity of CH4:

CH4 = 0.16 - 0.048 = 0.112M

Molarity of H2O:

H20 = 0.08 - 0.048 = 0.032M

The equilibrium constant is given by:

$K_{c} = \frac{[CO][H_{2}]^{3} }{[CH_{4}][H_{2}O ] }$

$K_{c}$ = $\frac{0.048.0.144^{3} }{0.112.0.032}$

$K_{c}$ = 2. $10^{-2}$

The concentration equilibrium constant for the process is $K_{c}$ = 2. $10^{-2}$ .

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3 years ago

In the reaction below, if 5.71 g of sulfur is reacted with 10.0 g of oxygen, how many grams of sulfur trioxide will be produced?

yawa3891 [41]

Answer:

14.3 g SO₃

Explanation:

2S + 3O₂ → 2SO₃

First, find the limiting reactant. To do that, calculate the mass of oxygen needed to react with all the sulfur.

5.71 g S × (1 mol S / 32 g S) = 0.178 mol S

0.178 mol S × (3 mol O₂ / 2 mol S) = 0.268 mol O₂

0.268 mol O₂ × (32 g O₂ / mol O₂) = 8.57 g O₂

There are 10.0 g of O₂, so there's enough oxygen. The limiting reactant is therefore sulfur.

Use the mass of sulfur to calculate the mass of sulfur trioxide.

5.71 g S × (1 mol S / 32 g S) = 0.178 mol S

0.178 mol S × (2 mol SO₃ / 2 mol S) = 0.178 mol SO₃

0.178 mol SO₃ × (80 g SO₃ / mol SO₃) = 14.3 g SO₃

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3 years ago

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Answer:

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Starch: It is formed by the condensation of amylose and amylopectin. ...

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Explanation:

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3 years ago

Find the number of cations present in 7g of sodium prosphate Na=23 g/mol O=16g/mol P=31g/mol​

Find the number of cations present in 7g of sodium prosphate Na=23 g/mol O=16g/mol P=31g/mol