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3 years ago

What is the formula for Na+1 and SO3^-2

Chemistry

1 answer:

KATRIN_1 [288]3 years ago

8 0

Answer:

Na₂SO₃

Explanation:

We can use the criss-cross method to work out the formula.

1. Write the symbols of the anion and cation.

Na¹⁺ SO₃²⁻

2. Criss-cross the numbers of the charges to become the subscripts of the other ion (see image).

3. Write the formula with the new subscripts.

Na₂(SO₃)₁

4. Omit all subscripts that are 1.

The formula becomes Na₂SO₃.

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3 years ago

What is the molar mass of an unknown gas with a density of 4.95 g/L at 1.00 atm and 25.0 °C?

mestny [16]

Answer:

121 g/mol

Explanation:

To find the molar mass, you first need to calculate the number of moles. For this, you need to use the Ideal Gas Law. The equation looks like this:

PV = nRT

In this equation,

-----> P = pressure (atm)

-----> V = volume (L)

-----> n = moles

-----> R = constant (0.0821 L*atm/mol*K)

-----> T = temperature (K)

Because density is comparing the mass per 1 liter, I am assuming that the system has a volume of 1 L. Before you can plug the given values into the equation, you first need to convert Celsius to Kelvin.

P = 1.00 atm R = 0.0821 L*atm/mol*K

V = 1.00 L T = 25.0. °C + 273.15 = 298.15 K

n = ? moles

PV = nRT

(1.00 atm)(1.00L) = n(0.0821 L*atm/mol*K)(298.15 K)

1.00 = n(0.0821 L*atm/mol*K)(298.15 K)

1.00 = (24.478115)n

0.0409 = n

Now, we need to find the molar mass using the number of moles per liter (calculated) and the density.

0.0409 moles ? grams 4.95 grams
---------------------- x ------------------ = ------------------
1 L 1 mole 1 L

? g/mol = 121 g/mol

**note: I am not 100% confident on this answer

3 0

2 years ago

The mass of sodium chloride in (g) is 14.19 The volume of ammonia solution in (mL) is 36.15 Calculate the following: What is the

Svetach [21]

This is an incomplete question, here is a complete question.

A weighed amount of sodium chloride is completely dissolved in a measured volume of 4.00 M ammonia solution at ice temperature, and carbon dioxide is bubbled in. Assume that sodium bicarbonate is formed util the limiting reagent is entirely used up. the solubility of sodium bicarbonate in water at ice temperature is 0.75 mol per liter. Also assume that all the sodium bicarbonate precipitated is collected and converted quantitatively to sodium carbonate.

Data to be used for calculating the results

-The mass of sodium chloride in (g) is 14.19

-The volume of ammonia solution in (mL) is 36.15

Calculate the following: What is the theoretical yield of sodium bicarbonate in grams?

Answer : The theoretical yield of sodium bicarbonate in grams is, 20.4 grams.

Explanation :

First we have to calculate the moles of NaCl and $NH_3$ .

$\text{ Moles of }NaCl=\frac{\text{ Mass of }NaCl}{\text{ Molar mass of }NaCl}=\frac{14.19g}{58.5g/mole}=0.243moles$

$\text{ Moles of }NH_3=\text{ Concentration of }NH_3\times \text{ Volume of solution}=4.00M\times 0.3615L=1.446moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction will be:

$NH_3+NaCl+CO_2+H_2O\rightarrow NaHCO_3+NH_4Cl$

From the balanced reaction we conclude that

As, 1 mole of $NaCl$ react with 1 mole of $NH_3$

So, 0.243 mole of $NaCl$ react with 0.243 mole of $NH_3$

From this we conclude that, $NH_3$ is an excess reagent because the given moles are greater than the required moles and $NaCl$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $NaHCO_3$

From the reaction, we conclude that

As, 1 mole of $NaCl$ react to give 1 mole of $NaHCO_3$

So, 0.243 moles of $NaCl$ react to give 0.243 moles of $NaHCO_3$

Now we have to calculate the mass of $NaHCO_3$

$\text{ Mass of }NaHCO_3=\text{ Moles of }NaHCO_3\times \text{ Molar mass of }NaHCO_3$

Molar mass of sodium bicarbonate = 84 g/mol

$\text{ Mass of }NaHCO_3=(0.243moles)\times (84g/mole)=20.4g$

Thus, the theoretical yield of sodium bicarbonate in grams is, 20.4 grams.

4 0

3 years ago