How many atoms are in 2.85x10^3 g of gallium

odium carbonate (Na2CO3Na2CO3) is used to neutralize the sulfuric acid spill. How many kilograms of sodium carbonate must be add

Arte-miy333 [17]

Answer : The mass of sodium carbonate added to neutralize must be, $6.54\times 10^3kg$

Explanation :

First we have to calculate the moles of $H_2SO_4$ .

$\text{Moles of }H_2SO_4=\frac{\text{Mass of }H_2SO_4}{\text{Molar mass of }H_2SO_4}$

Given:

Molar mass of $H_2SO_4$ = 98 g/mole

Mass of $H_2SO_4$ = $6.05\times 10^3kg=6.05\times 10^6g$

Conversion used : (1 kg = 1000 g)

Now put all the given values in the above expression, we get:

$\text{Moles of }H_2SO_4=\frac{6.05\times 10^6g}{98g/mol}=6.17\times 10^4mol$

The moles of $H_2SO_4$ is, $6.17\times 10^4mol$

Now we have to calculate the moles of $Na_2CO_3$

The balanced neutralization reaction is:

$Na_2CO_3+H_2SO_4\rightarrow Na_2SO_4+H_2CO_3$

From the balanced chemical reaction we conclude that,

As, 1 mole of $H_2SO_4$ neutralizes 1 mole of $Na_2CO_3$

So, $6.17\times 10^4mol$ of $H_2SO_4$ neutralizes

Now we have to calculate the mass of $Na_2CO_3$

$\text{ Mass of }Na_2CO_3=\text{ Moles of }Na_2CO_3\times \text{ Molar mass of }Na_2CO_3$

Molar mass of $Na_2CO_3$ = 106 g/mole

$\text{ Mass of }Na_2CO_3=(6.17\times 10^4mol)\times (106g/mole)=6.54\times 10^6g=6.54\times 10^3kg$

Thus, the mass of sodium carbonate added to neutralize must be, $6.54\times 10^3kg$

3 0

4 years ago

What volume of air is needed to burn an entire 55-L (approximately 15-gal) tank of gasoline? Assume that the gasoline is pure oc

White raven [17]

Answer:

The volume of air required is 527,686.25L.

Explanation:

When the question says "burn", it refers to a combustion reaction, where a substance (in this case octane) reacts with oxygen to produce carbon dioxide and water.

Step 1: Write a balanced equation

Considering it is a combustion reaction, the balanced equation is:

C₈H₁₈ + 12.5 O₂ ⇄ 8 CO₂ + 9 H₂O

In this step, we start balancing elements that are present only in one compound on each side of the equation, namely, carbon and hydrogen.

To finish, it is important to count that there are the same number of atoms on both sides of the equation. In this case there are 8 atoms of Carbon, 18 atoms of Hydrogen and 25 atoms of Oxygen, so it is balanced.

Step 2: Find out the mass of C₈H₁₈

Since the balanced equation gives us information about the mass of C₈H₁₈ involved in the reaction, we need to find out how many grams we have.

The info we have is:

55 L of gasoline (assuming gasoline to be pure octane).
The density of octane is 0,70 g/cm³

Density relates mass and volume, so we can find out how many grams are represented by 55 L. Since the units used are different, first we need to convert liters into cm³. We use the conversión factor 1 L = 1000 cm³.

$55L.\frac{1000cm^{3} }{1L} =55000cm^{3}$

Since density = mass/volume, we can solve for mass:

$mass = density.volume=0.70\frac{g}{cm^{3} } .55,000cm^{3} =38,500g$

Step 3: Establish the theoretical relationship between the mass of octane and the moles of oxygen.

This relationship comes from the balanced equation.

For octane:

molar mass of C₈H₁₈ = molar mass of C . 8 + molar mass of H . 18 =

12 g/mol . 8 + 1g/mol . 18 = 114 g/mol

According to the balanced equation reacts 1 mol of octane, which means 114 grams of it.

For oxygen:

According to the balanced equation, 12.5 moles of oxygen react.

Then, the relationship is 114 g octane : 12.5 moles of oxygen

Step 4: Use the theorethical relationship to find the moles of oxygen that reacted

We use the mass of octane found in step 2 and apply the proper conversión factor.

$38,500g (octane) . \frac{12.5mol (oxygen)}{114g (octane)} = 4,221.49mol(oxygen)$

Step 5: Find out the volume of oxygen.

We know that 1 mol of any gas at room temperature occupies about 25 L. Then,

$4,221.49mol(oxygen).\frac{25L(oxygen)}{1mol(oxygen)} =105,537.25 L (oxygen)$

Step 6: Look the volume of air that contains such amount of oxygen

Given oxygen represents 20% of air, we can use that relationship to find the volume of air.

$105,537.25L(oxygen).\frac{100L(air)}{20L(oxygen)} =527,686.25L (air)$

4 0

4 years ago

A sample of an unknown gas effuses in 14.4 min. An equal volume of H2 in the same apparatus under the same conditions effuses in

Elena-2011 [213]

Answer:- molar mass of the unknown gas is 71.5 gram per mol.

Solution:- From Graham's law of effusion rates, the rate of effusion of a gas is inversely proportional to the square root of it's molar mass.

When we compare the effusion rates of two gases then the formula for Graham's law is:

$\frac{rate_1}{rate_2}=\sqrt{\frac{M_2}{M_1}}$

In this formula, V stands for volume and M stands for molar mass

Rate is volume effused per unit time. Since, the volumes are same, the formula could be written as:

$\frac{t_2}{t_1}=\sqrt{\frac{M_2}{M_1}}$

let's say in formula, subscript 1 is for hydrogen gas and 2 is for the unknown gas.

Molar mass of hydrogen is 2.02 grams per mol and the time taken to effuse it is 2.42 min. The time taken to effuse the unknown gas is 14.4 min and we are asked to calculate it's molar mass. let's plug in the values in the formula:

$\frac{14.4}{2.42}=\sqrt{\frac{M_2}{2.02}}$

$5.95=\sqrt{\frac{M_2}{2.02}}$

doing squares to both sides:

$35.4=\frac{M_2}{2.02}$

$M_2=35.4*2.02$

$M_2=71.5$

So, the molar mass of the unknown gas is 71.5 grams per mol.

4 0

3 years ago

Which of the following is the correct model of C6H14?

asambeis [7]

Answer:

This question is incomplete

Explanation:

This question is incomplete because of the absence of options. However, the compound C₆H₁₄ is hexane. Hexane is a member of saturated hydrocarbons (homologous series) called alkanes (with the general formula CₙH₂ₙ₊₂). The structure for an hexane is shown below

H H H H H H

I I I I I I

H - C - C - C - C - C - C - H

I I I I I I

H H H H H H

which can also be written as

CH₃CH₂CH₂CH₂CH₂CH₃

3 0

3 years ago

Read 2 more answers

Write a formula for the ionic compound that forms from each pair of elements. sodium and sulfur

larisa86 [58]

Na: +1 (sodium)
S: -2 (sulfur)
=> Na2S

5 0

3 years ago