What volume of 1. 50m nacl contains 3. 40 moles of nacl?.

Convert 1663.5 g to mg

WINSTONCH [101]

Answer:

It will be 1663500mg

Explanation:

4 0

3 years ago

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Fritz Haber, a German chemist, discovered a way to synthesize ammonia gas (NH3) by combining hydrogen and nitrogen gases accordi

Westkost [7]

1) Write the balanced equation to state the molar ratios:

<span>3H2(g) + N2(g) → 2NH3(g)

=> molar ratios = 3 mol H2 : 1 mol N2 : 2 mol NH3

What volume of nitrogen is needed to produce 250.0 L of ammonia gas at STP?

First, convert the 250.0 L of NH3 to number of moles at STP .

Use the fact that 1 mole of gas at STP occupies 22.4 L

=> 250.0 L * 1mol/22.4 L = 11.16 L

Second, use the molar ratio to find the number of moles of N2 that produces 11.16 L of NH3

=> 11.16 L NH3 * [1 mol N2 / 2 mol NH3] = 5.58 mol N2

Third, convert 5.58 mol N2 into liters at STP

=> 5.58 mol N2 * [22.4 L/mol] = 124.99 liters

Answer: 124,99 liters

What volume of hydrogen is needed to produce 2.50 mol NH3 at STP?

First, find the number of moles of H2 that produce 2.50 mol by using the molar ratios:

2.50 mol NH3 * [3mol H2 / 2 mol NH3] = 3.75 mol H2

Second, convert the number of moles to liters of gas at STP:

3.75 mol * 22.4 L/mol = 84 liters of H2

Answer: 84 liters

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3 years ago

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What is the answer? plz give me the right answer which you are sure of .

alisha [4.7K]

Mention four reasons why the poll ordinance failed

8 0

2 years ago

Assuming that an acetic acid solution is 12% by mass and that the density of the solution is 1.00 g/mL, what volume of 1 M NaOH

Doss [256]

Explanation:

Let us assume that total mass of the solution is 100 g. And, as it is given that acetic acid solution is 12% by mass which means that mass of acetic acid is 12 g and 88 g is the water.

Now, calculate the number of moles of acetic acid as its molar mass is 60 g/mol.

No. of moles = $\frac{mass}{\text{molar mass}}$

= $\frac{12 g}{60 g/mol}$

= 0.2 mol

Molarity of acetic acid is calculated as follows.

Density = $\frac{mass}{volume}$

1 g/ml = $\frac{100 g}{volume}$

volume = 100 ml

Hence, molarity = $\frac{\text{no. of moles}}{volume}$

= $\frac{0.2 mol}{0.1 L}$

= 2 mol/l

As reaction equation for the given reaction is as follows.

$NaOH + CH_{3}COOH \rightarrow CH_{3}COONa + H_{2}O$

So, moles of NaOH = moles of acetic acid

Let us suppose that moles of NaOH are "x".

$x \times 1 M = 10 mL \times 2 M$ (as 1 L = 1000 ml)

x = 20 L

Thus, we can conclude that volume of NaOH required is 20 ml.

6 0

3 years ago

Burning 12.00 g of an oxoacid produces 17.95 g of carbon dioxide and 4.87 g of water. Consider that 0.25

Veseljchak [2.6K]

Answer: The molecular formula will be $C_6H_6O_6$

Explanation:

Mass of $CO_2$ = 17.95 g

Mass of $H_2O$ = 4.87 g

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44g of carbon dioxide, 12 g of carbon is contained.

So, in 17.95 g of carbon dioxide, = $\frac{12}{44}\times 17.95=4.89g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18g of water, 2 g of hydrogen is contained.

So, in 4.87 g of water, = $\frac{2}{18}\times 4.87=0.541g$ of hydrogen will be contained.

Mass of oxygen in the compound = (12.00) - (4.89+0.541) = 6.57 g

Mass of C = 4.89 g

Mass of H = 0.541 g

Mass of O = 6.57 g

Step 1 : convert given masses into moles.

Moles of C = $\frac{\text{ given mass of C}}{\text{ molar mass of C}}= \frac{4.89g}{12g/mole}=0.407moles$

Moles of H= $\frac{\text{ given mass of H}}{\text{ molar mass of H}}= \frac{0.541g}{1g/mole}=0.541moles$

Moles of O= $\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{6.57g}{16g/mole}=0.410moles$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For C = $\frac{0.407}{0.407}=1$

For H = $\frac{0.541}{0.407}=1$

For O = $\frac{0.410}{0.407}=1$

The ratio of C : H : O = 1: 1 : 1

Hence the empirical formula is $CHO$ .

Hence the empirical formula is $CHO$

The empirical weight of $CHO$ = 1(12)+1(1)+1(16)= 29 g.

If 0.25 moles has mass of 44.0 g

Thus 1 mole has mass of = $\frac{44.0}{0.25}\times 1=176g$

Thus molecular mass is 176 g

Now we have to calculate the molecular formula.

$n=\frac{\text{Molecular weight }}{\text{Equivalent weight}}=\frac{176g}{29g}=6$

The molecular formula will be= $6\times CHO=C_6H_6O_6$

5 0

2 years ago