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ANEK [815]
4 years ago
14

The density of a solution of sulfuric acid is 1.29 g/cm3 and it is 38.1% acid by mass. What volume of the sulfuric acid solution

is needed to supply 163 g of sulfuric acid? 1 cm3 = 1 mL 1. 428 g 2. 80.1 mL 3. 552 mL 4. 252 mL 5. 48.1 mL 6. 8010 mL 7. 332 mL 8. 0.00397 mL
Chemistry
1 answer:
pychu [463]4 years ago
6 0

Answer : The correct option is, (7) 332 mL

Explanation : Given,

Density of solution of sulfuric acid = 1.29g/cm^3=1.29g/mL

38.1 % acid by mass that means 38.1 grams of sulfuric acid present in 100 grams of solution of sulfuric acid.

Now we have to calculate the mass of solution of sulfuric acid.

As, 38.1 grams of sulfuric acid present in 100 grams of solution of sulfuric acid

So, 163 grams of sulfuric acid present in \frac{163}{38.1}\times 100=428.95 grams of solution of sulfuric acid

Now we have to calculate the volume of sulfuric acid solution.

Density=\frac{Mass}{Volume}

1.29g/mL=\frac{428.95g}{Volume}

Volume=332mL

Therefore, the volume of sulfuric acid solution needed is 332 mL.

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5 0
3 years ago
Using the phase diagram for water, what phase is water at 0.01 degrees Celsius and 0.0060 atmospheres?
anzhelika [568]
<span>Answer: at 0.01 °C and 0.0060 atm the three phases (solid, liquid and gas)
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</span><span>Explanation:
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</span><span>1) Water at 0.0060 atm and 0.01° C is at its triple point.
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2) The triple point is the point in the phase diagram at which the three physical states coexist: gas, liquid and solid.

3) That means that water can freeze and boil at the same time. In fact they can happen any of the six changes of phase: freezing (liquid to solid), melting (solid to liquid), evaporation (liquid to gas), condensation (gas to liquid), sublimation (solid to gas), and deposition (gas to solid).

The phase diagram is unique for any substance, meaning that it is different for different substances: the normal boiling and melting points are different.
8 0
4 years ago
In a combustion experiment, it was found that 12.096 g of hydrogen molecules combined with 96.000 g of oxygen molecules to form
Mkey [24]
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E = Δmc²
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5 0
3 years ago
1.12g H2 is allowed to react with 9.60 g N2, producing 1.23 g NH3.
andriy [413]

Answer:

A. m_{NH_3}^{theo} =1.50gNH_3

B. Y=82.2\%

Explanation:

Hello!

In this case, since the undergoing chemical reaction between nitrogen and hydrogen is:

N_2+3H_2\rightarrow 2NH_3

Thus we proceed as follows:

A. Here, we first need to compute the moles of ammonia yielded by each reactant, in order to identify the limiting one:

n_{NH_3}^{by \ H_2}=1.12gH_2*\frac{1molH_2}{2.02gH_2}*\frac{2molNH_3}{3molH_2}=0.370molNH_3\\\\  n_{NH_3}^{by \ N_2}=1.23gN_2*\frac{1molN_2}{28.02gN_2}*\frac{2molNH_3}{1molN_2}=0.0878molNH_3

Thus, since nitrogen yields the fewest moles of ammonia, we realize it is the limiting reactant, so the theoretical yield, in grams, of ammonia is:

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B. Finally, since the actual yield of ammonia is 1.23, the percent yield turns out:

Y=\frac{1.23gNH_3}{1.50gNH_3} *100\%\\\\Y=82.2\%

Best regards!

5 0
3 years ago
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garri49 [273]

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3.) H^{+}+Cl^-\rightarrow HCl

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The lewis-dot structure image is shown below.

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