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2 years ago

If 7.50 g of the unknown compound contained 0.250 mol of c and 0.500 mol of h, how many moles of oxygen, o, were in the sample?

1 answer:

3 0

It would be 100.60 because we're adding like to the atmosphere it connects to lubricator and then it turns so that's 100.60

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A sample of hydrogen gas was collected over water at 21Âºc and 685 mmhg. the volume of the container was 7.80 l. calculate the m

murzikaleks [220]

To determine the mass of the hydrogen gas that was collected, we calculate for the moles of hydrogen gas from the conditions given. In order to do this, we need an equation which would relate pressure, volume and temperature. For simplicity, we assume the gas is an ideal gas so we use the equation PV = nRT where P is the pressure, V is the volume, n is the number of moles of the gas, T is the temperature and R is the universal gas constant. We calculate as follows:

PV = nRT
n = PV / RT
n = (18.6/760) (7.80) / 0.08205 ( 21 + 273.15)
n = 0.0079 mol

Mass = 0.0079 mol ( 18.02 g / mol ) = 0.1425 g H2

8 0

2 years ago

0.9775 grams of an unknown compound is dissolved in 50.0 ml of water. Initially the water temperature is 22.3 degrees Celsius. A

elena-14-01-66 [18.8K]

Answer:

The enthlapy of solution is -55.23 kJ/mol.

Explanation:

Mass of water = m

Density of water = 1 g/mL

Volume of water = 50.0 mL

m = Density of water × Volume of water = 1 g/mL × 50.0 mL=50.0 g

Change in temperature of the water ,ΔT= 27.0°C - 22.3°C = 4.7°C

Heat capacity of water,c =4.186 J/g°C

Heat gained by the water when an unknown compound is dissolved be Q

Q= mcΔT

$Q=50.0 g\times 4.186 J/g^oC\times 4.7^oC=983.71 J$

heat released when 0.9775 grams of an unknown compound is dissolved in water will be same as that heat gained by water.

Q'=-Q

Q'= -983.71 J =-0.98371 kJ

Moles of unknown compound = $\frac{0.9975 g}{56 g/mol}=0.01781 mol$

The enthlapy of solution :

$\frac{Q'}{moles}$

$=\frac{-0.98371 kJ}{0.01781 mol}=-55.23 kJ/mol$

The enthlapy of solution is -55.23 kJ/mol.

8 0

2 years ago

During an experiment, the percent yield of calcium chloride from a reaction was 85.22%. Theoretically, the expected amount shoul

devlian [24]

The formula to find yield is

(Actual Yield)/(Theorectical Yield) x100

Just do the math.

85.22% x 113 = 96.2986

Convert it to 3 significant figures

Ans: 96.3g

7 0

2 years ago

Read 2 more answers

(g) On the graph in part (d) , carefully draw a curve that shows the results of the second titration, in which the student titra

atroni [7]

Answer:

Here's what I get

Explanation:

(g) Titration curves

I can't draw two curves on the same graph, but I can draw two separate curves for you.

The graph in part (d) had an equivalence point at 20 mL.

In the second titration, the NaOH was twice as concentrated, so the volume to equivalence point would be half as much — 10 mL.

The two titration curves are below.

(h) Evidence of reaction

HCl and NaOH are both colourless.

They don't evolve a gas or form a precipitate when they react.

The student probably noticed that the Erlenmeyer flask warmed up — a sign of a chemical change.

4 0

2 years ago

Can 750 mL of water dissolve 0.60 mol of gold(III) chloride (AuCl3)?

nydimaria [60]

Answer:

Yes

Explanation:

1. Mass of 0.60 mol of AuCl₃

$\text{Mass} = \text{0.60 mol} \times \dfrac{\text{303.33 g}}{\text{1 mol}} = \text{184 g}$

2. Mass of AuCl₃ in 750 mL

The solubility of AuCl₃ is 68 g/100 mL.

In 750 mL of water, you can dissolve

$\text{Mass of AuCl}_{3} = \text{750 mL} \times \dfrac{\text{68 g}}{\text{100 mL}} = \text{510 g AlCl}_{3}$

∴ Yes, 750 mL of water can dissolve 0.60 mol of AuCl₃.

3 0

2 years ago