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Gnoma [55]
3 years ago
8

2. Arnie is going to dissolve 2.65g of zinc using 6.0M hydrochloric acid. A) what volume of the acid will he need to dissolve al

l the zinc? B) He wants to collect the gas produced by the reaction in a container, determine the volume of container he will need if the pressure is 628 mmHg and temperature is 23ᵒC. C) What is the mass of the zinc chloride he will produce? You will need a balanced equation and a gas law for this question.
Chemistry
1 answer:
fenix001 [56]3 years ago
5 0

Answer:

Explanation:

Zn + 2HCl = ZnCl₂ + H₂

A ) mole of Zn = 2.65 / 65

= .04 mol

mole of HCl required = .04 x 2 mol

.08 mol

If v be the volume required

v x 6 = .08

v = .0133 liter

= 13.3 cc

B )

volume of gas at NTP :

moles of gas  obtained = .04 moles

= 22.4 x .04 liter

= .896 liter

we have to find this volume at given temperature and pressure

\frac{P_1V_1}{T_1} =\frac{P_2V_2}{T_2}

\frac{760\times.896}{273} =\frac{628\times V_2}{296}

V_2 = 1.175 liter.

C )

.04 mole of zinc chloride will be produced

mol weight of zinc chloride

= 65 + 35.5 x 2

= 136 gm

.04 mole = 136 x .04

= 5.44 gm

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4 0
3 years ago
2)
vodomira [7]

Answer:

Final volume 30.513 L.

Explanation:

According to general gas equation:

P₁V₁/T₁ = P₂V₂/T₂

Given data:

Initial volume = 17 L

Initial pressure = 2.3 atm

Initial temperature = 299 K

Final temperature = 350 K

Final volume = ?

Final pressure = 1.5 atm

Solution:

P₁V₁/T₁ = P₂V₂/T₂

V₂ = P₁V₁ T₂/ T₁ P₂  

V₂ = 2.3 atm × 17 L × 350 K / 299 K × 1.5 atm

V₂  =  13685 atm .L.  K  / 448.5 K . atm

V₂  = 30.513 L

6 0
2 years ago
What is it called when a substance changes from a vapor to a liquid
skad [1K]
The answer is CONDENSATION.
4 0
3 years ago
A chemical reaction takes place inside a flask submerged in a water bath. The water bath contains 8.10kg of water at 33.9 degree
lions [1.4K]

Answer:

The new temperature of the water bath 32.0°C.

Explanation:

Mass of water in water bath ,m= 8.10 kg = 8100 g ( 1kg = 1000g)

Initial temperature of the water = T_1=33.9^oC=33.9+273K=306.9 K

Final temperature of the water = T_2

Specific heat capacity of water under these conditions =  c = 4.18 J/gK

Amount of energy lost by water = -Q = -69.0 kJ = -69.0 × 1000 J

( 1kJ=1000 J)

Q=m\times c\times \Delta T=m\times c\times (T_2-T_1)

-69.0\times 1000 J=8100 g\times 4.18 J/g K\times (T_2-306.9 K)

-69,000.0 J=8100 g\times 4.18 J/g K\times (T_2-306.9 K)

T_2=304.86 K=304.86 -273^oC=31.86^oC\approx 32.0^oC

The new temperature of the water bath 32.0°C.

5 0
3 years ago
c3H8 + 5O2 = 3CO2 + 4H2O When 44.0 grams of propane (C3H8) under goes complete combustion, how many grams of water will be produ
Umnica [9.8K]

<u>Answer:</u> 72 grams of water will be produced.

<u>Explanation:</u>

To calculate the number of moles, we use the formula:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}    ....(1)

Mass of propane = 44 grams

Molar mass of propane = 44 grams

Putting values in above equation, we get:

\text{moles of propane}=\frac{44g}{44g/mol}=1mole

For the reaction of combustion reaction of propane, the equation follows:

C_3H_8+5O_2\rightarrow 3CO_2+4H_2O

By Stoichiometry of the reaction,

1 mole of propane produces 4 moles of water.

So, 1 mole of propane will produce = \frac{1}{1}\times 4=4moles of water.

Now, to calculate the amount of water, we use equation 1, we get:

Molar mass of water = 18 g/mol

4mol=\frac{\text{Mass of water}}{18g/mol}

Mass of water produced = 72 grams

Hence,  72 grams of water will be produced.

6 0
3 years ago
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