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gavmur [86]
3 years ago
13

The following reaction is done at

Chemistry
2 answers:
sergejj [24]3 years ago
4 0

Answer : The volume of H_ gas produced is, 305.8 L

Explanation :

First we have to calculate the moles of Ca.

\text{Moles of Ca}=\frac{\text{Mass of Ca}}{\text{Molar mass of Ca}}

Molar mass of Ca = 40 g/mol

\text{Moles of Ca}=\frac{500.0g}{40g/mol}=12.5mol

Now we have to calculate the moles of H_2 gas.

The balanced chemical reaction is:

Ca(s)+2HCl(aq)\rightarrow CaCl_2(aq)+H_2(g)

From the balanced chemical reaction we conclude that,

As, 1 mole of Ca react to give 1 mole of H_2 gas

So, 12.5 mole of Ca react to give 12.5 mole of H_2 gas

Now we have to calculate the volume of H_ gas.

Using ideal gas equation:

PV=nRT

where,

P = Pressure of H_2 gas = 1.0 atm

V = Volume of H_2 gas = ?

n = number of moles H_2 = 12.5 mole

R = Gas constant = 0.0821L.atm/mol.K

T = Temperature of H_2 gas = 25^oC=273+25=298K

Putting values in above equation, we get:

1.0atm\times V=12.5mole\times (0.0821L.atm/mol.K)\times 298K

V=305.8L

Thus, the volume of H_ gas produced is, 305.8 L

bagirrra123 [75]3 years ago
3 0

Answer: 280dm3

Explanation:Please see attachment for explanation

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B- Candle wax melts when it is heated.

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What kind of substances can you not heat on a Bunsen burner?
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Answer:

Substances which are non heat-resistant and volatile organic liquid with flammable vapor.

Explanation:

Bunsen Burner are used as a source of fire in the laboratory for heating up substances. Extra care however should be taken during heating in order to prevent fire or other forms of accidents.

Heating of mind heat resistant substances should be frowned at as the substances may get heated up and melt thereby exposing the liquid substances which may be flammable to fire thereby causing fire outbreak.

Heating of volatile organic liquid with flammable vapor should also be discouraged to prevent fire accidents.

3 0
3 years ago
g 32.53 g of a solid is heated to 100.oC and added to 50.0 g of water in a coffee cup calorimeter and the contents are allowed t
Radda [10]

Answer:

0.886 J/g.°C

Explanation:

Step 1: Calculate the heat absorbed by the water

We will use the following expression

Q = c × m × ΔT

where,

  • Q: heat
  • c: specific heat capacity
  • m: mass
  • ΔT: change in the temperature

Q(water) = c(water) × m(water) × ΔT(water)

Q(water) = 4.184 J/g.°C × 50.0 g × (34.4 °C - 25.36 °C) = 1.89 × 10³ J

According to the law of conservation of energy, the sum of the energy lost by the solid and the energy absorbed by the water is zero.

Q(water) + Q(solid) = 0

Q(solid) = -Q(water) =  -1.89 × 10³ J

Step 2: Calculate the specific heat capacity of the solid

We will use the following expression.

Q(solid) = c(solid) × m(solid) × ΔT(solid)

c(solid) = Q(solid) / m(solid) × ΔT(solid)

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8 0
3 years ago
| | | |<br> | | | |<br> | | | |<br> VVV
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Answer:the answer would be a because that's what it would talk about and is the correct answer

Explanation:yes it is correct just answer it trust me

7 0
4 years ago
You have 363 mL of a 1.25M potassium chloride solution, but you need to make a 0.50M potassium chloride solution. How many milli
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Answer:- 544.5 mL of water need to be added.

Solution:- It is a dilution problem. The equation used for solving this type of problems is:

M_1V_1=M_2V_2

where, M_1 is initial molarity and  M_2 is the molarity after dilution. Similarly,  V_1 is the volume before dilution and  V_2 is the volume after dilution.

Let's plug in the values in the equation:

1.25M(363mL)=0.50M(V_2)

V_2=\frac{1.25M(363mL)}{0.50M}

V_2=907.5mL

Volume of water added = 907.5mL - 363mL  = 544.5 mL

So, 544.5 mL of water are need to be added to the original solution for dilution.

3 0
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