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3 years ago

BaCl2(aq) + Na2CO3(aq) BaCO3(s) + NaCl(aq

1 answer:

7 0

This is the equation balanced:

BaCl2(aq) + Na2CO3(aq) = BaCO3(s) + 2 NaCl(aq)

Then the coefficient in front of Na Cl is 2.

Answer: 2

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How much heat, in joules, must be added to a 75.0-g iron block with a specific heat of 0.449 J/g °C to increase its temperature

svp [43]

Answer:

50,849.25 Joules

Explanation:

The amount of heat, Q, required to raise the temperature of a body with mass, m, and specific heat capacity, c is given by:

Q = mcΔT, where ΔT represents the change in temperature.

In the case of the iron block:

m = 75 g

c = 0.449 J/g °C

ΔT = 1535 - 25 = 1510 °C

Therefore,

Q = 75 g x 0.449 J/g °C x 1510 °C

= 50,849.25 Joules

Hence, 50,849.25 Joules of heat must be added to a 75.0-g iron block with a specific heat of 0.449 J/g °C to increase its temperature from 25 °C to its melting temperature of 1535 °C

5 0

2 years ago

During which process is liquid changed to a gas within the water cycle?

stepladder [879]

Answer:

evaporation.

evaporation is the process by which water changes from liquid to a gas or vapor.

7 0

3 years ago

3-chloropropylamine what is?

marta [7]

Answer:

a chemical compound

6 0

1 year ago

A 1.68 g sample of water is injected into a closed evacuated 5.3 liter flask at 65°C. What percent (by mass) of the water will b

Angelina_Jolie [31]

Answer:

50.4 % of the water will be vapor

Explanation:

Step 1: Data given

Mass of water = 1.68 grams

volume of the flask = 5.3 L

Temperature = 65°C

Vapor pressure of water at 65°C = 187.5 mmHg = 0.2467 atm

Step 2: Calculate moles of H2O

p*V=n*R*T

⇒ p = the pressure of water = 0.2467 atm

⇒ V = the volume of the flask = 5.3 L

⇒ n = moles of water

⇒ R = gas constant = 0.08206 L*atm/ K*mol

⇒ T = the temperature = 65°C = 338 Kelvin

n = (p*V)/(R*T)

n = (0.2467 * 5.3) /(0.08206* 338)

n = 0.047 moles

Step 3: Calculate mass of water

Mass of water = moles of water * molar mass of water

Mass of water = 0.047 moles *18.02 g/mol

Mass of water = 0.84694 grams

Step 4: Calculate the percent of water vaporized

% = (0.84694 grams/1.68 grams) *100%

% = 50.4%

50.4 % of the water will be vapor

5 0

3 years ago

How many grams of XeF6 are required to react with 0.579 L of hydrogen gas at 4.46 atm and 45°C in the reaction shown below?

natka813 [3]

Answer:

8.1433 g of XeF₆ are required.

Explanation:

Balanced chemical equation;

XeF₆ (s) + 3H₂ (g) → Xe (g) + 6HF (g)

Given data:

Volume of hydrogen = 0.579 L

Pressure = 4.46 atm

Temperature = 45 °C (45+273= 318 k)

Solution:

First of all we will calculate the moles of hydrogen

PV = nRT

n = PV/ RT

n = 4.46 atm × 0.579 L / 0.0821 atm. dm³. mol⁻¹. K⁻¹ × 318 K

n = 2.6 atm . L / 26.12 atm. dm³. mol⁻¹

n = 0.0995 mol

Mass of hydrogen:

Mass = moles × molar mass

Mass = 0.0995 mol × 2.016 g/mol

Mass = 0.2006 g

Now we will compare the moles of hydrogen with XeF₆ from balance chemical equation.

H₂ : XeF₆

3 : 1

0.0995 : 1/3× 0.0995 = 0.0332 mol

Now we will calculate the mass of XeF₆.

Mass = moles × molar mass

Mass = 0.0332 mol × 245.28 g/mol

Mass = 8.1433 g

4 0

3 years ago