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yawa3891 [41]
3 years ago
14

What’s the molar mass of Rb3n

Chemistry
1 answer:
Salsk061 [2.6K]3 years ago
4 0

Answer:

147.473 g/mol

Explanation:

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A act of law to remove
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3 years ago
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A balanced chemical reaction obeys the law of...
Anit [1.1K]

Answer: (C) conservation of matter

Solution: Law of conservation of matter or mass states that' total mass of the reactants should always be equal to the total mass of the product that is the total mass is remained conserved in a chemical reaction.

A balanced chemical equation always follow this law.

For example:

2H_2(g)+O_2(g)\rightarrow 2H_2O(l)

Mass of hydrogen = 1 g/mol

Mass of Oxygen = 16 g/mol

Total mass on the reactants = 2(2×1)+(2×16)= 36g/mol

Total mass on the product side = 2[(2×1) +16] = 36 g/mol

As,

Mass on reactant side = Mass on the product side

Therefore, a balanced chemical reaction follows Law of Conservation of mass.

4 0
3 years ago
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Kepler’s investigations of the movement of the planets explained which of the following?
morpeh [17]

Answer:

The earth‘s orbit

Explanation:

8 0
3 years ago
A tank of gas has partial pressures of nitrogen and oxygen equal to 1.61 × 104 kPa and 4.34 × 103 kPa , respectively.What is t
Alchen [17]

Answer:

20.44\times 10^3 kPa is the total pressure of the tank.

Explanation:

Partial pressures of nitrogen = p_{N_2}=1.61\times 10^4 kPa

Partial pressure of oxygen =  p_{O_2}=4.34\times 10^3 kPa

Total pressure of gases in the tank = P

Applying Dalton's law of partial pressures :

P=p_{N_2}+p_{O_2}=1.61\times 10^4 kPa+4.34\times 10^3 kPa

P=20.44\times 10^3 kPa

20.44\times 10^3 kPa is the total pressure of the tank.

7 0
3 years ago
Given that Δ H ∘ f [ Br ( g ) ] = 111.9 kJ ⋅ mol − 1 Δ H ∘ f [ C ( g ) ] = 716.7 kJ ⋅ mol − 1 Δ H ∘ f [ CBr 4 ( g ) ] = 29.4 kJ
JulsSmile [24]

Answer:

283.725 kJ ⋅ mol − 1

Explanation:

C(s) + 2Br2(g) ⇒ CBr4(g) , Δ H ∘ = 29.4 kJ ⋅ mol − 1

\frac{1}{2}Br2(g) ⇒ Br(g) ,  Δ H ∘ = 111.9 kJ ⋅ mol − 1

C(s) ⇒ C(g) ,  Δ H ∘ = 716.7 kJ ⋅ mol − 1

4*eqn(2) + eqn(3) ⇒ 2Br2(g) + C(s) ⇒ 4 Br(g) + C(g) , Δ H ∘ = 1164.3 kJ ⋅ mol − 1

eqn(1) - eqn(4) ⇒ 4 Br(g) + C(g) ⇒ CBr4(g) , Δ H ∘ = -1134.9 kJ ⋅ mol − 1

so,

   average bond enthalpy is \frac{1134.9}{4} = 283.725 kJ ⋅ mol − 1

4 0
3 years ago
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