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strojnjashka [21]
3 years ago
9

Someone sprays perfume in another room. It can be smelled a few minutes later.

Chemistry
2 answers:
raketka [301]3 years ago
4 0

D is the right answer

Molodets [167]3 years ago
4 0
The answer would be letter D
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Which has a greater amount of particles, 1.00 mole of hydrogen (H) or 1.00 moles of oxygen(0)?
Bas_tet [7]

Answer:

Both have the same amount of particles.

Explanation:

From Avogadro's hypothesis, we understood that 1 mole of any substance contains 6.02×10²³ particles.

This implies that 1 mole of Hydrogen contains 6.02×10²³ particles. Also, 1 mole of oxygen contains 6.02×10²³ particles.

Thus, 1 mole of Hydrogen and 1 mole of oxygen contains the same number of particles.

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
How many particles would be found in a 89.2 g sample of Na?
julsineya [31]
One mole of a substance contains 6.02 × 10∧23 particles. Thus we first convert  89.2 g to moles. 1 mole of sodium contains 23 g
Hence 89.2 g = 89.2 / 23 g = 3.878 moles
Therefore, 3.878 × 6.02×10∧23 particles= 23.346 × 10∧23 particles
Hence 89.2 g of sodium contains 2.335 ×10∧24 particles 
5 0
3 years ago
Which of the following demonstrates a double-replacement reaction?
Volgvan
A: Na3PO4 + MnCl2 > Mn3(PO4)2 + NaCl
6 0
3 years ago
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Please help me out i'm new here but i really need help for science (if you need help with any hw like ela i can help!) (50 point
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6 0
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