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Kobotan [32]
3 years ago
11

Are these correct ?! Please help !!!!!

Chemistry
1 answer:
julsineya [31]3 years ago
6 0
All part B is right.
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How many grams of diphosphorus pentoxide result if 100.0 g of phosphorus are combined with sufficient oxygen?
gregori [183]
Answer is: 230 g.
Chemical reaction: P₄ + 5O₂ → 2P₂O₅.
m(P₄) = 100 g.
M(P₄) = 4 · 31 g/mol = 124 g/mol.
n(P₄) = m(P₄) ÷ M(P₄) = 100g ÷ 124g/mol = 0,806 mol.
From reaction: n(P₄) : n(P₂O5) = 1 : 2.
n(P₂O₅) = 1,612 mol.
m(P₂O₅) = 1,612 mol · 142g/mol = 230g.
M - molar mass.
n - amount of substance.
3 0
3 years ago
Which statement correctly describes a chemical equilibrium? it must take place in an open system the mass of the reactants and t
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In your choices, the best answer is the mass of the reactants and the mass of the products are no equal. The chemical equilibrium can take place in a close system and can not be affected by catalyst and is a reversible reaction. The best describe should be the concentration of reactants and products are constant.
6 0
3 years ago
Read 2 more answers
The pk1, pk2, and pkr for the amino acid glutamate are 2.1, 9.5, and 4.1, respectively. at ph 11.0, glutamate would be charged p
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8 0
4 years ago
In your own words what is equilibrium?
fenix001 [56]

Answer:

when the forward and reverse reactions occur at equal rates.

chemical reaction is in equilibrium when the concentrations of reactants and products are constant - their ratio does not vary.

8 0
3 years ago
Calculate δ h for the reaction:no (g) + o2 (g) ↔ no2 (g). given: 2o3(g) ↔ 3o2(g) δh=-426 kj o2(g) ↔ 2o(g) δh=+ 490 kj no(g) + o3
maks197457 [2]
To calculate the <span>δ h, we must balance first the reaction: 

NO + 0.5O2 -----> NO2

Then we write all the reactions,

2O3 -----> 3O2    </span><span>δ h = -426 kj        eq. (1)

O2 -----> 2O    </span><span>δ h = 490 kj             eq. (2)

NO + O3 -----> NO2 + O2    </span><span>δ h = -200 kj          eq. (3)


We divide eq. (1) by 2, we get

</span>O3 -----> 1.5O2    δ h = -213  kj             eq. (4)

Then, we subtract eq. (3) by eq. (4) 

NO + O3 ----->  NO2 + O2   δ h = -200 kj
-       (O3 -----> 1.5 O2         δ h = -213  kj)
NO -----> NO2 - 0.5O2        δ h = 13  kj               eq. (5)


eq. (2) divided by -2. (Note: Dividing or multiplying by negative number reverses the reaction)

O -----> 0.5O2  <span>δ h = -245  kj         eq. (6)
</span>
Add eq. (6) to eq. (5), we get

NO -----> NO2 - 0.5O2        δ h = 13  kj 
+  O -----> 0.5O2                 δ h = -245  kj
NO + O ----> NO2               δ h = -232 kj

<em>ANSWER:</em> <em>NO + O ----> NO2               δ h = -232 kj</em>


4 0
3 years ago
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