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slava [35]
3 years ago
11

What's the ratio of atoms of each element in the compound ammonium monohydrogen phosphate (NH4)2HPO4?

Chemistry
1 answer:
GREYUIT [131]3 years ago
4 0
So there are 2 N, 9 H, 1 P and 4 O. so N:H:P:O is 2:9:1:4? i think
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The combustion of ethene in the presence of excess oxygen yields carbon dioxide and water: c2h4 (g) + 3o2 (g) → 2co2 (g) + 2h2o
meriva

Answer:

\boxed{-267.5}

Explanation:

You can calculate the entropy change of a reaction by using the standard molar entropies of reactants and products.

The formula is

\Delta_{r} S^{\circ} = \sum_n {nS_{\text{products}}^{\circ} - \sum_{m} {mS_{\text{reactants}}^{\circ}}}

The equation for the reaction is

                        C₂H₄(g) + 3O₂(g) ⟶ 2CO₂(g) + 2H₂O(ℓ)

ΔS°/J·K⁻¹mol⁻¹   219.5      205.0         213.6         69.9

\Delta_{r} S^{\circ} = (2\times213.6 + 2\times69.9) - (1\times219.5 + 3\times205.0)\\\\= 567.0 - 834.5 = \boxed{-267.5 \text{ J}\cdot\text{K}^{-1} \text{mol}^{-1}}

3 0
3 years ago
Which of the following statements is true?
VikaD [51]

The answer is likely to be A

7 0
3 years ago
Read 2 more answers
C10H22 + O2 -- CO2 + H2O balance
Tasya [4]

Answer:

1,31÷2 =10,11

Explanation:

c10h22+31÷2o2=10co2+11h2o

5 0
3 years ago
4NH3 + 3O2 --> 2N2 + 6H2O
larisa86 [58]

The balanced reaction is:

4NH3 + 3O2 --> 2N2 + 6H2O

 <span>We are given the amount of reactants to be used for the reaction. This will be the starting point of our calculation.</span>

83.7g of O2 ( 1 mol / 32 g) = 2.62 mol O2

2.81 moles of NH3

From the balanced reaction, we have a 4:3 ratio of the reactants. The limiting reactant would be oxygen. We will use the amount for oxygen for further calculations.

<span>2.62 mol O2</span><span> (6 mol H2O  / 3 mol O2) (18.02 g H2O / 1 mol H2O) = 94.42 g H2O</span>

8 0
2 years ago
Enthalpy of <br><br> CH4(g) + 2NO2(g) -&gt; N2(g) + CO2(g) + 2H2O(l)
stira [4]

Answer:

-177.9 kJ.

Explanation:

Use Hess's law. Ca(s) + CO2(g) + 1/2O2(g) → CaCO3(s) ΔH = -812.8 kJ 2Ca(s) + O2(g) → 2CaO(s) ΔH = -1269.8 kJ We need to get rid of the Ca and O2 in the equations, so we need to change the equations so that they're on both sides so they "cancel" out, similar to a system of equations. I changed the second equation. Ca(s) + CO2(g) + 1/2O2(g) → CaCO3(s) ΔH = -812.8 kJ 2CaO(s) → 2Ca(s) + O2(g) ΔH = +1269.8 kJ The sign changes in the second equation above since the reaction changed direction. Next, we need to multiply the first equation by two in order to get the coefficients of the Ca and O2 to match those in the second equation. We also multiply the enthalpy of the first equation by 2. 2Ca(s) + 2CO2(g) + O2(g) → 2CaCO3(s) ΔH = -1625.6 kJ 2CaO(s) → 2Ca(s) + O2(g) ΔH = +1269.8 kJ Now we add the two equations. The O2 and 2Ca "cancel" since they're on opposite sides of the arrow. Think of it more mathematically. We add the two enthalpies and get 2CaO(s) + 2CO2(g) → 2CaCO3(s) and ΔH = -355.8 kJ. Finally divide by two to get the given equation: CaO(s) + CO2(g) → CaCO3(s) and ΔH = -177.9 kJ.

5 0
2 years ago
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