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SCORPION-xisa [38]
3 years ago
14

In the reaction below, how many grams of h2o(g) are produced when 2.1 grams o2(g) are consumed? c4h6(g) + o2(g) → co2(g) + h2o(g

)
Chemistry
2 answers:
Sergio [31]3 years ago
4 0
<span>0.54 g
       
Lookup the atomic weights of oxygen and hydrogen. Atomic weight oxygen = 15.999 Atomic weight hydrogen = 1.00794 Calculate molar masses Molar mass O2 = 2 * 15.999 = 31.998 g/mol Molar mass H2O = 2 * 1.00794 + 15.999 = 18.01488 g/mol Now determine how many moles of O2 you have 2.1 g / 31.998 g/mol = 0.065629102 mol Balance the equation so you get as the balanced equation, 2 C4H6(g) + 11 O2(g) → 8 CO2(g) + 6 H2O(g) Looking at the balanced equation, for every 11 moles of O2 consumed, 5 moles of H2O are produced. So divide the number of moles of O2 you have by 11, then multiply by 5, giving: 0.065629102 mol / 11 * 5 = 0.02983141 mol And finally, multiply by the molar mass of H2O, so 0.02983141 mol * 18.01488 g/mol = 0.53740927 g Round the results to 2 significant figures getting 0.54 g</span>
amm18123 years ago
3 0
The balanced equation that illustrates the reaction is:
2C4H6 + 11O2 ......> 8CO2 + 6H2O 

number of moles = mass / molar mass 
number of moles of oxygen = 2.1 / 32 = 0.065625 moles

Now, from the balanced equation, we can note that:
11 moles of oxygen are required to produce 6 moles of water.
Therefore:
0.065625 moles of oxygen will produce:
(0.065625*6) / 11 = 0.03579 moles of water

number of moles = mass / molar mass
mass = number of moles * molar mass
mass of water = 0.03579 * 18 = 0.644 grams
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What is the empirical formula of a compound composed of 30.5 g potassium (k) and 6.24 g oxygen (o)?
VLD [36.1K]
K5O2

convert grams to moles, divide both by the smallest mole mass, multiply that until hole.

30.5 g K ÷ 39.10 = .78 mol
6.24 g O ÷ 16 = .39 mol
.78 mol ÷ .39 mol = 2.5
.39 mol ÷ .39 mol = 1
2.5 x 2 = 5
1 x 2 = 2
K5O2

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3 years ago
The measure of how closely a set of measurements are to each other is the best definition for which of the following terms. a er
ahrayia [7]
D. precision. At first glance you can mark out A and B because the answers does not relate to the question. If question had said "what do you call it when the measurement is close to the actual answer" then you would have picked C. So that leaves you D. precision.
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3 years ago
1. Which of the following is a correctly written thermochemical equation?
WITCHER [35]

Explanation:

1. Thermochemical equation is balance stoichiometric chemical equation written with the phases of the reactants and products in the brackets along with the enthalpy change of the reaction.

The given correct thermochemical reactions are:

Fe(s)+O_2(g)\rightarrow Fe_2O_3(s),\Delta H = 3,926 kJ&#10;

C_3H_8(g)+5O_2 (g)\rightarrow 3CO_2 (g)+4H_2O(l),\Delta H= 2,220 kJ/mol

2. Phase change affect the value of the enthalpy change of the thermochemical equation. This is because change in phase is accompanied by change in energy. For example:

H_2O(s)\rightarrow H_2O(g),\Delta H_{s}=51.1 kJ/mol

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In both reaction phase of water is changing with change in energy of enthalpy of reaction.

7 0
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Read 2 more answers
Chamber 1 and Chamber 2 have equal volumes of 1.0L and are assumed to be rigid containers. The chambers are connected by a valve
vitfil [10]
1) At tne same temperature and with the same volume, initially the chamber 1 has the dobule of moles of gas  than the chamber 2, so the pressure in the chamber 1 ( call it p1) is the double of the pressure of chamber 2 (p2)

=> p1 = 2 p2

Which is easy to demonstrate using ideal gas equation:

p1 = nRT/V = 2.0 mol * RT / 1 liter

p2 = nRT/V = 1.0 mol * RT / 1 liter

=> p1 / p2 = 2.0 / 1.0 = 2 => p1 = 2 * p2

2) Assuming that when the valve is opened there is not change in temperature, there will be 1.00 + 2.00 moles of gas in a volumen of 2 liters.

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p = nRT/V = 3.0 mol * RT / 2liter

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p / p1 = (3/2) / 2 = 3/4 => p = (3/4)p1

So, the answer is that the pressure in the chamber 1 decreases to 3/4 its original pressure.

You can also see how the pressure in chamber 2 changes:

p / p2 = (3/2) / 1 = 3/2, which means that the pressure in the chamber 2 decreases to 3/2 of its original pressure.
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Answer:

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3 0
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