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Nataliya [291]
3 years ago
14

What coefficient will Oy have after balancing this equation?

Chemistry
1 answer:
Gelneren [198K]3 years ago
8 0

la respuesta es la letra C. 8

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WHAT IS THE PERCENT BY VOLUME OF ETHANOL IN A SOLUTION THAT CONTAINS 35 mL ETHANOL IN 115 mL OF WATER?
Dmitry [639]
We need to first add both of the solution volumes together 35+115=150. Now we can divide the volume of the ethanol by the total volume 35/150=.233. To double check we can multiply the total volume by the percentage of ethanol by volume we got as a solution 150x.233=35. So the percentage by volume of ethanol in the solution is .233x100=23.3%.
3 0
3 years ago
You calculate that your semester average in history is 97.5. When you get your report card, your average is 96. What was the per
andrew11 [14]
Ok percent error is abs(calculated-actual)/actual(100%)

So 1.5/96 *100%
5 0
3 years ago
Read 2 more answers
What is the difference between a heterogeneous mixture and homogeneous mixture?
Fiesta28 [93]
A homogeneous mixture has the same uniform appearance and composition throughout


A heterogeneous mixture consists of visibly different substances or phases.
7 0
3 years ago
A sample of gas contains 0.1800 mol of CO(g) and 0.1800 mol of NO(g) and occupies a volume of 23.2 L. The following reaction tak
baherus [9]

Answer:

The volume of the sample is 17.4L

Explanation:

The reaction that occurs requires the same amount of CO and NO. As the moles added of both reactants are the same you don't have any limiting reactant. The only thing we need is the reaction where 4 moles of gases (2mol CO + 2mol NO) produce 3 moles of gases (2mol CO2 + 1mol N2). The moles produced are:

0.1800mol + 0.1800mol reactants =

0.3600mol reactant * (3mol products / 4mol reactants) = 0.2700 moles products.

Using Avogadro's law (States the moles of a gas are directly proportional to its pressure under constant temperature and pressure) we can find the volume of the products:

V1n2 = V2n1

<em>Where V is volume and n moles of 1, initial state and 2, final state of the gas</em>

Replacing:

V1 = 23.2L

n2 = 0.2700 moles

V2 = ??

n1 = 0.3600 moles

23.2L*0.2700mol = V2*0.3600moles

17.4L = V2

<h3>The volume of the sample is 17.4L</h3>
8 0
2 years ago
A 1.250-g sample of benzoic acid, C7H6O2, was placed in a combustion bomb. The bomb was filled with an excess of oxygen at high
Degger [83]

Answer:

3224 kJ/mol

Explanation:

The combustion of benzoic acid occurs as follows:

C₇H₆O₂ + 13/2O₂ → 7CO₂ + 3H₂O + dE

The change in temperature in the reaction is the change due the energy released, that is:

3.256K * (10.134kJ / K) = 33.00kJ are released when 1.250g reacts

To find the heat released per mole we have to find the moles of benzoic acid:

<em>Moles benzoic acid -Molar mass: 122.12g/mol-:</em>

1.250g * (1mol / 122.12g) = 0.0102 moles

<em />

The dE combustion per mole of benzoic acid is:

33.00kJ / 0.0102moles =

<em>3224 kJ/mol </em>

4 0
3 years ago
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