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2 years ago

2. Which of the following is true about the total number of reactants and the total number of

Chemistry

2 answers:

Soloha48 [4]2 years ago

8 0

9 moles of reactants chemically change into 11 moles of product.

Explanation:

In the reaction above, 9 moles of reactants chemically change into 11 mole of products. The coefficients in a reaction is the number of moles of the reacting atoms .

For example 8O₂ depicts 8 moles of two oxygen atoms.
The number of moles is a unit for quantifying particles.
You can liken it to a dozen, gross or a score.
Since a mole of a substance contains avogadro number of particles. We can relate the number of moles to other parameters.

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laila [671]2 years ago

8 0

Answer:

A) 9 moles of reactants chemically change into 11 moles of product.

Explanation:

C5H12(l) + 8O2(g) -> 5CO2(g) + 6H2O(g)

The first step is to check if the equation is indeed a balanced one before proceeding to compare the moles.

This is done by comparing the number of atoms on the reactant side with that in the product side, if equal the reaction is indeed a balanced one.

Number of moles of Carbon atoms

Reactant = 5

Product = 5

Number of moles of Hydrogen atoms

Reactant = 12

Product = 12

Number of moles of Oxygen atoms

Reactant = 16

Product = 10 + 6 = 16

Since the equation is balanced, we can now compare the moles.

From the reactant side we have a total number of 9 (1 + 8 ) moles reacting to yield a total of 11 ( 5 + 6 ) moles of product.

This alone leads us to take option A as our answer.

Let us see how other options are wrong however;

Option B) It cannot be grams as what chemical equations shows us the molar (number of moles) relationship between elememts/molecules/compounds/ions etc. The gram relationship can be calculated however if the molar mass of the compounds is taken into consideration.

Option C) This is wrong again due to same reasons mentioned for option B.

Option C) In the compound C5H12 alone, there are 17 (5 C and 12 H) atoms. Hence this option is also wrong.

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State and explain the relative change in the pH and in the buffer-component concentration ratio, [NaA]/[HA], for each of the fol

melisa1 [442]

When pure HA is added to the buffer, the buffer component ratio and the pH decrease.

<h3>State and explain the relative change in the pH and in the buffer-component concentration ratio, [NaA]/[HA] for the dissolve of pure HA in the buffer.</h3>

When pure HA is added to the buffer, the buffer component ratio and the pH decrease. The added HA increases the concentrations of NA and HA. However, there is a greater relative increase in the concentration of HA. Hence, the ratio of [NaA]/[HA] decreases, causing the solution to become more acidic.

The capacity of a buffer to withstand pH change is measured. The concentration of the buffer's components namely, the acid and its conjugate base determine this ability. Greater buffer capacity is associated with higher buffer concentration.

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1 year ago

Which image shows 2-hexyne?

MaRussiya [10]

Answer: C

Explanation:

A. Shows 3-Hexyne (NOT 2-HEXYNE)

B. Shows 7 carbons (too many) (NOT 2-HEXYNE)

C. Shows a triple bond (yne) and 6 carbons and it's on the second carbon (2-HEXYNE)

D. Shows two substitent on the second carbon but the triple bond is on the 3rd carbon so it's 2,2-dimethyl-3-heptyne (NOT 2-HEXYNE)

4 0

2 years ago

A beaker with 1.60×102 mL of an acetic acid buffer with a pH of 5.000 is sitting on a benchtop. The total molarity of acid and c

stich3 [128]

Answer:

The pH will change 0.16 ( from 5.00 to 4.84)

Explanation:

Step 1: Data given

volume of acetic acid buffer = 160 mL

The total molarity of acid and conjugate base in this buffer is 0.100 M

A student adds 7.10 mL of a 0.460 M HCl solution to the beaker.

The pKa of acetic acid is 4.740

pH = 5.00

Step 2: Calculate concentration of acid

Consider x = concentration acid

Consider y = concentration conjugate base

x + y = 0.100

5.00 = 4.740 + log y/x

5.00 - 4.740 = log y/x

0.26 = log y/x

10^0.26 =1.82 = y/x

1.82 x = y

Since x+y = 0.100

x + 1.82 x = 0.100

2.82 x = 0.100

x =0.0355 M = concentration acid

Step 3: Calculate concentration of conjugate base

y = 0.100 - x

0.100 - 0.0355 =0.0645 M= concentration conjugate base

Step 4: Calculate moles of acid

Moles = volume * molarity

moles acid = 0.160 L * 0.0355 M= 0.00568 moles

Step 5: Calculate moles of conjugate base

moles conjugate base = 0.0645 M * 0.160 L=0.01032 moles

Step 6: Calculate moles HCl

moles HCl = 7.10 * 10^-3 L * 0.460 M=0.003266 moles

Step 7: Calculate new moles

A- + H+ = HA

moles conjugate base = 0.01032 - 0.003266 =0.007054 moles

moles acid = 0.00568 + 0.003266=0.008946 moles

Step 8: Calculate the total volume

total volume = 160 + 7.10 = 167.1 mL = 0.1671 L

Step 9: Calculate the concentration of the acid

concentration acid = 0.008946/ 0.1671 =0.0535 M

Step 10: Calculate the concentration of conjugate base

concentration conjugate base = 0.007054/ 0.1671 =0.0422 M

Step 11: Calculate the pH

pH = 4.740 + log 0.0535/ 0.0422=4.84

change pH = 5.00 - 4.84=0.16

The pH will change 0.16

5 0

3 years ago

If 50.75 g of a gas occupies 10.0 L at STP, 129.3 g of the gas will occupy __________ L at STP.

Vlad [161]

If 50.75 g of a gas occupies 10.0 L at STP, 129.3 g of the gas will occupy 25.48 L at STP.

<h3>How to calculate volume?</h3>

The volume of a gas at STP can be calculated using the direct proportion method.

According to this question, 50.75 g of a gas occupies 10.0 L at STP, then 129.3g of the same gas will occupy the following:

= 129.3 × 10/50.75

= 25.48L

Therefore, if 50.75 g of a gas occupies 10.0 L at STP, 129.3 g of the gas will occupy 25.48 L at STP.

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1 year ago

A student puts 0.020 mol of methyl methanoate into an empty and rigid 1.0 L vessel at 450 K. The pressure is measured to be 0.74

stellarik [79]

Explanation:

Starting moles of ethanol acid = 0.020 mol

At the equilibrium 50 % of the ethanol acid molecules reacted

∴ Moles of ethanol acid reacted = 0.020 mol * 50 %/100 %

= 0.010 mol

Moles of ethanol acid remain = 0.020 mol + 0.010 mol = 0.010 mol

Moles of the product $(CH3COOH)^{2}$ gas formed are calculated as

0.010 mol CH3COOH * 1 mol $(CH3COOH)^{2}$ / 2 mol CH3COOH

= 0.005 mol $(CH3COOH)^{2}$

Therefore at the equilibrium total moles of gas present in the vessel are 0.010 mol CH3COOH and 0.005 mol $(CH3COOH)^{2}$

That is total gas moles at equilibrium = 0.010 mol + 0.005 mol = 0.015 mol

Now Calculate the pressure :

0.020 mol gas has pressure of 0.74 atm therefore at the same condition what will be the pressure exerted by 0.015 mol gas

P1/n1 = P2/n2

P2 = P1*n2 / n1

= 0.74 atm * 0.015 mol / 0.020 mol

= 0.555 atm

4 0

3 years ago