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sergejj [24]
2 years ago
7

How many mol of C7H16 would you have if you have 76.36 grams? Give your answer to 2 decimal spaces.

Chemistry
1 answer:
tensa zangetsu [6.8K]2 years ago
7 0

0.761 mol of C_7H_{16}  would you have if you have 76.36 grams.

<h3>What is a mole?</h3>

A mole is defined as 6.02214076 × 10^{23}of some chemical unit, be it atoms, molecules, ions, or others. The mole is a convenient unit to use because of the great number of atoms, molecules, or others in any substance.

Given data:

Mass=76.36 grams

Moles = \frac{mass}{molar \;mass}

Moles = \frac{76.36 grams}{100.21 g/mol}

Moles = 0.761

Hence,  0.761 mol of  C_7H_{16} would you have if you have 76.36 grams.

Learn more about moles here:

brainly.com/question/8455949

#SPJ1

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Pipettes are used when measuring the volumes of liquids with high degree of precision. The following volumes were obtained durin
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Answer:

0.075

Explanation:

First obtain the mean of the measurement;

Mean = 10.15 + 9.95 + 9.99 + 10.02/4 = 10.03

Then obtain d^2= (mean-score)^2 for each score;

(10.15-10.03)^2 = 0.0144

(9.95-10.03)^2 = 0.0064

(9.99-10.03)^2 = 0.0016

(10.02-10.03)^2= 0.0001

∑d^2= 0.0144 + 0.0064 + 0.0016 + 0.0001

∑d^2= 0.0225

Variance = ∑d^2/N = 0.0225/4 = 0.005625

Standard deviation= √0.005625

Standard deviation= 0.075

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3 years ago
Which of these is a ball and stick model?
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Answer:

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You have at your disposal 3 1-pound bags of various pure salts that dissolve readily in water. You can add one of the bags to a
Sedaia [141]

Answer:

This question appears incomplete

Explanation:

However, it should be noted that addition of soluble salts generally lowers the freezing point of water hence after the addition, water will no longer freeze at 0°C but lower.

Soluble salts tend to form more ions in water, it is these ions that are responsible for interfering with the hydrogen bonds hence lowering the freezing. Thus, (since each bag are of the same weight) <u>the bag that contains the salt that ionizes more in water will lower the freezing point by the greatest amount</u>.

NOTE: Different weight of the salts could lead to more ions been formed in the water by some salts as against the other.

6 0
4 years ago
If I contain 3 moles of gas in a container with a volume of 60 liters and at a
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Answer:    n = 3.0 moles

V = 60.0 L

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From PV = nRT, you can find P

P = nRT/V = (3.0 mol)(0.0821 L-atm/K-mol)(400 K)/60.0L

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3 years ago
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Which of the following possess the greatest concentration of hydroxide ions?
jek_recluse [69]

Answer : The correct option is (d) a solution of 0.10 M NaOH

Explanation :

<u>(a) a solution of pH 3.0</u>

First we have to calculate the pOH.

pH+pOH=14\\\\pOH=14-pH\\\\pOH=14-3.0=11

Now we have to calculate the OH^- concentration.

pOH=-\log [OH^-]

11=-\log [OH^-]

[OH^-]=1.0\times 10^{-11}M

Thus, the OH^- concentration is, 1.0\times 10^{-11}M

<u>(b) a solution of 0.10 M NH_3</u>

As we know that 1 mole of NH_3 is a weak base. So, in a solution it will not dissociates completely.

So, the OH^- concentration will be less than 0.10 M

<u>(c) a solution with a pOH of 12.</u>

We have to calculate the OH^- concentration.

pOH=-\log [OH^-]

12=-\log [OH^-]

[OH^-]=1.0\times 10^{-12}M

Thus, the OH^- concentration is, 1.0\times 10^{-12}M

<u>(d) a solution of 0.10 M NaOH</u>

As we know that NaOH is a strong base. So, it dissociates to give Na^+ ion and OH^- ion.

So, 0.10 M of NaOH in a solution dissociates to give 0.10 M of Na^+ ion and 0.10 M of OH^- ion.

Thus, the OH^- concentration is, 0.10 M

<u>(e) a 1\times 10^{-4}M solution of HNO_2</u>

As we know that 1 mole of HNO_2 in a solution dissociates to give 1 mole of H^+ ion and 1 mole of NO_2^- ion.

So, 1\times 10^{-4}M of HNO_2 in a solution dissociates to give 1\times 10^{-4}M of H^+ ion and 1\times 10^{-4}M of NO_2^- ion.

The concentration of H^+ ion is 1\times 10^{-4}M

First we have to calculate the pH.

pH=-\log [H^+]

pH=-\log (1.0\times 10^{-4})

pH=4

Now we have to calculate the pOH.

pH+pOH=14\\\\pOH=14-pH\\\\pOH=14-4=10

Now we have to calculate the OH^- concentration.

pOH=-\log [OH^-]

10=-\log [OH^-]

[OH^-]=1.0\times 10^{-10}M

Thus, the OH^- concentration is, 1.0\times 10^{-10}M

From this we conclude that, a solution of 0.10 M NaOH possess the greatest concentration of hydroxide ions.

Hence, the correct option is (d)

3 0
3 years ago
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