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ValentinkaMS [17]
3 years ago
11

What is the density of motor oil in g/mL if 1.25 quart (qt) weighs 2.25 pounds (lb)?

Chemistry
1 answer:
Doss [256]3 years ago
5 0

Answer:

0.86 g/mL

Explanation:

In order to solve the problem, you have to know the formula for "Density." Density refers to<em> </em><em>"mass per unit volume." </em><em>It's formula is:</em>

  • <em>Density (ρ) = </em>\frac{mass}{Volume}<em />

Next, we have to convert the units of measurement in accordance to what is being asked in the problem. It is asking for "g/mL," which means we have to convert 1.25 quarts<em> (qt.) </em>to mL and 2.25 pounds<em> (lbs.)</em> to grams.

Let's solve.

  • 1 quart = 946.35 mL

1.25 qt. x \frac{946.35 mL}{1 qt.} =<em> 1,182.94 mL</em>

  • 1 pound = 453.592 grams

2.25 pounds x \frac{453.592 grams}{1 pound} = <em>1,020.58 grams</em>

Now that we converted them to their appropriate units of measurement, we can now solve for density.

  • <em>ρ = </em>\frac{mass}{Volume}<em />
  • <em>ρ = </em>\frac{1,020.58 grams}{1,182.94 mL}<em />
  • <em>ρ = 0.86 g/mL</em>

<em>The density of the motor oil is 0.86 g/mL</em>

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What is the conjugate acid of HCO3- ?OH-H2CO3CO32-H2OH3O+
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<u>Answer:</u> The conjugate acid of HCO_3^- is H_2CO_3

<u>Explanation:</u>

According to the Bronsted-Lowry conjugate acid-base theory:

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To form a conjugate acid of HCO_3^-, this compound will accept one proton to form H_2CO_3

The chemical equation for the formation of conjugate acid follows:

HCO_3^-+H^+\rightarrow H_2CO_3

The conjugate acid formed is named as carbonic acid.

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3 years ago
Draw the cis and trans isomers of 2-butene, ch3chchch3. show all hydrogen atoms.
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The cis-trans isomers are shown in the picture. As you can see, in the cis isomer, the methane functional group are both in the same side. Same as well with the hydrogen atoms. On the other hand, these functional groups are opposite to each other in the trans isomer.

4 0
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Read 2 more answers
You are given 10ml (M) 20 Naoh solution in a conical flask and asked to titrate with (M) 20 Hcl and (M) 20 H2so4 separately. cal
Solnce55 [7]

Answer:

n_{HCl}=0.2molHCl\\n_{H_2SO_4}=0.1molH_2SO_4

Explanation:

Hello!

In this case, since the reactions between NaOH and the acids are:

NaOH+HCl\rightarrow NaCl+H_2O\\\\2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O

Whereas we can see the 1:1 and 2:1 mole ratios between NaOH and HCl and H2SO4 respectively. In such a way, at the equivalence point we realize that:

n_{HCl}=n_{NaOH}=V_{NaOH}M_{NaOH}=0.01L*20mol/L=0.2molHCl\\\\2n_{H_2SO_4}=n_{NaOH}\\\\n_{H_2SO_4}=\frac{1}{2} V_{NaOH}M_{NaOH}=\frac{0.01L*20mol/L}{2} =0.1molH_2SO_4

Best regards!

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Answer:

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