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worty [1.4K]
3 years ago
15

The density of benzene at 15 ∘C∘C is 0.8787 g/mL. Calculate the mass of 0.1700 LL of benzene at this temperature. Express your a

nswer using four significant figures.
Chemistry
1 answer:
SVETLANKA909090 [29]3 years ago
5 0

Answer:

Mass of benzene is: 149.3 g

Explanation:

Let's use density to calculate mass.

Density = Mass / Volume

Mass = Density . volume

Be careful, because density is in g/mL and the volume is in L. So let's convert the L to mL: 0.170 L . 1000 mL / 1L = 170 mL

0.8787 g/mL . 170 mL = Mass of benzene

Mass of benzene is: 149.3 g

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I believe the answer is compound B may have a lower molecular weight compared to compound A. 
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onsider the following reaction: CaCN2 + 3 H2O → CaCO3 + 2 NH3 105.0 g CaCN2 and 78.0 g H2O are reacted. Assuming 100% efficiency
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Answer : The excess reactant is, H_2O

The leftover amount of excess reagent is, 7.2 grams.

Solution : Given,

Mass of CaCN_2 = 105.0 g

Mass of H_2O = 78.0 g

Molar mass of CaCN_2 = 80.11 g/mole

Molar mass of H_2O = 18 g/mole

Molar mass of CaCO_3 = 100.09 g/mole

First we have to calculate the moles of CaCN_2 and H_2O.

\text{ Moles of }CaCN_2=\frac{\text{ Mass of }CaCN_2}{\text{ Molar mass of }CaCN_2}=\frac{105.0g}{80.11g/mole}=1.31moles

\text{ Moles of }H_2O=\frac{\text{ Mass of }H_2O}{\text{ Molar mass of }H_2O}=\frac{78.0g}{18g/mole}=4.33moles

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

CaCN_2+3H_2O\rightarrow CaCO_3+2NH_3

From the balanced reaction we conclude that

As, 1 mole of CaCN_2 react with 3 mole of H_2O

So, 1.31 moles of CaCN_2 react with 1.31\times 3=3.93 moles of H_2O

From this we conclude that, H_2O is an excess reagent because the given moles are greater than the required moles and CaCN_2 is a limiting reagent and it limits the formation of product.

Left moles of excess reactant = 4.33 - 3.93 = 0.4 moles

Now we have to calculate the mass of excess reactant.

\text{ Mass of excess reactant}=\text{ Moles of excess reactant}\times \text{ Molar mass of excess reactant}(H_2O)

\text{ Mass of excess reactant}=(0.4moles)\times (18g/mole)=7.2g

Thus, the leftover amount of excess reagent is, 7.2 grams.

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In a titration of hno3, you add a few drops of phenolphthalein indicator to 50.00 ml of acid in a flask. You quickly add 20.00 m
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Answer:

The HNO3 solution has a concentration of 0.07 M

Explanation:

<u>Step 1:</u> find a balanced equation

HNO3 (aq) + NaOH (aq) → NaNO3 (aq) + H2O (l)

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