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olganol [36]
3 years ago
13

about 70 million tons of paper are used in the UnitedStates. If the U.S. population is about 301.6 million people, how many kilo

grams of paper are used by the averatge U.S. citizen?
Chemistry
1 answer:
tensa zangetsu [6.8K]3 years ago
4 0

Answer : The amount of paper used by average United States citizen are, 210.47 kilograms.

Explanation : Given,

Total mass of paper used in the United States = 70 million tons

Total number of people in the United States = 301.6 million

First we have to calculate the amount of paper used by average United States citizen.

\text{Amount of paper used}=\frac{\text{Total mass of paper used}}{\text{Total number of people}}

\text{Amount of paper used}=\frac{70\text{ million tons}}{301.6\text{million}}

\text{Amount of paper used}=0.232\text{ tons}

Now we have to convert amount of paper from tons to kilogram.

Conversion used :

1 ton = 907.185 kg

As, 1 ton = 907.185 kg

So, 0.232 ton = \frac{0.232ton}{1ton}\times 907.185kg=210.47kg

Therefore, the amount of paper used by average United States citizen are, 210.47 kilograms.

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When 1.82 mole of HCL reacts with excess MnO2, how many moles of Cl2 form?
liubo4ka [24]
The balanced reaction is:

MnO2<span>(s) + 4HCl(aq) → Cl2(g) + MnCl2(aq) + 2H2O(l)
</span>
We are given the amount of hydrochloric acid to be used for the reaction. This will be the starting point for the calculations.

1.82 mol HCl ( 1 mol Cl2 / 4 mol HCl) = 0.46 mol Cl2

Therefore, 0.46 mol of chlorine gas is produced for the reaction of hydrochloric acid and manganese oxide.
6 0
4 years ago
At 25.0 ° C, a 10.00 L vessel is filled with 5.25 moles of Gas A and 7.05 moles of Gas B. What is the total pressure?
abruzzese [7]

Answer:

P = 30.1 atm

Explanation:

Given data:

Temperature of vessel = 25°C

Volume of vessel = 10.00 L

Moles in vessel = A + B = 5.25 mol + 7.05 mol = 12.3 moles

Total pressure inside vessel = ?

Solution:

The given problem will be solve by using general gas equation,

PV = nRT

P= Pressure

V = volume

n = number of moles

R = general gas constant = 0.0821 atm.L/ mol.K  

T = temperature in kelvin

Now we will convert the temperature.

25+273 = 298 K

P = nRT/V

P = 12.3 mol × 0.0821 atm.L/ mol.K × 298 K  / 10.00 L

P = 300.93 /  10.00 L

P = 30.1 atm

7 0
4 years ago
Given 7.20 g of butanoic acid and excess ethanol, how many grams of ethyl butyrate would be synthesized, assuming a complete 100
julsineya [31]

The reaction between butanoic acid and ethanol is:

CH3CH2CH2COOH + CH3CH2OH → CH3CH2CH2COOCH3CH2 + H2O

Based on the reaction stoichiometry:

1 mole of butanoic acid forms 1 mole of ethyl butyrate

Now,

Molar mass of Butanoic acid = 88.0 g/mol

Given mass of butanoic acid = 7.20 g

Therefore, # moles of butanoic acid reacted = 7.20/88.0 = 0.0818 moles

# moles of ethyl butyrate formed = 0.0818 moles

Molar mass of ethyl butyrate = 116 g/mol

Mass of ethyl butyrate synthesized = 0.0818 * 116 = 9.49 g



8 0
3 years ago
A forensic chemist is given a white powder for analysis. She dissolves 0.50 g of the substance in 8.0 g of benzene. The solution
Sholpan [36]

Answer:

The given compound cannot be cocaine.

Explanation:

The chemist can comment on the nature of compound being cocaine or not from the depression in freezing point.

Depression in freezing point of is related to molality as:

Depression in freezing point = Kf X molality

Where

Kf = cryoscopic constant = 4.90°C/m

depression in freezing point = normal freezing point - freezing point of solution

depression in freezing point = 5.5-3.9 = 1.6°C

1.6°C = 4.90 X molality

molality=\frac{1.6}{4.90} = 0.327 m

we know that:

molality=\frac{moles}{mass of solvent(kg)}=\frac{moles}{0.008kg}

therefore

moles = 0.327X0.008 = 0.00261 mol

moles=\frac{mass}{molarmass}

molarmass=\frac{mass}{moles}=\frac{0.50}{0.00261}= 191.57g/mol

The molar mass of cocaine is 303.353

So the given compound cannot be cocaine.

3 0
4 years ago
given that 32.0g sulphur contains 6.02×10^23 sulphur atoms,how many atoms are there in 2.70g of aluminum​
dangina [55]

Answer:

6.022 × 10²² atoms

Explanation:

Generally 1 mol of any element contains 6.02×10^23  atoms. The number 6.022 × 10²³ is known as Avogadro's number.

Mass of Aluminium = 2.70g

Molar mass = 27g/mol

Number of moles = Mass / Molar mass = 2.70 / 27 = 0.1 mol

1 mol =  6.022 × 10²³

0.1 mol = x

x = 6.022 × 10²³ * 0.1 = 6.022 × 10²² atoms

5 0
3 years ago
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