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AysviL [449]
3 years ago
12

How much has CO2 in the atmosphere increased since the Industrial Revolution?

Chemistry
1 answer:
leva [86]3 years ago
8 0

Answer:

The concentration of carbon dioxide in the atmosphere has increased more in the northern hemisphere where more fossil fuel burning occurs. Since the Industrial Revolution the concentration globally has increased by about 40 % .

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What is the Ph of a 3.9* 10^-8 M OH- solution
Maksim231197 [3]
Molarity = 3.9 * 10 ^ -8

pH = - log (3.9 * 10 ^ -8)

pH = - (-7.40893539)

pH = 7.409 = 7.4 ... Ans
3 0
3 years ago
When 15.0 g of fluorite (CaF₂) reacts with excess sulfuric acid, hydrogen fluoride gas is collected at 744 torr and 25.5°C. Soli
8090 [49]

The reaction equation is:  CaF₂ + H₂SO₄ → 2HF + CaSO₄

The molar ratio between fluorite and hydrogen fluoride is 1 : 1.

The moles of fluorite supplied are:

Moles = 15 / 78.07                            Moles = 0.200

The moles of hydrogen fluoride produced will be 0.2.Now, we may use the ideal gas equation to determine the temperature:

PV = nRT                                            T = PV/nR

T = (875 * 8.63) / (0.2 * 62.36)

T = 605.45K

The temperature will be 331.85 °C  which is  required to store the gas in an 8.63-L container at 875 torr.

To know more about ideal gas equation here

brainly.com/question/6776000

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3 0
2 years ago
PLEASE HELP, NO BOTS PLEASE
aleksandrvk [35]
Protons: 7
electrons: 7
neutrons: 7
4 0
2 years ago
Read 2 more answers
How many moles of Manganese there in are 5.76 x 10(15) atoms of Mn?
aksik [14]

Answer:

1. 9.57 × 10^-9 moles.

2. 7.38mol

Explanation:

1.) To find the number of moles there are in the number of particles in an atom, we divide the number of particles (nA) by Avagadro's constant (6.02 × 10^23)

Hence, to find the number of moles (n) of Manganese (Mn), we say:

5.76 x 10^15 atoms ÷ 6.02 × 10^23

5.76/6.02 × 10^(15-23)

= 0.957 × 10^-8

= 9.57 × 10^-9 moles.

2.) Mole = mass/molar mass

Molar mass of sodium chloride (NaCl) = 23 + 35.5

= 58.5g/mol

mole = 431.6 g ÷ 58.5g/mol

mole = 7.38mol

7 0
3 years ago
A vessel of volume 22.4 dm3 contains 2.0 mol h2 and 1.0 mol n2 at 273.15 k initially. all the h2 reacted with sufficient n2 to f
yanalaym [24]
Volume = 22.4 dm3
n = 2 mol of H2
n = 1 mol of N2
Temperature = 273.15
All H2 reacts
reaction
N2 + 3H2 = 2NH3
1:3 ratio
Calculation:
N2 initial - N2 reacted = Final N2
1 - 2*(1/3) = 0.3333 mol of N2 left
H2 = 0 left
NH3 formed = 2/3*1 = 2/3 = 0.666
Total mol:
0.3333 + 0.666 = 1 mol
Apply the equation : 
PV = nRT
P = nRT/V = 1*0.0082*(273.15)/(22.4) = 0.0999924 atm
PH2 = 0
PN2 = 1/3*0.0999924 = 0.0333308 atm
PNH3 = 2/3*0.0999924 = 0.0666616 atm
Answer is 0.0666616 atm
7 0
3 years ago
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