The solubility of gas in water is inversely proportional to the temperature. That means cool waters can hold more gases than hot waters. So when the oceans continue to warm all the green--houses gases present in oceanic waters will be released into to the atmosphere. This would further lead to the heating up of the planet. The global climate would keep changing and the temperature of the planet would increase further. Therefore, when the oceans continue to warm the amount of green-house gases cannot be sequestered by the oceans (as the temperatures are high) and so this would further enhance the greenhouse effect.
Answer:
I think yes
Explanation:
sorry i havent done these type a questions in a while
Answer:
Number of moles = 10.6 mol
Explanation:
Given data:
Molar mass of H = 1.008 g/mol
Molar mass of C = 12.01 g/mol
Molar mass of O = 16.00 g/mol
Mass of citric acid = 2.03 kg (2.03×1000 = 2030 g)
Number of moles of citric acid = ?
Solution:
Formula:
Number of moles = mass/molar mass
Now we will calculate the molar mass of citric acid:
C₆H₈O₇ = (12.01× 6) + (1.008×8) + (16.00×7)
C₆H₈O₇ = 72.06 + 8.064+112
C₆H₈O₇ = 192.124g/mol
Number of moles = 2030 g/ 192.124g/mol
Number of moles = 10.6 mol
The answer for the following problem is mentioned below.
<u><em>Therefore volume occupied by methane gas is 184.78 × 10^-3 liters </em></u>
Explanation:
Given:
mass of methane(
) = 272 grams
pressure (P) = 250 k Pa =250×10^3 Pa
temperature(t) = 54°C =54 + 273 = 327 K
Also given:
R = 8.31JK-1 mol-1 ,
Molar mass of methane() = 16.0 grams
We know;
According to the ideal gas equation,
<u><em>P × V = n × R × T</em></u>
here,
n = m÷M
n =272 ÷ 16
<u><em>n = 17 moles</em></u>
Therefore,
250×10^3 × V = 17 × 8.31 × 327
V = ( 17 × 8.31 × 327 ) ÷ ( 250×10^3 )
V = 184.78 × 10^-3 liters
<u><em>Therefore volume occupied by methane gas is 184.78 × 10^-3 liters </em></u>
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