Answer:
16.02 g
Explanation:
the balanced equation for the decomposition of CuCO₃ is as follows
CuCO₃ --> CuO + CO₂
molar ratio of CuCO₃ to CO₂ is 1:1
number of CuCO₃ moles decomposed - 45 g / 123.5 g/mol = 0.364 mol
according to the molar ratio
1 mol of CuCO₃ decomposes to form 1 mol of CO₂
therefore 0.364 mol of CuCO₃ decomposes to form 0.364 mol of CO₂
number of CO₂ moles produced - 0.364 mol
therefore mass of CO₂ produced - 0.364 mol x 44 g/mol = 16.02 g
16.02 g of CO₂ produced
Answer:
A region on top of a mountain is cooler than at the base.
Explanation:
Pressure and temperature have direct relationship with each other. With the decrease in pressure, the temperature decreases and vice versa. When the air rises in the atmosphere, the pressure starts to fall. The low pressure at the peak of the mountains tends to cause the fall in temperature. It is because of this reason that it is cooler at the top of the mountain while the temperature is less cool in the foothills.
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Answer</h2>
2.626984127 m
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Explanation:</h2><h2>
</h2>
You have to know the equation that relates wavelength, frequency, and velocity (it's like speed but a bit different).
v = f x λ
where:
v = velocity
f = frequency
λ = Wavelength
Rearrange to make λ subject:
λ = v / f
We've been given 331 as the speed, 126 as the frequency. Sub it into the equation:
331 / 126 = 2.626984127 m
Answer:
[HF]₀ = 0.125M
Explanation:
NaOH + HF => NaF + H₂O
Adding 20ml of 0.200M NaOH into 25ml of HF solution neutralizes 0.004 mole of HF leaving 0.004 mole NaF in 0.045L with 0.001M H⁺ at pH = 3. This is 0.089M NaF and 0.001M HF remaining.
=> 45ml of solution with pH = 3 and contains 0.089M NaF from titration becomes a common ion problem.
HF ⇄ H⁺ + F⁻
C(eq) [HF] 10⁻³M 0.089M (<= soln after adding 20ml 0.200M NaOH)
Ka = [H⁺][F⁻]/[HF]₀ => [HF]₀ = [H⁺][F⁻]/Ka
[HF]₀ = (0.001)(0.089)/(7.1 x 10⁻⁴) M = 0.125M
Answer: The volume of the oxygen gas at a pressure of 2.50 atm will be 1.44 L
At constant temperature, the volume of a fixed mass of gas is inversely proportional to the pressure it exerts, then
PV = c
Thus, if the pressure increases, the volume decreases, and if the pressure decreases, the volume increases.
It is not necessary to know the exact value of the constant c to be able to use this law since for a fixed amount of gas at constant temperature, it is satisfied that,
P₁V₁ = P₂V₂
Where P₁ and P₂ as well as V₁ and V₂ correspond to pressures and volumes for two different states of the gas in question.
In this case the first oxygen gas state corresponds to P₁ = 1.00 atm and V₁ = 3.60 L while the second state would be P₂ = 2.50 atm and V₂ = y. Substituting in the previous equation,
1.00 atm x 3.60 L = 2.50 atm x y
We cleared y to find V₂,
V₂ = y = 
= 1.44 L
Then, <u>the volume of the oxygen gas at a pressure of 2.50 atm will be 1.44 L</u>
