Answer:
-75 cm^3/min
Explanation:
Given from Boyle's law;
PV=C
From product rule;
VdP/dt + PdV/dt = dC/dt
but dC/dt = 0, V= 500 cm^3, P= 200kPa, dP= 30kPa/min
PdV/dt = dC/dt - VdP/dt
dV/dt = dC/dt - VdP/dt/ P
substituting values;
dV/dt = 0 - (500 * 30)/200
dV/dt = -75 cm^3/min
Answer:
Explanation:
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In this case, considering the given chemical reaction:
Thus, by applying the law of rate proportions, we can write:
Whereas the stoichiometric coefficients of reactants are negative due their disappearance and that of the product is positive due to its appearance. In such a way, when we relate the rate of disappearance of hydrogen gas to the rate of formation of hydrogen iodide, we obtain:
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Answer:
I think the second space is chemical bonds
The mass percent of hydrogen in CH₄O is 12.5%.
<h3>What is the mass percent?</h3>
Mass percent is the mass of the element divided by the mass of the compound or solute.
- Step 1: Calculate the mass of the compound.
mCH₄O = 1 mC + 4 mH + 1 mO = 1 (12.01 amu) + 4 (1.00 amu) + 1 (16.00 amu) = 32.01 amu
- Step 2: Calculate the mass of hydrogen in the compound.
mH in mCH₄O = 4 mH = 4 (1.00 amu) = 4.00 amu
- Step 3: Calculate the mass percent of hydrogen in the compound.
%H = (mH in mCH₄O / mCH₄O) × 100%
%H = 4.00 amu / 32.01 amu × 100% = 12.5%
The mass percent of hydrogen in CH₄O is 12.5%.
Learn more about mass percent here:brainly.com/question/4336659
Answer:
Explanation:
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