Answer:
2.22 g/L
Explanation:
There's a relationship using the ideal gas law between molar mass and density: 
, where MM is the molar mass, d is the density, R is the gas constant, T is the temperature, and P is the pressure.
We know from the problem that MM = 32.49 g/mol, T = 458 Kelvin, and P = 2.569 atm. The gas constant, R, in terms of the units atm and Kelvin is 0.08206. Let's substitute these values into the formula:
Solve for d:
d * 0.08206 * 458 K = 32.49 * 2.569
d = (32.49 * 2.569) / (0.08206 * 458 K) ≈ 2.22 g/L
The answer is thus 2.22 g/L.
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Answer:
The full amount (5.00 g) will be dissolved in 1 L of water at 25°C.
Explanation:
The molecular weight (MW) of Vanillin (C₈H₈O₃) is calculated from the chemical formula as follows:
MW(C₈H₈O₃) = (12 g/mol x 8) + (1 g/mol x 8) + (16 g/mol x 3) = 152 g/mol
If 0.070 mol of C₈H₈O₃ are soluble per liter of water at 25°C, the maximum mass that can be dissolved in 1 L is:
0.070 mol x 152 g/mol = 10.64 g
Since 5.00 g is lesser than the maximum amount that can be dissolved (10.64 g), the added amount will be completely dissolved in 1 L of water at 25°C.
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Answer:
c = 4
Explanation:
In general, for the reaction
a A + b B ⇒ c C + d D
the rate is given by:
rate = - 1/a ΔA/Δt = - 1/b ΔB/Δt = + 1/c ΔC/Δt = + 1/d ΔD/Δt
this is done so as to express the rate in a standarized way which is the same to all the reactants and products irrespective of their stoichiometric coefficients.
For this question in particular we know the coefficient of A and need to determine the coefficient c.
- 1/2 ΔA/Δt = + 1/c ΔC/Δt
- 1/2 (-0.0080 ) = + 1/c ( 0.0160 mol L⁻¹s⁻¹ )
0.0040 mol L⁻¹s⁻¹ c = 0.0160 mol L⁻¹s⁻¹
∴ c = 0.0160 / 0.0040 = 4
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