According to law of definite proportion, for a compound, elements always combine in fixed ratio by mass.
The formula of compound remains the same, let it be a_{x}b_{y} where, a and b are two different elements.
Since, the ratio of mass remains the same , calculate the ratio of masses of element a and b in both cases
\frac{a}{b}=\frac{15}{35}=\frac{10}{y}
rearranging,
y=\frac{10\times 35}{15}=23.3
Thus, mass of b produced will be 23.3 g.
3 is C i believe, and as for 4 minerals are solid, natural.. that's about all I can say...
Answer:
1.7 bar
Explanation:
We can use the <em>Ideal Gas Law</em> to calculate the individual gas pressure.
pV = nRT Divide both sides by V
p = (nRT)/V
Data: n = 1.7 × 10⁶ mol
R = 0.083 14 bar·L·K⁻¹mol⁻¹
T = 22 °C
V = 2.5 × 10⁷ L
Calculations:
(a) <em>Change the temperature to kelvins
</em>
T = (22 + 273.15) K
= 295.15 K
(b) Calculate the pressure
p = (1.7 × 10⁶ × 0.083 14 × 295.15)/(2.5× 10⁷)
= 1.7 bar
Answer:
Step 1;
q = w = -0.52571 kJ, ΔS = 0.876 J/K
Step 2
q = 0, w = ΔU = -7.5 kJ, ΔH = -5.00574 kJ
Explanation:
The given parameters are;
= 100 N·m
= 327 K
= 90 N·m
Step 1
For isothermal expansion, we have;
ΔU = ΔH = 0
w = n·R·T·ln(/) = 1 × 8.314 × 600.15 × ln(90/100) = -525.71
w ≈<em> -0.52571</em> kJ
At state 1, q = w = -0.52571 kJ
ΔS = -n·R·ln(/) = -1 × 8.314 × ln(90/100) ≈ 0.876
ΔS ≈ 0.876 J/K
Step 2
q = 0 for adiabatic process
ΔU = 25×(27 - 327) = -7,500
w = ΔU = <em>-7.5 kJ</em>
ΔH = ΔU + n·R·ΔT
ΔH = -7,500 + 8.3142 × 300 = -5,005.74
ΔH = ΔU = <em>-5.00574 kJ</em>
