The answer is 267.93 g
Molar mass of CaBr2 is the sum of atomic masses of Ca and Br:
Mr(CaBr2) = Ar(Ca) + 2Ar(Br)
Ar(Ca) = 40 g/mol
Ar(Br) = 79.9 g/mol
Mr(CaBr2) = 40 + 2 * 79.9 = 199.8 g/mol
The percentage of Br in CaBr2 is:
2Ar(Br) / Mr(CaBr2) * 100 = 2 * 79.9 / 199.8 * 100 = 79.98%
Now make a proportion:
x g in 79.98%
335 g in 100%
x : 79.98% = 335 g : 100%
x = 79.98% * 335 g : 100%
x = 267.93 g
It is going to be <span>Molar Volume
</span><span>3H2 + N2 --> 2NH3
</span><span> 54.1L*22.4 L/mol H2 , you can find mol of H2, then mol of NH3, and then L of NH3</span>
Answer:
a women standing in high heels
Explanation:
Answer:
0.861 L
Explanation:
We are given pressure, volume, and temperature, so let's apply the Combined Gas Law:
(P₁V₁)/T₁ = (P₂V₂)/T₂
Convert the temperatures to degrees Kelvin.
25.0°C -> 298 K, 100.0°C -> 373 K
Plug in the initial conditions on the left, then the final/new on the right, and solve for the unknown:
(165(2.5))/298 = (600(V₂))/373
V₂ = (165(2.5)(373))/(298(600))
V₂ = 0.861 L
Answer:
2.0202 grams
Explanation:
1.4% (m/v) glucose solution means: 1.4g glucose/100mL solution.
so ?g glucose = 144.3 mL soln
Now apply the conversion factor, and you have:
?g glucose = 144.3mL soln x (1.4g glucose/100mL soln).
so you have (144.3x1.4/100) g glucose= 2.0202 grams
