The answer is 34.1 mL.
Solution:
Assuming ideal behavior of gases, we can use the universal gas law equation
P1V1/T1 = P2V2/T2
The terms with subscripts of one represent the given initial values while for terms with subscripts of two represent the standard states which is the final condition.
At STP, P2 is 760.0torr and T2 is 0°C or 273.15K. Substituting the values to the ideal gas expression, we can now calculate for the volume V2 of the gas at STP:
(800.0torr * 34.2mL) / 288.15K = (760.0torr * V2) / 273.15K
V2 = (800.0torr * 34.2mL * 273.15K) / (288.15K * 760.0torr)
V2 = 34.1 mL
Answer:
D. 0.3 M
Explanation:
NH4SH (s) <--> NH3 (g) + H2S (g)
Initial concentration 0.085mol/0.25L 0 0
Change in concentration -0.2M +0.2 M +0.2M
Equilibrium 0.035mol/0.25 L=0.14M 0.2M 0.2M
concentration
Change in concentration (NH4SH) = (0.085-0.035)mol/0.25L =0.2M
K = [NH3]*[H2S]/[NH4SH] = 0.2M*0.2M/0.14M ≈ 0.29 M ≈ 0.3M
Answer:
1.70 g.cm⁻³
Solution:
Data Given;
Mass = 84.7 g
Volume = 49.6 cm³
Density = ?
Formula Used;
Density = Mass ÷ Volume
Putting values,
Density = 84.7 g ÷ 49.6 cm³
Density = 1.70 g.cm⁻³
Answer : The mass of 
needed are, 1.515 grams.
Explanation :
First we have to calculate the mole of 
.
Now we have to calculate the moles of .
The balanced chemical reaction will be,
![PbSO_4+Na_2CrO_4\rightarrow PbCrO_4+Na_2SO_4[tex]From the balanced chemical reaction, we conclude thatAs, 1 mole of [tex]PbCrO_4](https://tex.z-dn.net/?f=PbSO_4%2BNa_2CrO_4%5Crightarrow%20PbCrO_4%2BNa_2SO_4%5Btex%5D%3C%2Fp%3E%3Cp%3EFrom%20the%20balanced%20chemical%20reaction%2C%20we%20conclude%20that%3C%2Fp%3E%3Cp%3EAs%2C%201%20mole%20of%20%5Btex%5DPbCrO_4)
produced from 1 mole of
So, 0.005 mole of produced from 0.005 mole of
Now we have to calculate the mass of
Therefore, the mass of needed are, 1.515 grams.
Ba stays as Ba+2 and Cl stays as Cl-
