Answer:
0.43
Explanation:
divide the given mass by molar mass from the periodic table
Recall that density is Mass/Volume. We are given the mL of liquid which is volume so all we need is mass now. We are given the mass of the granulated cylinder both with and without the liquid, so if we subtract them, we can get the mass of the liquid by itself. So, 136.08-105.56= 30.52g. This is the mass of the liquid. We now have all we need to find the density. So, let’s plug these into the density formula. 30.52g/45.4mL= 0.672 g/mL. This is our final answer since the problem requests the answer in g/mL, but be careful, because some problems in the future may ask for g/L requiring unit conversions. Also note that 30.52 was 4 sigfigs and 45.4 was 3 sigfigs, and so dividing them required an answer that was 3 sigfigs as well, hence why the answer is in the thousandths place
Answer:
The volume of the gas sample at standard pressure is <u>819.5ml</u>
Explanation:
Solution Given:
let volume be V and temperature be T and pressure be P.
![V_1=250ml](https://tex.z-dn.net/?f=%20V_1%3D250ml)
![V_2=?](https://tex.z-dn.net/?f=%20V_2%3D%3F)
![P_{total}=735 mmhg](https://tex.z-dn.net/?f=%20P_%7Btotal%7D%3D735%20mmhg)
1 torr= 1 mmhg
42.2 torr=42.2 mmhg
so,
![P_{water}=42.2mmhg](https://tex.z-dn.net/?f=%20P_%7Bwater%7D%3D42.2mmhg)
![T_1=35°C=35+273=308 K](https://tex.z-dn.net/?f=%20T_1%3D35%C2%B0C%3D35%2B273%3D308%20K)
Now
firstly we need to find the pressure due to gas along by subtracting the vapor pressure of water.
![P_{gas}=P_{total}-P_{water}](https://tex.z-dn.net/?f=%20P_%7Bgas%7D%3DP_%7Btotal%7D-P_%7Bwater%7D%20)
=735-42.2=692.8 mmhg
Now
By using combined gas law equation:
![\frac{P_1*V_1}{T_1} =\frac{P_2*V_2}{T_2}](https://tex.z-dn.net/?f=%5Cfrac%7BP_1%2AV_1%7D%7BT_1%7D%20%3D%5Cfrac%7BP_2%2AV_2%7D%7BT_2%7D)
![V_2=\frac{P_1*}{P_2}*\frac{T_2}{T_1} *V_1](https://tex.z-dn.net/?f=V_2%3D%5Cfrac%7BP_1%2A%7D%7BP_2%7D%2A%5Cfrac%7BT_2%7D%7BT_1%7D%20%2AV_1)
![V_2=\frac{P_gas}{P_2}*\frac{T_2}{T_1} *V_1](https://tex.z-dn.net/?f=V_2%3D%5Cfrac%7BP_gas%7D%7BP_2%7D%2A%5Cfrac%7BT_2%7D%7BT_1%7D%20%2AV_1)
Here
are standard pressure and temperature respectively.
we have
![P_2=750mmhg \:and\: T_2=273K](https://tex.z-dn.net/?f=P_2%3D750mmhg%20%5C%3Aand%5C%3A%20T_2%3D273K)
Substituting value, we get
![V_2=\frac{692.8}{750}*\frac{273}{308} *250](https://tex.z-dn.net/?f=V_2%3D%5Cfrac%7B692.8%7D%7B750%7D%2A%5Cfrac%7B273%7D%7B308%7D%20%2A250)
![V_2= 819.51 ml](https://tex.z-dn.net/?f=V_2%3D%20819.51%20ml)
The mass of carbon contained in 2.25 g of potassium carbonate, K₂CO₃ is 0.196 g.
<h3>
Molecular mass of potassium carbonate</h3>
The molecular mass of potassium carbonate, K₂CO₃ is calculated as follows;
M = K₂CO₃
M = (39 x 2) + (12) + (16 x 3)
M = 138 g
mass of carbon in potassium carbonate, K₂CO₃ is = 12 g
The mass of carbon contained in 2.25 g of potassium carbonate, K₂CO₃ is calculated as follows;
138 g ------------ 12 g of carbon
2.25 g ------------ ?
= (2.25 x 12) / 138
= 0.196 g
Thus, the mass of carbon contained in 2.25 g of potassium carbonate, K₂CO₃ is 0.196 g.
Learn more about potassium carbonate here: brainly.com/question/27514966
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