Pangea was formed in the Paleozoic Era
Answer:
The density is 0.0187 g/L
Explanation:
First thing to do here is to calculate the Volume of 1 mole of CO2 using the ideal gas equation
Mathematically;
PV = nRT
thus V = nRT/P
what we have are;
n = 1 mole
R is the molar has constant = 0.082 L•atm•mol^-1•K^-1
P is the pressure = 0.0079 atm
T is temperature = 227 K
Substituting these values, we have;
V = nRT/P = (1 * 0.082 * 227)/0.0079
V = 2,356.20 dm^3
This means according to the parameters given in the question, the volume of 1 mole of carbon iv oxide is 2,356.20 dm^3
But this is not what we want to calculate
What we want to calculate is the density
Mathematically, we can calculate the density using the formula below;
density = molar mass/molar volume
Kindly recall that the molar mass of carbon iv oxide is 44 g/mol
Thus the density = 44/2356.20 = 0.018674136321195 which is approximately 0.0187 g/L
Answer:
False
Explanation:
It is wrong to claim that when the attraction between particles overcomes their motion, the particles will clump together to boil.
During boiling particles do no clump together, they tend to move apart more rapidly.
- Boiling occurs when the vapor pressure overcomes the ambient atmospheric pressure.
- The hotter part of the boiler close the heat source moves rapidly away because they have become less dense.
- The colder and denser part sinks and this interaction sets up a convection cell.
The correct answer is 13 I just did it bro
Answer:
0.85 mole
Explanation:
Step 1:
The balanced equation for the reaction of CaCl2 to produce CaCO3. This is illustrated below:
When CaCl2 react with Na2CO3, CaCO3 is produced according to the balanced equation:
CaCl2 + Na2CO3 -> CaCO3 + 2NaCl
Step 2:
Conversion of 85g of CaCO3 to mole. This is illustrated below:
Molar Mass of CaCO3 = 40 + 12 + (16x3) = 40 + 12 + 48 = 100g/mol
Mass of CaCO3 = 85g
Moles of CaCO3 =?
Number of mole = Mass /Molar Mass
Mole of CaCO3 = 85/100
Mole of caco= 0.85 mole
Step 3:
Determination of the number of mole of CaCl2 needed to produce 85g (i.e 0. 85 mole) of CaCO3.
This is illustrated below :
From the balanced equation above,
1 mole of CaCl2 reacted to produced 1 mole of CaCO3.
Therefore, 0.85 mole of CaCl2 will also react to produce 0.85 mole of CaCO3.
From the calculations made above, 0.85 mole of CaCl2 is needed to produce 85g of CaCO3
