Find the molar mass of CaCO3 then subtract the molar mass what it originally weighed and the loss of mass. Hopefully this works!
Answer:
5746.0 mL.
Explanation:
We can use the general law of ideal gas:<em> PV = nRT.</em>
where, P is the pressure of the gas in atm.
V is the volume of the gas in L.
n is the no. of moles of the gas in mol.
R is the general gas constant,
T is the temperature of the gas in K.
If n and P are constant, and have two different values of V and T:
<em>V₁T₂ = V₂T₁</em>
<em></em>
V₁ = 6193.0 mL, T₁ = 62.3°C + 273 = 335.3 K.
V₂ = ??? mL, T₂ = 38.1°C + 273 = 311.1 K.
<em>∴ V₂ = V₁T₂/T₁ </em>= (6193.0 mL)(311.1 K)/(335.3 K) = <em>5746.0 mL.</em>
Winter because the winter solstice is the point when the sun is at its lowest making it the coldest day of the year
Answer:
The mass of 2,50 moles of NaCl is 146, 25 g.
Explanation:
First we calculate the mass of 1 mol of NaCl, starting from the atomic weights of Na and Cl obtained from the periodic table. Then we calculate the mass of 2.50 moles of compound, making a simple rule of three:
Weight NaCl= Weight Na + Weight Cl= 23 g+ 35,5 g= 58, 5 g/ mol
1 mol ------ 58, 5 g
2,5 mol---x= (2,5 mol x 58, 5 g)/ 1 mol = <u>146, 25 g</u>
