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
1.5052g BaCl2.2H2O => 1.5052g / 274.25 g/mol = 0.0054884 mol
=> 0.0054884 mol Ba
<span>This means that at most 0.0054884 mol BaSO4 can form since Ba is the limiting reagent. </span>
<span>0.0054884 mol BaSO4 => 0.0054884 mol * 233.39 g/mol = 1.2809 g BaSO4</span>
Answer:
The answer to your question is C₂HO₃
Explanation:
Data
Hydrogen = 3.25%
Carbon = 19.36%
Oxygen = 77.39%
Process
1.- Write the percent as grams
Hydrogen = 3.25 g
Carbon = 19.36 g
Oxygen = 77.39 g
2.- Convert the grams to moles
1 g of H ----------------- 1 mol
3,25 g of H ------------- x
x = (3.25 x 1) / 1
x = 3.25 moles
12 g of C ---------------- 1 mol
19.36 g of C ---------- x
x = (19.36 x 1) / 12
x = 1.61 moles
16g of O --------------- 1 mol
77.39 g of O --------- x
x = (77.39 x 1)/16
x = 4.83
3.- Divide by the lowest number of moles
Carbon = 3.25/1.61 = 2
Hydrogen = 1.61/1.61 = 1
Oxygen = 4.83/1.61 = 3
4.- Write the empirical formula
C₂HO₃
Answer:
Explanation:
MnO₂(s) + 4 HCl(aq) = MnCl₂(aq) + 2 H₂O(l) + Cl₂
87 g 22.4 x 10³ mL
volume of given chlorine gas at NTP or at 760 Torr and 273 K
= 175 x ( 273 + 25 ) x 715 / (273 x 760 )
= 179.71 mL
22.4 x 10³ mL of chlorine requires 87 g of MnO₂
179.4 mL of chlorine will require 87 x 179.4 / 22.4 x 10³ g
= 696.77 x 10⁻³ g
= 696.77 mg .
I believe it is spontaneous generation
