First, we need to get the molar mass of:
KClO3 = 39.1 + 35.5 + 3*16 = 122.6 g/mol
KCl =39.1 + 35.5 = 74.6 g/mol
O2 = 16*2 = 32 g/mol
From the given equation we can see that:
every 2 moles of KClO3 gives 3 moles of O2
when mass = moles * molar mass
∴ the mass of KClO3 = (2mol of KClO3*122.6g/mol) = 245.2 g
and the mass of O2 then = 3 mol * 32g/mol = 96 g
so, 245.2 g of KClO3 gives 96 g of O2
A) 2.72 g of KClO3:
when 245.2 KClO3 gives → 96 g O2
2.72 g KClO3 gives → X
X = 2.72 g KClO3 * 96 g O2/245.2 KClO3
= 1.06 g of O2
B) 0.361 g KClO3:
when 245.2 g KClO3 gives → 96 g O2
0.361 g KClO3 gives → X
∴ X = 0.361g KClO3 * 96 g / 245.2 g
= 0.141 g of O2
C) 83.6 Kg KClO3:
when 245.2 g KClO3 gives → 96 g O2
83.6 Kg KClO3 gives → X
∴X = 83.6 Kg* 96 g O2 /245.2 g KClO3
= 32.7 Kg of O2
D) 22.4 mg of KClO3:
when 245.2 g KClO3 gives → 96 g O2
22.4 mg KClO3 gives → X
∴X = 22.4 mg * 96 g O2 / 245.2 g KClO3
= 8.8 mg of O2
Answer:
384.2 K
Explanation:
First we convert 27 °C to K:
- 27 °C + 273.16 = 300.16 K
With the absolute temperature we can use <em>Charles' law </em>to solve this problem. This law states that at constant pressure:
Where in this case:
We input the data:
300.16 K * 1600 m³ = T₂ * 1250 m³
And solve for T₂:
T₂ = 384.2 K
Use M x V = M' x V'
0.300 x V = 0.100 x 250
V = .......... ml
Answer:
Below
Explanation:
Salt solution, aka saline solution, is a homogeneous mixture. This means that the mixtures composition is completely uniform throughout. If you were to look at a saline solution, it would just look like plain water because the salt is dissolved (it looks the exact same in the entire mixture).
Something like bean soup would be an example of a heterogeneous mixture. You can see all the beans and vegetables as components of one mixture.
Hope this helps! Best of luck <3
