3.2 g KClO3
Explanation:
1.1 g C12H22O11 × (1 mol C12H22O11/342.3 g C12H22O11)
= 0.0032 mol C12H22O11
0.0032 mol C12H22O11 × (8 mol KClO3/1 mol C12H22O11)
= 0.026 mol KClO3
Therefore, the minimum amount of KClO3 needed is
0.026 mol KClO3 × (122.55 g KClO3/1 mol KClO3)
= 3.2 g KClO3
Answer:
The distance in kilometer that Ted will go if he runs at the same pace for 285 minutes is 60km
Step-by-step explanation:
It was given that Ted Ted runs 20 km in 95 minutes, so for us to know how many kilometers he runs at the same pace for 285 minutes by using proportion.
For Ted to run 20 km in 95 minutes.

where given distance= 20km
Given time =95minutes
Speed=20/95 .............eqn(1)
To calculate the distance in kilometers Ted will go if he runs at the same pace for 285 minutes, then
Speed= K/225 ...............eqn(2)
Let us represent distance with 'K'
Given time 225minutes
Equating eqn(1) and eqn(2)
Hence the proportion model that Ted could use in modelling this situation is this;

Cross multiply
K×95=20×225

The distance in kilometer that Ted will go if
he runs at the same pace for 285 minutes is 60km
<span>Molar mass (MM) of benzene C6H6
C = 6 * 12 = 72u
H = 6 * 1 = 6u
MM C6H6 = 72 + 6 = 78 g / mol
Benzene - Molar Mass = 78 g --------- 1 mol
Of A Mix has 468 g -------------- x
78x = 468
X = 468/78
X = 6 moles
Molar mass (MM) of Hydrochloric Acid HCl
H = 1 * 1 = 1u
CI = 1 * 35 = 35u
MM HCl = 1 + 35 = 36 g / mol
Hydrochloric Acid - Molar Mass = 36 g ---------- 1 mol
Of A Mix has 72 g ------------ y
36y = 72
Y = 72/36
Y = 2 moles
Thus, a mixture has a total of 8 moles (6 mol + 2 mol).
Dividing One Mole Amount of Each Substance by the Number of Total Mole Amounts,
Then we will obtain a Molar Fraction of Each:
Molar fraction make benzene = (6/8) simplify 2 = 3/4
Molar Fraction to make Hydrochloric Acid = (2/8) = simplify 2 = 1/4
Note:. The sum of the molar fractions of the always give goes 1, we have: 3/4 + 1/4 = 1
ANSWER:
</span>
Nitrogen pure
carbon is mixture
Your protons are correct but it’s 28 neutrons not 27!