Hey there!
Magnesium chlorate: Mg(ClO₃)₂
Find molar mass.
Mg: 1 x 24.305 = 24.305
Cl: 2 x 35.453 = 70.906
O: 6 x 16 = 96
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191.211 g/mol
We have 187.54 grams.
187.54 ÷ 191.211 = 0.9808
There are 0.9808 moles in 187.54 grams of magnesium chlorate.
Hope this helps!
The answer would be 4,000
Because 474 is gonna go down but if it was 500 it will go up
Answer:
It takes 178.4 years to drop the activity from 15.0 Ci to 0.25 Ci
Explanation:
Step 1: Data given
Half-life time = 30.2 years
Initial activity = 15.0 Ci
Final activity = 0.250 Ci
Step 2: Calculate the time needed
N / No = e^(-0.693 * t / T1/2)
⇒ with N = The activity after dropped = 0.250 Ci
⇒ with N0 = the initial activity = 15.0
⇒ with t= the time (in years) needed to drop the activity from 15.0 to 0.250
⇒ with t1/2 = the half- life time is 30.2 years
0.250 / 15.0 = e^(-0.693 * t / 30.2)
ln(0.250/15.0) = (-0.693 * t / 30.2)
-4.09 * 30.2 = -0.693t
-123.65 = -0.693t
t= 178.4 years
It takes 178.4 years to drop the activity from 15.0 Ci to 0.25 Ci
Answer:
5.0 moles.
Explanation:
- The balanced equation of the reaction between hydrogen and oxgen to produce water is:
<em>H₂ + 1/2O₂ → H₂O.</em>
- It is clear that 1.0 mole of H₂ reacts with 0.5 mole of O₂ to produce 1.0 mole of H₂O.
∴ 2.5 mole of O₂ will react with 5.0 moles of H₂.
- The limiting reactant is O₂ and H₂ is found in excess (1.0 mole more).
From the stichiometry: 2.5 mole of O₂ will react with 5.0 moles of H₂ to produce 5.0 moles of H₂O.
<em>∴ 5.0 moles of water are produced when 6.0 moles of hydrogen gas react with 2.5 moles of oxygen gas</em>
Answer:
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I hope this answer helps