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WARRIOR [948]
2 years ago
6

2MG+O2 ...........2MGO

Chemistry
1 answer:
Galina-37 [17]2 years ago
5 0
Mg + 1/2 O2 → MgO

1 mol = 24 g of Mg

X mol = 12 g of Mg

x = 0.5 moles of Mg

Mg :MgO = 1:1 (coefficient from equations using mole ratio)

So

0.5 moles of MgO

1 mol MgO = (24+16) g = 40 g

0.5 moles of MgO = 0.5 × 40

= 20 g of MgO produced
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For water to change from a liquid to solid, thermal energy needs to be <br><br> .
telo118 [61]

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6 0
2 years ago
An aqueous 0.300 M glucose solution is prepared with a total volume of 0.150 L. The molecular weight of
kherson [118]

Answer:

We need 8.11 grams of glucose for this solution

Explanation:

Step 1: Data given

Molarity of the glucose solution = 0.300 M

Total volume = 0.150 L

The molecular weight of glucose = 180.16 g/mol

Step 2: Calculate moles of glucose in the solution

Moles glucose = molarity solution * volume

Moles glucose = 0.300 M * 0.150 L

Moles glucose = 0.045 moles glucose

Step 3: Calculate mass of glucose

MAss glucose = moles glucose* molecular weight of glucose

MAss glucose = 0.045 moles * 180.16 g/mol

MAss glucose = 8.11 grams

We need 8.11 grams of glucose for this solution

6 0
3 years ago
What is the maximum amount in moles of P2O5P2O5 that can theoretically be made from 112 gg of O2O2 and excess phosphorus
Nastasia [14]

Answer:

n_{P_2O_5}^{max}=1.4molP_2O_5

Explanation:

Hello!

In this case, since the reaction between phosphorous and oxygen to form diphosphorous pentoxide is:

2P+\frac{5}{2}O_2\rightarrow P_2O_5

Thus, since phosphorous is in excess and oxygen and diphosphorous pentoxide are in a 5/2:1 mole ratio, we can compute the maximum moles of product as shown below:

n_{P_2O_5}^{max}=112 gO_2*\frac{1molO_2}{32.00gO_2}*\frac{1molP_2O_5}{5/2molO_2}\\\\  n_{P_2O_5}^{max}=1.4molP_2O_5

Best regards!

5 0
2 years ago
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Arte-miy333 [17]
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5 0
3 years ago
Read 2 more answers
How many atoms are in 123 g of calcium
Keith_Richards [23]

Answer:

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1 mole = 6.022 * 10^23 atoms, so 3.075 moles Ca= 18.51*10^23 atoms

Explanation:

5 0
2 years ago
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