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chubhunter [2.5K]
3 years ago
5

In the following reaction, what coefficient will be written before the oxygen molecule (O2) to balance the chemical equation?

Chemistry
1 answer:
zmey [24]3 years ago
8 0
 RESPOSTA E 8, PARA BALANCEAR O OXIGENIO 
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How many moles in 68.1 grams of Cu(OH)2
Andrew [12]

To do this, you would first add together the molar mass of all involved elements, to find how many grams are in a mole of Cu(OH)2. Keep in mind, the molar mass is equal to the atomic mass of an element in grams. For example the molar mass of copper (Cu) would be 63.55 (with 2 sig. figs.)

Therefore, now we add together the mass of all elements involved.

Cu: (63.55)+O2(15.99x2=31.98)+H2(1.01x2=2.02)

63.55+31.98+2.02= 97.55g per mole of Cu(OH)2.

Now, divide what we have by how much it takes to get a mole of the stuff.

68.1/97.55= 0.698mol Cu(OH)2


6 0
3 years ago
Please help!!!! Thank you
kolezko [41]
The answer is “C“ since the earth rotation makes it seem like the stars and sun is moving when in reality we are moving ฅ^•ﻌ•^ฅ
6 0
3 years ago
Write dissociation equations for the following soluble salts dissolving in water. Then draw a particle - level representation fo
zhuklara [117]

Answer:

The dissociation equations for NaBr gives Na+ and Br-

The dissociation equations for ZnCl2 gives Zn2+ and 2 Cl-  

Explanation:

The following pictures shows that the dissociation of one particle of NaBr produces one particle of Na+ (sodium cation) and one particle of Br- (bromine anion).

The dissociation of one particle of ZnCl2 produces one particle of Zn+2 (Zinc cation) and two particles of Cl- (chlorine anion).

8 0
4 years ago
5. Given 18.5 grams of CHA and 24.0 grams of Oz the following reaction occurs. Calculate the number of grams of
ehidna [41]

Answer:

16.5 g of CO₂ are produced by 18.5 of methane and 24 g of O₂

Explanation:

This is a reaction of combustion:

CH₄ +  2O₂  →  CO₂  + 2H₂O

1 mol of methane react to 2 moles of oxygen in order to produce 1 mol of carbon dioxide and 2 moles of water.

First of all, we determine the moles of each reactant:

18.5 g . 1mol/ 16g = 1.15 moles of methane

24 g . 1mol / 32g = 0.75 moles of oxygen

Now, we determine the limiting reactant:

1 mol of methane reacts to 2 moles of oyxgen

1.15 moles of methane may react to (1.15 . 2) /1 = 2.3 moles.

We only have 0.75 moles of O₂ and we need 2.3, that's why the oxygen is the limiting reagent, because we do not have enough oxygen for the reaction.

2 moles of O₂ produce 1 mol of CO₂

Then 0.75 moles may produce (0.75 . 1) /2 = 0.375 moles of CO₂

We convert the moles to mass → 0.375 mol . 44g /mol = 16.5 g

7 0
3 years ago
A 9.75 L 9.75 L container holds a mixture of two gases at 41 ° C. 41 °C. The partial pressures of gas A and gas B, respectively,
Triss [41]

Answer:

The total  pressure P = 1.642 atm

Explanation:

From the dalton's law

Total pressure of the mixture is

P = P_{A} + P_{B} + P_{C} ----- (1)

P_A = 0.419 atm

P_B = 0.589 atm

P_C = \frac{nRT}{V}

n = 0.24 mole

R = 0.0821 \frac{L.Atm}{K.mol}

T = 41 °c = 314 K

P_C = \frac{(0.24)(0.0821)(314)}{9.75}

P_C = 0.634 atm

From equation (1)

P = 0.419 + 0.589 + 0.634

P = 1.642 atm

Thus the total  pressure P = 1.642 atm

3 0
3 years ago
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