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serious [3.7K]
3 years ago
14

Use coefficients to increase the atoms on each side. Check to make sure you have the same number of each type of atom on each si

de. Count the atoms on each side. Identify the atoms on each side.
Chemistry
2 answers:
tigry1 [53]3 years ago
8 0

Answer:

balancing

Explanation:

NemiM [27]3 years ago
4 0

Answer:

Balancing an equation

Explanation:

These steps describes the processes which are involved in balancing a chemical equation.

In balancing the a chemical equation, the law of conservation of mass is strictly adhered to. It suggests that in a chemical reaction, atoms are neither created nor destroyed, but they combine with one another.

  • Therefore, the number atoms at start and end of a reaction must be the same.
  • You can use coefficients to increase the atoms on each side.
  • Also, check to makes sure the number of atoms on both sides are the same.
  • Count the atoms on each side of the expression.
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If the mole fraction of nitric acid (HNO3) in an aqueous solution is 0.275, what is the percent by mass of HNO3?
jeka94
Molar mass of    N    = 14 g/molMolar mass of   O2    = 32 g/molAdding both masses = 46 g/molActual molar mass/ Empirical molar mass = 138.02 / 46 = 3Now multiplying this co effecient with empirical fomula NO2 = 3(NO2)                                                                                             = N3O6So according to above explanation,D) N3O6, is the correct answer.

8 0
3 years ago
Gaseous ethane will react with gaseous oxygen to produce gaseous carbon dioxide and gaseous water . Suppose 2.7 g of ethane is m
Bond [772]

Answer:

m_{H_2O}=4.86gH_2O

Explanation:

Hello,

In this case, the described chemical reaction is:

C_2H_6+\frac{7}{2} O_2\rightarrow 2CO_2+3H_2O

Thus, for the given reacting masses, we must identify the limiting reactant for us to determine the maximum mass of water that could be produced, therefore, we proceed to compute the available moles of ethane:

n_{C_2H_6}=2.7gC_2H_6*\frac{1molC_2H_6}{30gC_2H_6} =0.09molC_2H_6

Next, we compute the moles of ethane consumed by 13.0 grams of oxygen by using the 1:7/2 molar ratio between them:

n_{C_2H_6}^{consumed\ by \ O_2}=13.0gO_2*\frac{1molO_2}{32gO_2}*\frac{1molC_2H_6}{\frac{7}{2} molO_2}=0.116molC_2H_6

Thus, we notice there are less available moles of ethane, for that reason, it is the limiting reactant, thereby, the maximum amount of water is computed by considering the 1:3 molar ratio between ethane and water:

m_{H_2O}=0.09molC_2H_6*\frac{3molH_2O}{1molC_2H_6} *\frac{18gH_2O}{1molH_2O} \\\\m_{H_2O}=4.86gH_2O

Best regards.

3 0
4 years ago
what is the molecular formula for a compound that is 39.99% carbon, 6.73% hydrogen, and 53.28% oxygen and has a molar mass of 18
4vir4ik [10]

Answer:

IDEK

Explanation:

5 0
3 years ago
Barium nitrate and sodium sulfate solutions can be used to precipitate barium sulfate. How many grams
-Dominant- [34]

Answer:

40.8g of sodium sulfate must be added

Explanation:

The reaction of barium nitrate, Ba(NO₃)₂ with sodium sulfate, Na₂SO₄ is:

Ba(NO₃)₂ + Na₂SO₄ → 2 NaNO₃ + BaSO₄(s)

That means, for a complete reaction of an amount of barium nitrate you must add the same amount in moles of sodium sulfate. To solve this problem we need to convert the mass of barium nitrate to moles = Moles of sodium sulfate that must be added:

<em>Moles Ba(NO₃)₂ -Molar mass: 261.3g/mol-:</em>

75g * (1mol / 261.3g) = 0.287 moles = Moles Na₂SO₄

<em>Mass Na₂SO₄ -Molar mass: 142.04g/mol-:</em>

0.287 moles * (142.04g / mol) =

<h3>40.8g of sodium sulfate must be added</h3>
7 0
3 years ago
During the process of diluting a solution, ________.
lana [24]
A kfjfjdnsndmxkkdnndndnd
4 0
3 years ago
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