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mixas84 [53]
2 years ago
15

Is the chemical equation balanced? Yes or no

Chemistry
2 answers:
Strike441 [17]2 years ago
6 0
I think it is not so no
irinina [24]2 years ago
5 0

Answer:

<em><u>Yes the equation is balanced </u></em>

Explanation:

<em>There</em><em> </em><em>is</em><em> </em><em>2Na</em><em> </em><em>and</em><em> </em><em>2F</em><em> </em><em>on</em><em> </em><em>both</em><em> </em><em>sides</em><em> </em><em>which</em><em> </em><em>means</em><em> </em><em>LHS</em><em>=</em><em>RHS</em><em>.</em><em> </em><em>Hence</em><em>,</em><em>its</em><em> </em><em>balanced</em><em>.</em><em> </em>

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What is the volume of 8.80 g of CH4 gas at STP?
Ann [662]

Answer:

12.32 L.

Explanation:

The following data were obtained from the question:

Mass of CH4 = 8.80 g

Volume of CH4 =?

Next, we shall determine the number of mole in 8.80 g of CH4. This can be obtained as follow:

Mass of CH4 = 8.80 g

Molar mass of CH4 = 12 + (1×4) = 12 + 4 = 16 g/mol

Mole of CH4 =?

Mole = mass/Molar mass

Mole of CH4 = 8.80 / 16

Mole of CH4 = 0.55 mole.

Finally, we shall determine the volume of the gas at stp as illustrated below:

1 mole of a gas occupies 22.4 L at stp.

Therefore, 0.55 mole of CH4 will occupy = 0.55 × 22.4 = 12.32 L.

Thus, 8.80 g of CH4 occupies 12.32 L at STP.

6 0
3 years ago
An aqueous sodium acetate, NaC2H3O2 , solution is made by dissolving 0.395 mol NaC2H3O2 in 0.505 kg of water. Calculate the mola
liq [111]

<u>Answer:</u> The molality of NaC_2H_3O_2 solution is 0.782 m

<u>Explanation:</u>

Molality is defined as the amount of solute expressed in the number of moles present per kilogram of solvent. The units of molarity are mol/kg. The formula used to calculate molality:

\text{Molality of solution}=\frac{\text{Moles of solute}}{\text{Mass of solvent (in kg)}} .....(1)

Given values:

Moles of NaC_2H_3O_2 = 0.395 mol

Mass of solvent (water) = 0.505 kg

Putting values in equation 1, we get:

\text{Molality of }NaC_2H_3O_2=\frac{0.395mol}{0.505kg}\\\\\text{Molality of }NaC_2H_3O_2=0.782m

Hence, the molality of NaC_2H_3O_2 solution is 0.782 m

8 0
3 years ago
Deuterium is a relatively uncommon form of hydrogen, but can be created from what common source?.
Elena-2011 [213]

Deuterium is a relatively uncommon form of hydrogen, but can be created from water.

  • Heavy hydrogen commonly  known as deuterium
  • stable isotopes of hydrogen
  • gets its name from the Greek word deuterons means second.
  • has only one proton and one neutron
  • nucleus of the hydrogen's deuterium atom is known as a deuteron containing one proton and one neutron.
  • Deuterium forms chemical bonds that are stronger than regular hydrogen
  • gas deuterium is colorless
  • Deuterated water is used in Magnetic Resonance Spectroscopy.
  • used in the determination of the isotopologue of various organic compounds.
  • used in Infrared Spectroscopy.

To know more about Deuterium visit : brainly.com/question/27870183

#SPJ4

8 0
2 years ago
What is the mass of oxygen in 3.34 g of potassium permanganate?
3241004551 [841]

Answer:

1.35 g

Explanation:

Data Given:

mass of Potassium Permagnate (KMnO₄) = 3.34 g

Mass of Oxygen: ?

Solution:

First find the percentage composition of Oxygen in Potassium Permagnate (KMnO₄)

So,

Molar Mass of KMnO₄ = 39 + 55 + 4(16)

Molar Mass of KMnO₄ = 158 g/mol

Calculate the mole percent composition of  Oxygen in Potassium Permagnate (KMnO₄).

Mass contributed by Oxygen (O) = 4 (16) = 64 g

Since the percentage of compound is 100

So,

                        Percent of Oxygen (O) = 64 / 158 x 100

                        Percent of Oxygen (O) = 40.5 %

It means that for ever gram of Potassium Permagnate (KMnO₄) there is 0.405 g of Oxygen (O) is present.

So,

for the 3.34 grams of Potassium Permagnate (KMnO₄) the mass of Oxygen will be

                  mass of Oxygen (O) = 0.405 x 3.34 g

                  mass of Oxygen (O) = 1.35 g

5 0
3 years ago
wo reactions and their equilibrium constants are given. A + 2 B − ⇀ ↽ − 2 C K 1 = 2.57 2 C − ⇀ ↽ − D K 2 = 0.226 A+2B↽−−⇀2CK1=2.
Papessa [141]

<u>Answer:</u> The value of equilibrium constant for the net reaction is 11.37

<u>Explanation:</u>

The given chemical equations follows:

<u>Equation 1:</u>  A+2B\xrightarrow[]{K_1} 2C

<u>Equation 2:</u>  2C\xrightarrow[]{K_2} D

The net equation follows:

D\xrightarrow[]{K} A+2B

As, the net reaction is the result of the addition of first equation and the reverse of second equation. So, the equilibrium constant for the net reaction will be the multiplication of first equilibrium constant and the inverse of second equilibrium constant.

The value of equilibrium constant for net reaction is:

K=K_1\times \frac{1}{K_2}

We are given:

K_1=2.57

K_2=0.226

Putting values in above equation, we get:

K=2.57\times \frac{1}{0.226}=11.37

Hence, the value of equilibrium constant for the net reaction is 11.37

6 0
3 years ago
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