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stepladder [879]
3 years ago
12

IS THIS A BALANCE EQUATION MgCl2 + Na2O -- > MgO + 2 NaCl

Chemistry
1 answer:
lorasvet [3.4K]3 years ago
7 0
Yes this is balances. 1 Mg on each side
2Cl on each side
2 Na on each side
1 O on each side

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CH, (g) + 20 (g) → CO(g) + 2H 0(g)
galben [10]
<h3>Answer:</h3>

126.14 g

<h3>Explanation:</h3>
  • Complete combustion of hydrocarbons yields water and carbon dioxide.
  • Methane is a hydrocarbon in the homologous series known as alkanes.
  • Methane undergoes combustion in air to produce water and carbon dioxide according to the equation below.

CH₄ (g) + 2O₂ (g) → CO₂(g) + 2H₂O(g)

<u>We are given;</u>

  • 3.5 moles of Methane

We are required to determine the mass of H₂O produced

<h3>Step 1: Moles of H₂O produced</h3>
  • From the equation 1 mole of methane undergoes combustion to produce 2 mole of H₂O.

Therefore, Moles of methane = Moles of H₂O × 2

Hence, Moles of H₂O = 7.0 moles

<h3>Step 2: Mass of H₂O produced </h3>

We know that; mass = Moles × Molar mass

Molar mass of water = 18.02 g/mol

Therefore;

Mass of water = 7.0 moles × 18.02 g/mol

                       = 126.14 g

Thus, the mass of water produced is 126.14 g

6 0
3 years ago
Which of the following statements describes a basketball under high pressure (fully inflated) compared to a basketball under low
Lera25 [3.4K]

In a fully inflated basketball, there are more particles of gas.

Answer: Option D.

<u>Explanation:</u>

When a ball is fully inflated, it means it has a lot of air in it. With a lot of air in it, it means that there a lot of gas particles in that ball. With the large amount of gas particles in it, there is a lot of collision among these particles inside the ball.

This collision leads to the pressure on the greater side compared to the ball which has less air in it and is deflated because it has less particles of gas inside.

8 0
3 years ago
Both the sweatshirt and the balloon are negatively charged. They are most
Makovka662 [10]

Answer:

B attract

Explanation:

There will be statication

8 0
2 years ago
A certain first-order reaction has a rate constant of 2.15×10−2 s−1 at 20 ∘C. What is the value of k at 55 ∘C if Ea = 72.0 kJ/mo
adell [148]

Answer:

k_2=0.504s^{-1}

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to calculate the rate constant at 55 °C by using the temperature-variable version of the Arrhenius equation:

ln(\frac{k_2}{k_1} )=-\frac{Ea}{R}(\frac{1}{T_2} -\frac{1}{T_1} )

Thus, we plug in the temperatures, activation energy and universal constant of gases in consistent units to obtain:

ln(\frac{k_2}{0.0215s^{-1}} )=-\frac{72000\frac{J}{mol}}{8.3145\frac{J}{mol*K}}(\frac{1}{55+273} -\frac{1}{20+273} ) \\\\ln(\frac{k_2}{0.0215s^{-1}} )=3.154\\\\k_2=0.0215s^{-1}exp(3.154)\\\\k_2=0.504s^{-1}

Regards!

3 0
3 years ago
Looking at a potential energy diagram what is the name given to represent the amount of energy given off during an exothermal re
harkovskaia [24]

Answer:

Released energy

Explanation:

8 0
3 years ago
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