I will give Brainliest and 30 points to whoever answers first!

Hi can someone hellp this is grade 7

saw5 [17]

Explanation:

option B kinetic energy

hope this helps you!

3 0

2 years ago

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What is the name for a star that has two shells of hydrogen and helium undergoing fusion?

nasty-shy [4]

Answer:

I am not so sure but I think it is A.

Explanation:

If i got it wrong sorry.... :(

6 0

3 years ago

1. Sodium and water react according to the following equation. If 31.5g of sodium are added to excess water, how many liters of

Vika [28.1K]

<h3> Question 1</h3>

The number of liter of hydrogen gas that are formed at STP is

15.344 L

calculation

2 Na +2 H₂O → 2NaOH +H₂

calculate the moles of Na

moles =mass/molar mass

from periodic table the molar mass of Na = 23 g/mol

moles = 31.5 g /23 g /mol = 1.37 moles

From equation above the mole ratio of Na: H2 is 2:1

therefore the moles of H2 =1.37 x1/2 =0.685 moles

At STP 1 moles of a gas = 22.4 L

0.685 moles = ? L

by cross multiplication

=[ (0.685 moles x 22.4 L) / 1 mole =15.344 L

<h3> Question 2</h3>

The percent yield = 84.77%

calculation

4 Na + O₂ → 2 Na₂O

%yield =actual yield / theoretical yield x 100

Actual yield= 61.8 g

The theoretical yield is calculated as below

Step 1: find the moles of Na

moles = mass /molar mass

from periodic table the molar mass of Na = 23 g/mol

moles = 54.1 g /23 g/mol =2.352 moles

Step 2: use the mole ratio to determine the moles of Na₂O

Na:Na₂O is 4: 2 therefore the moles of Na₂O = 2.352 x 2/4 =1.176 moles

Step 3: find the theoretical mass of Na₂O

mass = moles x molar mass

The molar mass of Na₂O = [(23 x 2 ) + 16] =62 g/mol

mass = 1.176 moles x 62 g/mol = 72.9 g

% yield is therefore = 61.8 g /72.9 x 100 = 84.77%

8 0

3 years ago

A 1.00 liter container holds a mixture of 0.52 mg of He and 2.05 mg of Ne at 25oC. Determine the partial pressures of He and Ne

Ymorist [56]

Answer:

pHe = 3.2 × 10⁻³ atm

pNe = 2.5 × 10⁻³ atm

P = 5.7 × 10⁻³ atm

Explanation:

Given data

Volume = 1.00 L

Temperature = 25°C + 273 = 298 K

mHe = 0.52 mg = 0.52 × 10⁻³ g

mNe = 2.05 mg = 2.05 × 10⁻³ g

The molar mass of He is 4.00 g/mol. The moles of He are:

0.52 × 10⁻³ g × (1 mol / 4.00 g) = 1.3 × 10⁻⁴ mol

We can find the partial pressure of He using the ideal gas equation.

P × V = n × R × T

P × 1.00 L = 1.3 × 10⁻⁴ mol × (0.082 atm.L/mol.K) × 298 K

P = 3.2 × 10⁻³ atm

The molar mass of Ne is 20.18 g/mol. The moles of Ne are:

2.05 × 10⁻³ g × (1 mol / 20.18 g) = 1.02 × 10⁻⁴ mol

We can find the partial pressure of Ne using the ideal gas equation.

P × V = n × R × T

P × 1.00 L = 1.02 × 10⁻⁴ mol × (0.082 atm.L/mol.K) × 298 K

P = 2.5 × 10⁻³ atm

The total pressure is the sum of the partial pressures.

P = 3.2 × 10⁻³ atm + 2.5 × 10⁻³ atm = 5.7 × 10⁻³ atm

6 0

3 years ago

What change would shift the equilibrium system to the left?

SVEN [57.7K]

Answer:

Adding more of gas C to the system

Explanation:

Le Châtelier's principle states that when there is an dynamic equilibrium, and this equilibrium is disturbed by an external factor, the equilibrium will be shifted in the direction that can cancel the effect of the external factor to reattain the equilibrium.

1) Adding more of gas C to the system:

Adding more C gas will increase the concentration of the products side. So, the reaction will be shifted to the left to attain the equilibrium again.

2) Heating the system:

Heating the system will increase the concentration of the reactants side as the reaction is endothermic. so, the reaction will be shifted to the right to attain the equilibrium again.

3) Increasing the volume:

has no effect since the no. of moles of gases is the same in both reactants and products sides.

4) Removing some of gas C from the system:

Removing some of gas C from the system will decrease the concentration of the products side. So, the reaction will be shifted to the right to attain the equilibrium again.

So, the right choice is: Adding more of gas C to the system.

3 0

3 years ago

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