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3 years ago

BRAINLIEST TO CORRECT what would the cook time for 2 half-lives of the popcorn?

Chemistry

1 answer:

Svetlanka [38]3 years ago

3 0

Answer:

Pop for two minutes in the microwave, and enjoy the perfect balance of buttery and salty taste in every bite. Try this delicious salty snack for backyard barbeques, movie nights, birthday parties or an office snack. Choose ACT II Butter Popcorn for the best value in popcorn.

Explanation:

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What body of water has the largest tidal range in meters

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3 years ago

What volume of nitrogen (n2) would be completely consumed in the reaction with 30.80 g of

Shtirlitz [24]

The answer is 285.33g nitrogen would be completely consumed in the reaction with 30.80 g of hydrogen gas.

A mole is defined as 6.02214076 × 10²³ atoms, molecules, ions, or other chemical units.

Write a balanced equation.

Calculate the moles of H₂ in 30.8 g.

Calculate the moles of N₂ required to react with H₂.

Calculate the mass of N₂.

Calculate the initial mass of N₂.

Start with a balanced equation.

N₂ + 3H₂ --> 2NH₃

Calculate the moles of H₂ in 30.8 g.

n = m/M; where n = moles, m = mass, and M = molar mass.

M(H₂) = 1.008 g/mol

n(H₂) = (30.8 g)/(1.008 g/mol) = 30.56 mol H₂

Calculate the moles of N₂ required to react with 30.56 mol H₂ , using the mole ratio between H₂ and N₂ in the balanced equation.

30.56 mol H₂ × 1 mol N₂/3 mol H₂ = 10.18 mol N₂

Calculate the mass of N₂ in 10.18 mol.

m = n × M

M(N₂) = 2 × 14.007 g/mol N = 28.014 g/mol N₂

m(N₂) = 10.18 mol × 28.014 g/mol = 285.33g N₂

Therefore 285.33g nitrogen would be completely consumed in the reaction with 30.80 g of hydrogen gas.

To know more about mole

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5 0

2 years ago

Upon adding solid potassium hydroxide pellets to water the following reaction takes place: KOH(s) → KOH(aq) + 43 kJ/mol Answer t

nekit [7.7K]

Answer:

a) Warmer

b) Exothermic

c) -10.71 kJ

Explanation:

The reaction:

KOH(s) → KOH(aq) + 43 kJ/mol

It is an exothermic reaction since the reaction liberates 43 kJ per mol of KOH dissolved.

Hence, the dissolution of potassium hydroxide pellets to water provokes that the beaker gets warmer for being an exothermic reaction.

The enthalpy change for the dissolution of 14 g of KOH is:

$n = \frac{m}{M}$

<u>Where:</u>

m: is the mass of KOH = 14 g

M: is the molar mass = 56.1056 g/mol

$n = \frac{m}{M} = \frac{14 g}{56.1056 g/mol} = 0.249 mol$

The enthalpy change is:

$\Delta H = -43 \frac{kJ}{mol}*0.249 mol = -10.71 kJ$

The minus sign of 43 is because the reaction is exothermic.

I hope it helps you!

5 0

3 years ago