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avanturin [10]
3 years ago
15

PLEASE HELP!!!!! NEED HELP ASAP!!!!!

Chemistry
1 answer:
Anarel [89]3 years ago
3 0

Answer:

1.BaCO3 2.MgBr2 3.aluminum and oxygen 4.Potassium chloride

Explanation:

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Each student in a class placed a 2.00 g sample of a mixture of Cu and Al in a beaker and placed the beaker in a fume hood. The s
Aleks [24]

Answer:

Percentage mass of copper in the sample = 32%

Explanation:

Equation of the reaction producing Cu(NO₃) is given below:

Cu(s)+ 4HNO₃(aq) ---> Cu(NO₃)(aq) + 2NO₂(g) + 2H₂O(l)

From the equation of reaction, 1 mole of Cu(NO₃) is produced from 1 mole of copper. Therefore, 0.010 moles of Cu(NO₃) will be produced from 0.010 mole of copper.

Molar mass of copper = 64 g/mol

mass of copper = number of moles * molar mass

mass of copper = 0.01 mol * 64 g/mol = 0.64 g

Percentage by mass of copper in the 2.00 g sample = (0.64/2.00) * 100%

Percentage mass of copper in the sample = 32%

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3 years ago
How many moles of mg3n2 would be produced when 180.0 g of mg reacts with 85.0 g of n2?
aalyn [17]
2.47 moles are produce if 85g of N2 react with 180g of Mg
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Name a common greenhouse gas with only hydrogen and oxygen
AVprozaik [17]
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Which of the primary component of natural gas? ​
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The answer is Methane!
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Calculate the fraction of atoms in a sample of argon gas at 400 K that have an energy of 10.0 kJ or greater.
nikdorinn [45]

Answer:

The answer to this can be arrived at by clculating the mole fraction of atoms higher than the activation energy of 10.0 kJ by pluging in the values given into the Arrhenius equation. The answer to this is 20.22 moles of Argon have energy equal to or greater than 10.0 kJ

Explanation:

From Arrhenius equation showing the temperature dependence of reaction rates.

K = Ae^{\frac{Ea}{RT} } where

k = rate constant

A = Frequency or pre-exponential factor

Ea   =       energy of activation

R = The universal gas constant

T = Kelvin absolute temperature

we have

f = e^{\frac{Ea}{RT} }

Where

f = fraction of collision with energy higher than the activation energy

Ea = activation energy = 10.0kJ = 10000J

R = universal gas constant = 8.31 J/mol.K

T = Absolute temperature in Kelvin = 400K

In the Arrhenius equation k = Ae^(-Ea/RT), the factor A is the frequency factor and the component e^(-Ea/RT) is the portion of possible collisions with high enough energy for a reaction to occur at the a specified temperature  

Plugging in the values into the equation relating f to activation energy we get

f = e^{\frac{10000J}{(8.31J/((mol)(K)))(400K)} } or f = e^{3.01} = 20.22 moles of argon have an energy of 10.0 kJ or greater

5 0
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