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Ber [7]
3 years ago
7

Convert 30°C into kelvin

Chemistry
1 answer:
lakkis [162]3 years ago
4 0

Answer:303.15

Explanation:

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(will give brainliest) show your work. How many grams of Copper(I) nitrate, CuNO3 are required to produce 88.0 grams of aluminum
ValentinkaMS [17]

Based on the stoichiometry of the reaction, 156.114 g of CuNO3 are required to produce 88.0 grams of aluminum nitrate, Al(NO3)3.

<h3>What is stoichiometry of a reaction?</h3>

The stoichiometry of a reaction is the molar ratio in which reactants combine to form products.

The stoichiometry of the reaction shows that 6 moles of copper (i) nitrate produces 2 moles of aluminium nitrate.

molar mass of Copper(I) nitrate, CuNO3 = 126 g

molar mass of aluminum nitrate, Al(NO3)3 = 213 g

88.0 g of aluminum nitrate, Al(NO3)3 = 88.0/213 moles = 0.413 moles

0.413 moles of Al(NO3)3 will be produced by 0.413 ×6/3 = 1.239 moles of CuNO3

Mass of 1.239 moles of CuNO3 = 1.239 × 126 = 156.114 g of CuNO3

Therefore, 156.114 g of CuNO3 are required to produce 88.0 grams of aluminum nitrate, Al(NO3)3.

Learn more about stoichiometry at: brainly.com/question/16060223

Therefore, 156.114 g of CuNO3

4 0
1 year ago
How many moles are in 187.54 grams of magnesium chlorate?
Svetlanka [38]

Hey there!

Magnesium chlorate: Mg(ClO₃)₂

Find molar mass.

Mg: 1 x 24.305 = 24.305

Cl: 2 x 35.453 = 70.906

O: 6 x 16 = 96

------------------------------------

                      191.211 g/mol

We have 187.54 grams.

187.54 ÷ 191.211 = 0.9808

There are 0.9808 moles in 187.54 grams of magnesium chlorate.

Hope this helps!

3 0
3 years ago
you produce 115 grams of water in a combustion reaction. how much methane do you need in grams for this to react?
lora16 [44]

Answer:about 2 grams of methane

Explanation:

idk just got this right

6 0
3 years ago
One of the steps in processing some metals is smelting. what does smelting do for a metal?
vichka [17]
Hey there,
Smelting's main use is to produce a base metal from its ore

Hope this helps :))

~Top
8 0
3 years ago
Suppose that a catalyst lowers the activation barrier of a reaction from 125kJ/mol to 55kJ/mol125⁢kJ/mol to 55kJ/mol. By what fa
sineoko [7]

Answer:

The factor of increasing reaction rate is 1,85x10¹².

Explanation:

Using arrhenius formula:

k = A e^\frac{-E_{a}}{RT}

Where k is rate constant; A is frecuency factor; Eₐ is activation energy; R is gas constant (0,008134 kJ/molK); T is temperature 25°C = 298,15K

Thus, replacing for an activation energy of 125 kJ/mol assuming A as 1:

k = 1,25x10⁻²²

When activation energy is 55kJ/mol:

k = 2,31x10⁻¹⁰

Thus, the factor of increasing reaction rate is:

2,31x10⁻¹⁰/1,25x10⁻²² =<em> 1,85x10¹²</em>

<em></em>

I hope it helps!

8 0
3 years ago
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