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solniwko [45]
3 years ago
15

How many moles of N2O5 are needed to produce 7.90 g of NO2?

Chemistry
1 answer:
Roman55 [17]3 years ago
4 0

Answer:

0.085 moles of  N₂O₅ are needed

Explanation:

Given data:

Mass of NO₂ produces = 7.90 g

Moles of N₂O₅ needed = ?

Solution:

2N₂O₅       →     4NO₂  + O₂

Number of moles of NO₂ produced :

Number of moles = mass/ molar mass

Number of moles = 7.90 g/ 46 g/mol

Number of moles = 0.17 mol

now we will compare the moles of NO₂   with N₂O₅.

                NO₂          :       N₂O₅

                  4            :          2

                0.17          :         2/4×0.17 = 0.085 mol

Thus, 0.085 moles of  N₂O₅ are needed.

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2.50 g of Zn is added to 5.00 g of HCl
Debora [2.8K]
Zn+2HCl ----> 2ZnCl2 + H2

For 2.50 g of Zn

Mass per mol = 2.50/molar mass of Zn = 2.50/65.38 = 0.0382 g/mol
There are two moles of ZnCl2 and total mass = 2*0.0382*molar mass of ZnCl2 = 2*0.0382*136.286 = 10.42 g

For 2 g of HCl

Mass per mol = 2/2*molar mass of HCl = 2/ (2*36.46) = 0.0274 g/mol
For the two moles of ZnCl2, mass produced = 2*0.0274*136.286 = 7.48 g

It can be noted that 2 g of HCl produced less amount of ZnCl and thus it is the limiting reagent.
5 0
3 years ago
14. As the temperature of the reaction is increased,
AfilCa [17]
Reactant molecules collide more frequently and with greater energy per collision
3 0
3 years ago
A student sets up an experiment according to the table below. What question was the student most likely investigating?
user100 [1]

Answer:

88

Explanation:

7 0
3 years ago
Do anyone know how to do question B
Over [174]

Answer:

a) IUPAC Names:

                   1) (<em>trans</em>)-but-2-ene

                   2) (<em>cis</em>)-but-2-ene

                   3) but-1-ene

b) Balance Equation:

                       C₄H₁₀O + H₃PO₄   →   C₄H₈ + H₂O + H₃PO₄

As H₃PO₄ is catalyst and remains unchanged so we can also write as,

                                    C₄H₁₀O   →   C₄H₈ + H₂O

c) Rule:

           When more than one alkene products are possible then the one thermodynamically stable is favored. Thermodynamically more substituted alkenes are stable. Furthermore, trans alkenes are more stable than cis alkenes. Hence, in our case the major product is trans alkene followed by cis. The minor alkene is the 1-butene as it is less substituted.

d) C is not Geometrical Isomer:

        For any alkene to demonstrate geometrical isomerism it is important that there must be two different geminal substituents attached to both carbon atoms. In 1-butene one carbon has same geminal substituents (i.e H atoms). Hence, it can not give geometrical isomers.

7 0
2 years ago
Pls help this my final 20 points!!!! Be right
Anastaziya [24]

Answer:

I think it's B

Explanation:

I hope it helps

5 0
2 years ago
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