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Diano4ka-milaya [45]
3 years ago
15

Zn + 2 HCl --> ZnCl2 + H2 How many moles of hydrochloric acid (HCl) are used (assume a complete reaction) if 8.3 moles of zin

c chloride (ZnCl) are produced? A 8.3 mol ZnCl2 B None of these answers C 8.3 mol HCl D 4.15 mol HCl E 16.6 mol HCl
Chemistry
1 answer:
uysha [10]3 years ago
4 0

Answer:

E. 16.6 mol HCl

Explanation:

The equation for the reaction is;

Zn + 2 HCl --> ZnCl2 + H2

From the reaction 2 moles of HCl produces 1 mole of ZnCl2

Therefore; 8.3 moles of ZnCl2 will be produced by;

  = 8.3 moles ×2

  = 16.6 Moles of HCl

  Therefore;  E.  16.6 mol HCl

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a solution is prepared by dissolving 55g of CaCl2 into 300g of water. if the Kf of water is -1.86 c/m, shat is the freezing poin
Zepler [3.9K]

Answer:

Relation between , molality and temperature is as follows.

                T =

It is also known as depression between freezing point where, i is the Van't Hoff factor.

Let us assume that there is 100% dissociation. Hence, the value of i for these given species will be as follows.

         i for  = 3

          i for glucose = 1

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Depression in freezing point will have a negative sign. Therefore, d

depression in freezing point for the given species is as follows.

       

                 =

       

                  =

     

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Therefore, we can conclude that given species are arranged according to their freezing point depression with the least depression first as follows.

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Explanation:

3 0
3 years ago
Question
madam [21]

Answer:

The specific heat of the metal is 2.09899 J/g℃.

Explanation:

Given,

For Metal sample,

mass = 13 grams

T = 73°C

For Water sample,

mass = 60 grams

T = 22°C.

When the metal sample and water sample are mixed,

The addition of metal increases the temperature of the water, as the metal is at higher temperature, and the  addition of water decreases the temperature of metal. Therefore, heat lost by metal is equal to the heat gained by water.

Since, heat lost by metal is equal to the heat gained by water,

Qlost = Qgain

However,

Q = (mass) (ΔT) (Cp)

(mass) (ΔT) (Cp) = (mass) (ΔT) (Cp)

After mixing both samples, their temperature changes to 27°C.

It implies that , water sample temperature changed from  22°C to 27°C and metal sample temperature changed from 73°C to 27°C.

Since, Specific heat of water = 4.184 J/g°C

Let Cp be the specific heat of the metal.

Substituting values,

(13)(73°C - 27°C)(Cp) = (60)(27°C - 22℃)(4.184)

By solving, we get Cp =

Therefore, specific heat of the metal sample is 2.09899 J/g℃.

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3 years ago
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Answer:

A)The spring scale has a high level of precision and a low level of accuracy.

Explanation:

Hope it works for u guys

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