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tamaranim1 [39]
3 years ago
11

Determine the amount in grams of KCl that exists in 20.3 g of a solution that contains 1.14 % KCl by mass 1.14 g KCl

Chemistry
1 answer:
Fiesta28 [93]3 years ago
3 0

The mass of KCl that exists in 20.3 g of solution containing 1.14% by mass of KCl is 0.23 g

From the question given above, the following data were obtained:

Mass of solution = 20.3 g

Percentage of KCl = 1.14% = 1.14 / 100 = 0.0114

<h3>Mass of KCl =?</h3>

Percentage = \frac{mass of solute}{mass of solution}   * 100\\\\0.0114 = \frac{mass of KCl}{20.3} \\\\

Cross multiply

Mass of KCl = 0.0114 × 20.3

<h3>Mass of KCl = 0.23 g</h3>

Therefore, the mass of KCl in the solution is 0.23 g

Learn more: brainly.com/question/14876861

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13. How many moles of zinc are in 25.00 g Zn?
topjm [15]
0.3824 moles of Zinc
4 0
3 years ago
The balanced redox reactions for the sequential reduction of vanadium are given below.
Minchanka [31]

Answer : 0.0392 grams of Zn metal would be required to completely reduced the vanadium.

Explanation :

Let us rewrite the given equations again.

2VO_{2}^{+} (aq)+ 4H^{+}(aq) + Zn (s)\rightarrow 2VO^{2+}(aq)+Zn^{2+}+2H2O(l)2VO^{2+}(aq)+ 4H^{+}(aq) + Zn (s)\rightarrow 2V^{3+} (aq)+Zn^{2+}+2H2O(l)

2V^{3+} (aq)+ Zn (s)\rightarrow 2V^{2+}(aq)+Zn^{2+}(aq)

On adding above equations, we get the following combined equation.

2VO_{2}^{+} (aq)+ 8H^{+} (aq) + 3Zn (s)\rightarrow 2V^{2+}(aq)+3Zn^{2+}(aq)+4H_{2}O(l)

We have 12.1 mL of 0.033 M solution of VO₂⁺.

Let us find the moles of VO₂⁺ from this information.

12.1 mL \times \frac{1L}{1000mL}\times \frac{0.033mol}{L}=0.0003993mol NO_{2}^{+}

From the combined equation, we can see that the mole ratio of VO₂⁺ to Zn is 2:3.

Let us use this as a conversion factor to find the moles of Zn.

0.0003993mol NO_{2}^{+}\times \frac{3mol Zn}{2molNO_{2}^{+}}=0.00059895mol Zn

Let us convert the moles of Zn to grams of Zn using molar mass of Zn.

Molar mass of Zn is 65.38 g/mol.

0.00059895mol Zn\times \frac{65.38gZn}{1molZn}=0.0392gZn

We need 0.0392 grams of Zn metal to completely reduce vanadium.

6 0
3 years ago
The student ran out of time and did not do the second heating. Explain how this error would affect the calculation for the numbe
Ber [7]

The error caused will lower the number of moles when he did not do the second heating.This question is related to the empirical formula.

<h3>What is Empirical formula ?</h3>

A formula giving the proportions of the elements present in a compound but not the actual numbers or arrangement of atoms.

The empirical formula is the simplest ratio of atoms in a given compound.

we know that heating is used to eliminate most of the water molecules which are attached to the compound in the question, it's given that the second heating was not performed by the students.

This means that there are still some water molecules attached to the compound. This will imply that higher mass will be calculated for the compound because the water molecules are also considered.

So now, if we look at the calculation for the number of moles of water in this, we can say that this will be lower than the actual amount.

This is because there are some unaccounted moles of water which are still attached to the compound and these will be accounted in the compound mass rather than the waters. Therefore, the number of moles will be lower.

Hence, The error caused will lower the number of moles when he did not do the second heating.

Learn more about Empirical formula here ;

https://brainly.in/question/32699671

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6 0
2 years ago
60 points please help me i will appreciate it!
VARVARA [1.3K]

Answer:

This is a pretty straightforward example of how an ideal gas law problem looks like.

Your strategy here will be to use the ideal gas law to find the pressure of the gas, but not before making sure that the units given to you match those used by the universal gas constant.

So, the ideal gas law equation looks like this

∣

∣

∣

∣

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

a

a

P

V

=

n

R

T

a

a

∣

∣

−−−−−−−−−−−−−−−

Here you have

P

- the pressure of the gas

V

- the volume it occupies

n

- the number of moles of gas

R

- the universal gas constant, usually given as

0.0821

atm

⋅

L

mol

⋅

K

T

- the absolute temperature of the gas

Take a look at the units given to you for the volume and temperature of the gas and compare them with the ones used in the expression of

R

.

a

a

a

a

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Need

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Have

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Liters, L

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Liters, L

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Kelvin, K

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Celsius,

∘

C

a

a

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×

Notice that the temperature of the gas must be expressed in Kelvin in order to work, so make sure that you convert it before plugging it into the ideal gas law equation

∣

∣

∣

∣

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

a

a

T

[

K

]

=

t

[

∘

C

]

+

273.15

a

a

∣

∣

−−−−−−−−−−−−−−−−−−−−−−−−

Rearrange the ideal gas law equation to solve for

P

P

V

=

n

R

T

⇒

P

=

n

R

T

V

Plug in your values to find

P

=

0.325

moles

⋅

0.0821

atm

⋅

L

mol

⋅

K

⋅

(

35

+

273.15

)

K

4.08

L

P

=

∣

∣

∣

∣

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

a

a

2.0 atm

a

a

∣

∣

−−−−−−−−−−−

The answer is rounded to two sig figs, the number of sig figs you have for the temperature of the gas.

6 0
2 years ago
Read 2 more answers
The chemical equation for a reaction between K2Cr2O7 and HCl is shown.
deff fn [24]

K2Cr2O7 + 14HCl → 2CrCl3 + 2KCl + 3Cl2 + 7H2O

the correct option is :

K2Cr2O7, because the oxidation number of Cr changes from +6 to +3.


<u>Oxidation number of Cr in K2Cr2O7 is:</u>

K2Cr2O7 = 2K + 2 Cr + 7 O

= 2(+1) + 2Cr + 7(-2)

= 2 + 2Cr -14

[total charge on K2Cr2O7 = 0], Hence;

2 + 2Cr -14 = 0

2Cr -12 = 0

2Cr = 12

Cr = 12/2

<u>Cr = +6</u>


<u>Oxidation number of Cr in CrCl3 is:</u>

CrCl3 = Cr + 3Cl = 0

Cr + 3(-1) = 0

Cr -3 = 0

<u>Cr = +3</u>

Hence Cr is changing its oxidation number from

+6 in K2Cr2O7 to +3 in CrCl3.

Since the oxidation number of Cr [ +6 → +3] is decreasing here,

Cr is getting reduced.

 

so K2Cr2O7 is an oxidizing agent,as it is getting itself reduced and oxidizes others.

3 0
3 years ago
Read 2 more answers
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