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Lunna [17]
2 years ago
12

How many ml of a 14.0 m nh3 stock solution are needed to prepare 200 ml of a 4.20 m dilute nh3 solution? hints how many ml of a

14.0 m stock solution are needed to prepare 200 ml of a 4.20 m dilute solution? 0.060 ml 60.0 ml 840 ml 667 ml?
Chemistry
1 answer:
Fudgin [204]2 years ago
5 0

To solve this we use the equation, 

M1V1 = M2V2

where M1 is the concentration of the stock solution, V1 is the volume of the stock solution, M2 is the concentration of the new solution and V2 is its volume.

14 M x V1 = 4.20 M x 200 mL

V1 = 60 mL needed of the concentrated solution
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The data in the table below were obtained for the reaction: 2clo2 (aq) + 2 oh- (aq) --> clo3- (aq) + clo2- (aq) + h2o (l) exp
SVETLANKA909090 [29]
Lets organise the data given in the question
                [ClO₂] (m)       [OH⁻] (m)        initial rate (m/s)
                  <span>0.060              0.030               0.0248
</span><span>                  0.020              0.030               0.00276
</span><span>                  0.020              0.090                0.00828
rate equation as follows 
rate = k [</span>ClO₂]ᵃ [OH⁻]ᵇ
where k - rate constant 
we need to find order with respect to ClO₂ therefore lets take the 2 equations where OH⁻ is constant.
1) 0.00276 = k [0.020]ᵃ[0.030]ᵇ
2) 0.0248 = k [0.060]ᵃ[0.030]ᵇ
divide first equation from the second
0.0248/0.00276 = [0.060/0.020]ᵇ
8.99 = 3ᵇ
8.99 rounded off to 9
9 = 3ᵇ
b = 2
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3 0
2 years ago
3. A mass of 0.15 ounces is equal to how many grams?
garik1379 [7]

Answer:

Option C = 4.25 g

Explanation:

Ounce and grams are unit of mass. Ounce is larger unit while gram is smaller unit. The one ounce is consist of 28.35 g or we can say that one ounce is equal to 28.35 g. In order to convert the given ounce value into grams the value is multiply with 28.35 g.

Given data:

Mass = 0.15 ounce

Mass in gram = ?

Solution:

One ounce is equal to 28.35 g, so

0.15 × 28.35 = 4.25 g

5 0
2 years ago
A food product is being frozen in a system capable of removing 6000 kJ of thermal energy. The product has a specifi c heat of 4
lapo4ka [179]

Answer : The exit temperature of the product is, -35.2^oC

Explanation :

Total heat = Heat lost by liquid + Latent heat of fusion + Heat lost by frozen

Q=m\times c_1\times (T_2-T_1)+m\times L_f+m\times c_2\times (T_4-T_3)

where,

Q =  Total heat = 6000 kJ

m = mass of product = 15 kg

c_1 = specific heat of liquid = 4kJ/kg^oC

L_f = latent heat of fusion = 275kJ/kg

c_2 = specific heat of frozen = 2.5kJ/kg^oC

T_1 = initial temperature of liquid = 2^oC

T_2 = final temperature of liquid = 10^oC

T_3 = initial temperature of frozen = ?

T_4 = final temperature of frozen = 2^oC

Now put all the given value in the above expression, we get:

6000kJ=[15kg\times 4kJ/kg^oC\times (10-2)^oC]+[15kg\times 275kJ/kg]+[15kg\times 2.5kJ/kg^oC\times (2-T_3)^oC]

T_3=-35.2^oC

Thus, the exit temperature of the product is, -35.2^oC

7 0
2 years ago
A 0.307-g sample of an unknown triprotic acid is titrated to the third equivalence point using 35.2 ml of 0.106 m naoh. calculat
Jet001 [13]
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2 years ago
How many seconds are required to produce 4.00 g of aluminum metal from the electrolysis of molten alcl3 with an electrical curre
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The correct answer is e. 3.57×10³
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F= Faraday's constant
I = electric current in'A'CA  C/s
t=(3×0.1483 mol ×96485 C/mol) /12(C15)
t=3577 second = 3.5 ×10³s
8 0
2 years ago
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