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Shtirlitz [24]
3 years ago
8

A mover packs books, CDs, and DVDs into a moving box. If the box contains 6.5 kg of books, 1.5 kg of CDs, and 2.0 kg of DVDs, wh

at is the percent by mass of each object in the box?
Chemistry
2 answers:
horsena [70]3 years ago
5 0

%M1 = 6.5 / 10 x 100 = 65%

%M2 = 1.5/10 x 100 = 15%

%M3 = 2.0/10 x 100 = 20%

mariarad [96]3 years ago
4 0
Given: 
<span>M1 = 6.5 kg of books
</span><span>M2 = 1.5 kg of CDs
</span><span>M3 = 2.0 kg of DVDs

Required: percent by mass of each object

Solution:
First, we calculate the total mass.

M = 6.5 kg + 1.5 kg + 2.0 kg =  10 kg

Percent by mass is calculated by getting the ration of the mass of an object and the total mass multiplied by 100 to get the percent.

%M1 = 6.5 / 10 x 100 = 65%
%M2 = 1.5/10 x 100 = 15%
%M3 = 2.0/10 x 100 = 20%</span>
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In the reaction H2SO4 + 2 NaOH -&gt; Na2SO4 + 2H2O, an equivalence point occurs when 29.43 mL of 0.1973 M NaOH is added to a 32.
Tamiku [17]
            moles NaOH = c · V = 0.1973 mmol/mL · 29.43 mL = 5.806539 mmol
            moles H2SO4 = 5.806539 mmol NaOH · 1 mmol H2SO4 / 2 mmol NaOH = 2.9032695 mmol
Hence
            [H2SO4]= n/V = 2.9032695 mmol / 32.42 mL = 0.08955 M
The answer to this question is  [H2SO4] = 0.08955 M

6 0
2 years ago
9
Svetllana [295]

Answer:

0.0025moles

Explanation:

Molarity of a solution (M) = number of moles (n) ÷ volume (V)

According to this question, to make 250 mL of a 0.01 M solution of CaCl, the following number of moles is needed:

Volume = 250mL = 250/1000 = 0.250Litres.

Using; molarity = n/V

0.01 = n/0.250

n = 0.0025

n = 2.5 × 10^-3 moles.

5 0
2 years ago
Some SbCl5 is allowed to dissociate into SbCl3 and Cl2 at 521 K. At equilibrium, [SbCl5] = 0.195 M, and [SbCl3] = [Cl2] = 6.98×1
Brilliant_brown [7]

Answer:

a) The equilibrium will shift in the right direction.

b) The new equilibrium concentrations after reestablishment of the equilibrium :

[SbCl_5]=(0.370-x) M=(0.370-0.0233) M=0.3467 M

[SbCl_3]=(6.98\times 10^{-2}+x) M=(6.98\times 10^{-2}+0.0233) M=0.0931 M

[Cl_2]=(6.98\times 10^{-2}+x) M=(6.98\times 10^{-2}+0.0233) M=0.0931 M

Explanation:

SbCl_5(g)\rightleftharpoons SbCl_3(g) + Cl_2(g)

a) Any change in the equilibrium is studied on the basis of Le-Chatelier's principle.

This principle states that if there is any change in the variables of the reaction, the equilibrium will shift in the direction to minimize the effect.

On increase in amount of reactant

SbCl_5(g)\rightleftharpoons SbCl_3(g) + Cl_2(g)

If the reactant is increased, according to the Le-Chatlier's principle, the equilibrium will shift in the direction where more product formation is taking place. As the number of moles of SbCl_5 is  increasing .So, the equilibrium will shift in the right direction.

b)

SbCl_5(g)\rightleftharpoons SbCl_3(g) + Cl_2(g)

Concentration of SbCl_5  = 0.195 M

Concentration of SbCl_3  = 6.98\times 10^{-2} M

Concentration of Cl_2  = 6.98\times 10^{-2} M

On adding more [SbCl_5 to 0.370 M at equilibrium :

SbCl_5(g)\rightleftharpoons SbCl_3(g) + Cl_2(g)

Initially

0.370 M         6.98\times 10^{-2}M    

At equilibrium:

(0.370-x)M   (6.98\times 10^{-2}+x)M  

The equilibrium constant of the reaction  = K_c

K_c=2.50\times 10^{-2}

The equilibrium expression is given as:

K_c=\frac{[SbCl_3][Cl_2]}{[SbCl_5]}

2.50\times 10^{-2}=\frac{(6.98\times 10^{-2}+x)M\times (6.98\times 10^{-2}+x)M}{(0.370-x) M}

On solving for x:

x = 0.0233 M

The new equilibrium concentrations after reestablishment of the equilibrium :

[SbCl_5]=(0.370-x) M=(0.370-0.0233) M=0.3467 M

[SbCl_3]=(6.98\times 10^{-2}+x) M=(6.98\times 10^{-2}+0.0233) M=0.0931 M

[Cl_2]=(6.98\times 10^{-2}+x) M=(6.98\times 10^{-2}+0.0233) M=0.0931 M

3 0
3 years ago
Try try try please for me please
Inga [223]

Answer:

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8 0
2 years ago
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Vaselesa [24]
The correct answer to this question would be heat energy

8 0
2 years ago
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