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Semmy [17]
3 years ago
11

A student used 10 mL water instead of 30mL for the extraction of the NaCl from the mixture. How would this affect the calculated

percentage of NaCl in the sample?
Chemistry
1 answer:
leonid [27]3 years ago
5 0
In this problem, the solute is sodium chloride, the salt while the solvent which comprise of a larger mass in the solution is water. When the solvent volume is decreased which is related to its mass, with the solute mass being held constant, then the concentration of salt increases. This is called concentration method. 
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How much work does the climber do on his pack if his pack weighs 90 N and he climbs to a height of 30 m?
Vladimir79 [104]

Answer:

2,700j

Explanation:

yupppppppp

8 0
3 years ago
Write a balanced equation to show the reaction of sulfurous acid with lithium hydroxide to form water and lithium sulfite
77julia77 [94]
H2SO3 + 2LiOH = 2H2O + Li2SO3
3 0
3 years ago
A. 14.8
Scilla [17]

Answer:

Explanation:

1)

Given data:

Initial volume  = 2.5 L

Initial temperature = 300 k

Final temperature = 80 k

Final volume = ?

Solution:

The given problem will be solve through the Charles Law.

According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.

Mathematical expression:

V₁/T₁ = V₂/T₂

V₁ = Initial volume

T₁ = Initial temperature

V₂ = Final volume  

T₂ = Final temperature

Now we will put the values in formula.

V₁/T₁ = V₂/T₂

V₂ = V₁T₂/T₁  

V₂ = 2.5 L × 80 K / 300 k

V₂ = 200 L.K / 300 K

V₂ = 0.67 L

2)

Given data:

Initial volume  = 752 mL

Initial temperature = 25.0°C (25+273 = 298 K)

Final temperature = 50.0°C(50+273 = 323 K)

Final volume = ?

Solution:

The given problem will be solve through the Charles Law.

According to this law, The volume of given amount of a gas is directly proportional to its temperature at constant number of moles and pressure.

Mathematical expression:

V₁/T₁ = V₂/T₂

V₁ = Initial volume

T₁ = Initial temperature

V₂ = Final volume  

T₂ = Final temperature

Now we will put the values in formula.

V₁/T₁ = V₂/T₂

V₂ = V₁T₂/T₁  

V₂ = 752 mL × 323 K / 298 k

V₂ = 242896 mL.K / 298 K

V₂ = 815.1  mL

3)

Given data:

Initial volume  = 2.75 L

Initial temperature = 20°C (20+273 = 293 K)

Final temperature = ?

Final volume = 2.46 L

Solution:

The given problem will be solve through the Charles Law.

V₁/T₁ = V₂/T₂

V₁ = Initial volume

T₁ = Initial temperature

V₂ = Final volume  

T₂ = Final temperature

Now we will put the values in formula.

V₁/T₁ = V₂/T₂

T₂  = V₂T₁/V₁

T₂ = 2.46 L × 293 K / 2.75  L

T₂ = 720.78 L.K /  2.75 L

T₂ = 262.1 K

4)

Given data:

Initial volume  = 1500 L

Initial temperature = 5°C (5+273 = 278 K)

Final temperature = 30 °C(30+273 = 303 K)

Final volume of heated air = ?

Solution:

The given problem will be solve through the Charles Law.

V₁/T₁ = V₂/T₂

V₁ = Initial volume

T₁ = Initial temperature

V₂ = Final volume  

T₂ = Final temperature

Now we will put the values in formula.

V₁/T₁ = V₂/T₂

V₂ = V₁T₂/T₁  

V₂ = 1500 L × 303 K / 278 k

V₂ = 454500 L.K / 278 K

V₂ = 1634.89  L

5)

Given data:

Initial volume = 15.5 L

Initial temperature = 20°C (20+273 = 293 K)

Final temperature = 7.0 °C(7.0+273 = 280 K)

Final volume of balloon = ?

Solution:

The given problem will be solve through the Charles Law.

V₁/T₁ = V₂/T₂

V₁ = Initial volume

T₁ = Initial temperature

V₂ = Final volume  

T₂ = Final temperature

Now we will put the values in formula.

V₁/T₁ = V₂/T₂

V₂ = V₁T₂/T₁  

V₂ = 15.5 L × 280 K / 293 k

V₂ = 4340 L.K / 293 K

V₂ = 14.8  L

6)

Given data:

Initial volume = 150 mL

Initial temperature = 23.5°C (23.5+273 = 296.5 K)

Final temperature = 72.5 °C(72.5+273 = 345.5 K)

Final volume of balloon = ?

Solution:

The given problem will be solve through the Charles Law.

V₁/T₁ = V₂/T₂

V₁ = Initial volume

T₁ = Initial temperature

V₂ = Final volume  

T₂ = Final temperature

Now we will put the values in formula.

V₁/T₁ = V₂/T₂

V₂ = V₁T₂/T₁  

V₂ = 150 mL × 345.5 K / 296.5 k

V₂ = 51825 mL.K / 296.5 K

V₂ = 174.79  mL

6 0
3 years ago
Calculate the energy for one mole of photons (an einstein) for light absorbed at 6.90 × 102 nm.
Kay [80]
We know that:
f * λ = c 
<span>and
E = h*f
so,rearranging:
 </span> 
<span>E/h = C/λ 
given:
h = </span><span>6.90 × 10^2 nm

</span><span>substituting the values:

E = </span> C/λ<span> x h  
(c is the speed of light;that is 3x10^8 m/s)

</span>    = 3x10^8 x 6.626x10^-34 / 6.90 x10^7m x 6.022x10^23
Hence,
using calculator:
we get,
 E = 1.69 x 10^5 Joules
5 0
4 years ago
Read 2 more answers
Calculate the ph at the equivalence point for the titration of 0.230 m methylamine (ch3nh2) with 0.230 m hcl. the kb of methylam
Vlad1618 [11]

Answer: The pH at the equivalence point for the titration will be 0.65.

Solution:

Let the concentration of [OH^-] be x

Initial concentration of [CH3NH_2], c = 0.230 M

        CH_3NH_2+H_2O\rightleftharpoons CH_3NH_3^++OH^-

at eq'm  c-x                         x                    x

Expression of K_b:

K_b=\frac{[CH_3NH_3^+][+OH^-]}{[CH_3NH_2]}=\frac{x\times x}{c-x}=\frac{x^2}{c-x}

Since ,methyl-amine is a weak base,c>>x so c-x\approx c.

K_b=\frac{x^2}{c}=5.0\times 10^{-4}=\frac{x^2}{0.230 M}

Solving for x, we get:

x=1.07\times 10^{-2} M

Given, HCl with 0.230 M , it dissociates fully in water which means [H^+] = 0.230 M

[OH^-]=[H^+] will result in neutral solution, since [OH^-]

Remaining [H^+] after neutralizing [OH^-]ions

[H^+]_{\text{left in solution}}=[H^+]-[OH^-]=0.230-1.07\times 10^{-2}=0.2193 M

pH=-log{[H^+]_{\text{left in solution}}=-log(0.2193)=0.65

The pH at the equivalence point for the titration will be 0.65.

8 0
4 years ago
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