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rjkz [21]
3 years ago
15

Solution remained colorless.

Chemistry
1 answer:
denis-greek [22]3 years ago
7 0
B because the others are not equal and if you add and then get the root it would be b
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(On the verge of tears)
zhenek [66]

Answer: The final molarity of a 20mL- 1.3M salt solution after it has been diluted with 100ml water is 0.22 M

Explanation:

According to the dilution law,

M_1V_1=M_2V_2

where,  

M_1 = molarity of stock solution = 1.3  M

V_1 = volume of stock solution = 20 ml  

M_1 = molarity of diluted solution = ?

V_1 = volume of  diluted solution = (20+100) ml = 120 ml

Putting in the values we get:

1.3\times 20=M_2\times 120

M_2=0.22M

Therefore  the final molarity of a 20mL- 1.3M salt solution after it has been diluted with 100ml water is 0.22 M

7 0
3 years ago
Is straining cooked pasta from water a physical or chemical separation
Leviafan [203]
It is a physical separation.

Only chemical separation involves the changing of substances and their bondings, which usually involves the use of heat or electricity. Meanwhile, physical changes does not require breaking the bonds of substances or changing a substance to another, common examples of physical separations are such as filtration or distillation.

Since straining pasta from water does not require breaking the pasta or water to other substances, plus heat and electricity is not involved, therefore this is a type of physical separation.
7 0
4 years ago
What is the freezing point of a solution in which 2.50 grams of sodium chloride are added to 230.0 ml of water
labwork [276]

The freezing point of a solution in which 2.5 grams of NaCl is added t0 230 ml of water is :  - 0.69°C

<h3>Determine the freezing point of the solution </h3>

First step : Calculate the molality of NaCl

molality =  ( 2.5 grams / 58.44 g/mol ) / ( 230 * 10⁻³ kg/ml )

              = 0.186  mol/kg

Next step : Calculate freezing point depression temperature

T = 2 * 0.186 * kf

where : kf = 1.86°c.kg/mole

Hence; T = 2 * 0.186 * 1.86 = 0.69°C

Freezing point of the solution

Freezing temperature of solvent - freezing point depression temperature

                                               0°C  -  0.69°C = - 0.69°C

Hence the Freezing temperature of the solution is  - 0.69°C

Learn more about The freezing point of a solution in which 2.5 grams of NaCl is added t0 230 ml of water is :  - 0.69°C

8 0
3 years ago
A popular classroom demonstration consists of filing notches into a new penny and soaking the penny in hydrochloric acid overnig
Ivanshal [37]

Answer:

HCl conc=2.2M

Explanation:

From the question we are told that:

Mass of zinc Mass_{z_n}=2.5g

Initial conc of HCl HConc_1=6.0M

Initial volume of HCl Hvol_1=20.0mL

Molar mass of zinc M_{Z_n}=65.38g/mol

Generally the equation for reaction  is mathematically given by

2HCL+Zn\implies ZnCl_2 +H_2

Generally the equation for moles of zinc m_{z_n} is mathematically given by

m_{z_n}=\frac{Mass_{Z_n}}{M_{Z_n}}

m_{z_n}=\frac{2.5g}{M_{65.38}}

m_{z_n}=0.038molZn

Generally the equation for moles of m_{HCl} is mathematically given by

m_{HCl}=HConc_1*Hvol_1

m_{HCl}=6.0M*20.0mL

m_{HCl}=6.0M*20.0*10^{-3}L

m_{HCl}=0.12\ mol\ HCl

Generally the reacted moles of HCl HCl_[reacted] is mathematically given by

Since

Zn:HCl =1:2

Therefore

HCl_[reacted]=0.038*\frac{2}{1}

HCl_[reacted]=0.076mol \ HCl

Generally the moles of HCl after Zn oxidization X is mathematically given by

x=0.12-0.076

x=0.044 mol HCl

Generally the conc of hydrochloride acid X is mathematically given by

X=\frac{0.044}{0.020}

X=2.2 M Hcl

Therefore Conc of HCl

HCl conc=2.2M

5 0
3 years ago
In collision and explosions, the total momentum before is______the total momentum after, ill mark the brainiest
oee [108]

Answer:

I think its less than

Explanation:

5 0
3 years ago
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