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Semenov [28]
3 years ago
14

Liquids that dissolve freely in one another in any proportion are

Chemistry
1 answer:
Alchen [17]3 years ago
8 0

dgdfasfdsafdsagdfggxcvxcvxczgdfsgsdfhgfghfjghfgfhjgfd

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In which state of matter are the particles able to move around one another
BartSMP [9]
C, because liquids as gas can move on there own at different points but solids are locked in place and do not move past one another like liquids and gases do.
3 0
3 years ago
When a 11.9 gram sample of copper metal is dropped into a glass of cold water, the temperature of the copper drops from 55°C to
Ugo [173]

Answer:

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Explanation:

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8 0
3 years ago
If 110g of copper sulphate is present in 550g of solution calculate the concentration of solution
Bingel [31]

Answer:

20%

Explanation:

mass by mass percentage of a solution =(mass of solute)/(mass of solution)

mass of solute=550g

therefore 110×100/550=20%

hope u will understand .:") credit to the owner

7 0
3 years ago
A student determines the molar mass of a compound by the method used in this experiment. She found the temperature of the ice wa
cluponka [151]

Answer:

139.94 grams of solute is present in  kilograms of water

Explanation:

Temperature of the ice water = T = 1.0°C

Temperature of the mixture = T_f =-3.0°C

Depression in freezing point \Delta T_f=T-T_f

\Delta T_f =1.0°C-( -3.0°C)=4°C

\Delta T_f=k_f\times m=k_f\times \frac{\text{Moles of solute}}{\text{mass of solvent (kg)}}

m = molality of the solution

k_f = Molal depression constant

4^oC=1.86^oC kg/mol\times \frac{\text{Moles of solute}}{0.0793 kg}

Mole of solute = 0.1705 moles

Mass of solute = 11.1 g

Moles of solute = 0.1705 mol=\frac{11.1 g}{M}

Molar mass of the solute= M

M = 65.088 g/mol

Molality of the solution = m = \frac{0.1705 mol}{0.0793 kg}

m\times M=\frac{0.1705 mol}{0.0793 kg}\times 65.088 g/mol=139.94g of solute /kg of water

139.94 grams of solute is present in  kilograms of water

3 0
3 years ago
If a gas is at a pressure of 46 mm Hg and temperature of 640 K, what would be the temperature if the pressure was raised to 760
Olegator [25]

Answer:

10573.9K

Explanation:

Using pressure law equation;

P1/T1 = P2/T2

Where;

P1 = initial pressure (mmHg)

P2 = final pressure (mmHg)

T1 = initial temperature (K)

T2 = final temperature (K)

According to the information provided in this question,

P1 = 46mmHg

P2 = 760mmHg

T1 = 640K

T2 = ?

Using P1/T1 = P2/T2

46/640 = 760/T2

Cross multiply

640 × 760 = 46 × T2

486400 = 46T2

T2 = 486400 ÷ 46

T2 = 10573.9K

8 0
3 years ago
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