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denis23 [38]
2 years ago
11

What is the change in the freezing point of water when 35.0 g of sucrose is

Chemistry
1 answer:
frez [133]2 years ago
3 0

From the calculations, we can see that, the change in the freezing point is -0.634°C.

<h3>What is freezing point?</h3>

The term freezing point refers to the temperature at which a liquid is changed to solid.

Given that;

ΔT = K  m i

Number of moles sucrose = 35.0 g/ 342.30 g/mol = 0.1 moles

molality =  0.1 moles/ 300.0 * 10^-3 Kg

= 0.33 m

Thus;

ΔT = -1.86°C/mol *  0.33 m * 1

= -0.634°C

Learn more about freezing point:brainly.com/question/3121416

#SPJ1

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Which scientist arranged the elements in order of increasing atomic number, rather than increasing atomic mass?
pentagon [3]

Answer:

Henry Moseley

Explanation:

Dmitry Mendeleef and Lothar Meyer proposed a periodic table based on the atomic mass.

They stated a periodic law expressed as "chemical properties of elements are a periodic function of their atomic weights".

But, Henry Moseley in 1900s re-stated periodic law by changing the basis of the law from atomic weight to atomic number.

The present periodic law is stated as "the properties of elements are a periodic function of their atomic number".

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Which determines the chemical properties of minerals?
azamat
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A 215-g sample of copper metal at some temperature is added to 26.6 g of water. The initial water temperature is 22.22 oC, and t
andrezito [222]

The initial temperature of the copper metal was 27.38 degrees.

Explanation:

Data given:

mass of the copper metal sample = 215 gram

mass of water = 26.6 grams

Initial temperature of water = 22.22 Degrees

Final temperature of water = 24.44 degrees

Specific heat capacity of water = 0.385 J/g°C

initial temperature of copper material , Ti=?

specific heat capacity of water = 4.186 joule/gram °C

from the principle of:

heat lost = heat gained

heat gained by water is given by:

q water = mcΔT

Putting the values in the equation:

qwater = 26.6 x 4.186 x (2.22)

qwater = 247.19 J

qcopper = 215 x 0.385 x (Ti-24.4)

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Now heat lost by metal = heat gained by water

82.77Ti - 2019.71 = 247.19

Ti = 27.38 degrees

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dlinn [17]

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Reactivity

Explanation:

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