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vladimir1956 [14]
3 years ago
7

in order to make tea, 32,000 J of energy were added to 100.0g of water. what was the temperature Chang of the water? ​

Chemistry
1 answer:
mina [271]3 years ago
5 0

Answer:

ΔT  = 76.5 °C

Explanation:

Given data:

Amount of water = 100.0 g

Energy needed = 32000 J

Change in temperature = ?

Solution,

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

Now we will put the values in formula.

Q = m.c. ΔT

ΔT  = Q / m.c

ΔT  = 32000 j/ 100.0 g × 4.184 j/g. °C

ΔT  =  32000 j / 418.4 j /°C

ΔT  = 76.5 °C

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Lead(II) nitrate is added slowly to a solution that is 0.0800 M in Cl− ions. Calculate the concentration of Pb2+ ions (in mol /
barxatty [35]

Answer:

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Ip>Ksp

Precipitation starts only when solution is supersaturated because solution become supersaturated then it does not stay in this form and precipitation starts itself only solution become saturated.

This usually happens when two solutions containing separate sources of cation and anion are mixed together and here also we are mixing lead (||)nitrate solution(source of lead(||)) into the Cl- solution.

Ip=[Pb^{2}][2Cl^-]^2=Ksp

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S=\sqrt[3]{\frac{Ksp}{4} }

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5 0
2 years ago
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kiruha [24]

Explanation:

1000ml \: contain \: 3 \: moles \\ 100 \: ml \: will \: contain \: ( \frac{100 \times 3}{1000} ) \: moles \\  = 0.3 \: moles \\ RFM = 95 \\ 1 \: mole \: weighs \: 95 \: g \\ 0.3 \: moles \: weigh \: ( \frac{(0.3 \times 95)}{1}  \: g \\  = 28.5 \: g \: of \: magnesium \: chloride

5 0
2 years ago
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