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3 years ago

How many grams of lead would be required to make 119 grams of water

Chemistry

2 answers:

Ludmilka [50]3 years ago

8 0

Answer:

(119 g H2O) / (18.01532 g H2O/mol) x (1 mol Pb / 2 mol H2O) x (207.21 g Pb/mol) = 684 g Pb

Explanation:

Strike441 [17]3 years ago

6 0

Answer:

684

Explanation:

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2 years ago

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3 years ago

How much would the temperature of 275 g of water increase if 36.5 kj of heat were added?

Andrew [12]

<u>Answer:</u> The temperature increase will be 31.70°C.

<u>Explanation:</u>

To calculate the increase in the temperature of the system, we use the equation:

$q=mc\Delta T$

where,

q = Heat absorbed = 36.5 kJ = 36500J

m = Mass of water = 275 g

c = Specific heat capacity of water = $4.186J/g^oC$

$\Delta T$ = change in temperature = ? °C

Putting values in above equation, we get:

$36500J=275\times 4.186J/g^oC\times \Delta T\\\\\Delta T=31.70^oC$

Hence, the temperature increase will be 31.70°C.

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3 years ago

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jolli1 [7]

Answer:

Explanation:

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3 years ago