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3 years ago

molar mass of beryllium (Be) molar mass of silicon (Si) molar mass of calcium (Ca) molar mass of rhodium (Rh)

Chemistry

2 answers:

tankabanditka [31]3 years ago

5 0

Beryllium (Be): 9.01

Silicon (Si): 28.09

Calcium (Ca): 40.08

Rhodium (Rh): 102.91

(Basically the below answers just rounded off a bit:)

Options C, then C, then C, and then C) Ijust found out they were all option C!

Neko [114]3 years ago

3 0

Molarmass of beryllium is 9.0
molar mass of silicon is 28.4
molar mass of calcium is 40.1
molar mass of rhodium is 103.

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When two solutions are mixed, the amount of heat released by solution 1 (liquid water) will be equal to the amount of heat absorbed by solution 2 (liquid water)

$Heat_{\text{absorbed}}=Heat_{\text{released}}$

The equation used to calculate heat released or absorbed follows:

$Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})$

$m_1\times c\times (T_{final}-T_1)=-[m_2\times c\times (T_{final}-T_2)]$ ......(1)

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$m_2$ = mass of solution 2 (liquid water) = 29.0 g

$T_{final}$ = final temperature = ?

$T_1$ = initial temperature of solution 1 = 25°C = [273 + 25] = 298 K

$T_2$ = initial temperature of solution 2 = 45°C = [273 + 45] = 318 K

c = specific heat of water= 4.18 J/g.K

Putting values in equation 1, we get:

$50.0\times 4.18\times (T_{final}-298)=-[29.0\times 4.18\times (T_{final}-318)]\\\\T_{final}=305.3K$

Converting this into degree Celsius, we use the conversion factor:

$T(K)=T(^oC)+273$

$305.3=T(^oC)+273\\T(^oC)=(305.3-273)=32.3^oC$

Hence, the final temperature of water is 32.3°C

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