3 years ago

(feso4.(Nh4) So4. 6H2o) +Kmno4+H2So4_Fe2(So4)3+K2So4+mnSo4+(Nh4)2So4+H2o

Chemistry

1 answer:

Semenov [28]3 years ago

7 0

Answer:

balancing the equation?

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12oz of water initially at 75oF is mixed with 20oz of water intiially at 140oF. What is the final temperature?

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$115.625^{\circ}\text{F}$

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$m_1$ = First mass of water = 12 oz

$m_2$ = Second mass of water = 20 oz

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$\Delta T_2$ = Temperature difference of the solution with respect to the second mass of water = $(T-75)^{\circ}\text{F}$

c = Specific heat of water

As heat gain and loss in the system is equal we have

$m_1c\Delta T_1=m_2c\Delta T_2\\\Rightarrow m_1\Delta T_1=m_2\Delta T_2\\\Rightarrow 12(T-75)=20(140-T)\\\Rightarrow 12T-900=2800-20T\\\Rightarrow 12T+20T=2800+900\\\Rightarrow 32T=3700\\\Rightarrow T=\dfrac{3700}{32}\\\Rightarrow T=115.625^{\circ}\text{F}$

The final temperature of the solution is $115.625^{\circ}\text{F}$ .

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2 years ago

(feso4.(Nh4) So4. 6H2o) +Kmno4+H2So4_Fe2(So4)3+K2So4+mnSo4+(Nh4)2So4+H2o​

(feso4.(Nh4) So4. 6H2o) +Kmno4+H2So4_Fe2(So4)3+K2So4+mnSo4+(Nh4)2So4+H2o