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3 years ago

Balance these equations :

Chemistry

1 answer:

vfiekz [6]3 years ago

6 0

Answer:

H2+O2_____2H20
2/8 S8_____2S8O3
2HgO_____2Hg+O2
Zn+2HCl____ZnCl2+H2
2Na+2H2O___2NaOH+H2
C10H16+4Cl2___5C+8HCl
4Fe+3O2____2Fe2O3
C6H12O6+2O2___2Co2+2H2O
Fe2O3+3H2____2Fe+3H2O
2Al+3FeO____Al2O3+3Fe

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This is someone else’s answer but it answers your question

Molarity is defined as the number of moles of solute dissolved in 1 L of solution.
KNO₃ - 2.0 M this means that 2.0 moles of KNO₃ are in 1 L of solution.
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Explanation:

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3 years ago

Read 2 more answers

devlian [24]

Answer:

The final temperature of both substances at thermal equilibrium is 17.3°C

Explanation:

To calculate the final temperature of both substances at thermal equilibrium -:

First , we calculate the heat of A1 cube as follows -

q= mSΔT

(where q = heat of the cube , m = mass of cube , S= specific heat of cube {0.902j/g°C}, T = Temperature )

Putting the values given in the question ,

$q= 32.5g\times0.902\j/g$ ° $C$ $\times(T_f-45.8$ ° $C)$

$29.315\times(T_f-45.8)J$

Now , calculate the heat of water -

q=mSΔT

Putting values from the question ,

$q=105.3g\times4.18j/g$ ° $C$ $\times (T_f-15.4$ ° $C)$

= $440.154\times(T_f-15.4)J$

Now ,

Heat lost by water A1= Heat gained by water [negative sign about heat lost]

$-29.315\times(T_f-45.8)J$ $=440.154\times(T_f-15.4)J$

$\frac{-(T_f-45.8)J}{T_f-15.4)J} =\frac{440.154}{29.315}=15.0$

$-T_f+45.8$ ° $C=15T_f-231.2$ ° $C$

$(45.8+231.2)$ ° $C$ = $16T_f$

$16T_f=277.03$ ° $C$

$T_f=\frac{277.03}{16}$ = 17.3°C

Therefore , the final temperature of both substances at thermal equilibrium is 17.3°C

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