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2 years ago

How much energy would need to be input to completely boil 17.3 moles of ice starting at 0 degrees * C and going to 100°C?

Chemistry

1 answer:

kolezko [41]2 years ago

4 0

Answer:

$Q = 934.911\,kJ$

Explanation:

The energy needed to boil 17.3 moles of ice is:

$Q = (17.3\,mol)\cdot \left[\left(40.5\,\frac{kJ}{mol} \right)+ \left(\frac{4.186}{1000}\,\frac{kJ}{g\cdot ^{\circ}C} \right)\cdot \left(\frac{18.015\,g }{1\,mol} \right)\cdot (100^{\circ}C-0^{\circ}C)+\left(6\,\frac{kJ}{mol}\right)\right]$ $Q = 934.911\,kJ$

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Alona [7]

Question 1:

(a) Sulfurous acid: H2SO3

Sulfuric acid: H2SO4

(b) Nitrous acid: H2NO2

Nitric acid: H2NO3

Question 2:

To calculate the pH, based on concentration of H+ ions, there is one formula:

$ph = - log_{10}(c(h + ))$

So the pH of this solution is

$- log_{10}( {10}^{ - 9} ) = 9$

(the solution is basic).

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2 years ago

What group is nickel in?<br> O2<br> 010<br> O 15<br> O4

Ronch [10]

Answer:

010

Explanation:

Group 10

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3 years ago

Calculate the mass of water produced when 7.26 g of butane reacts with excess oxygen

MaRussiya [10]

Answer:

11.3 g.

Explanation:

Hello there!

In this case, since the combustion of butane is:

$C_4H_{10}+\frac{13}{2} O_2\rightarrow 4CO_2+5H_2O$

Thus, since there is a 1:5 mole ratio between butane and water, we obtain the following mass of water:

$m_{H_2O}=7.26gC_4H_{10}*\frac{1molC_4H_{10}}{58.14gC_4H_{10}}*\frac{5molH_2O}{1molC_4H_{10}} *\frac{18.02gH_2O}{1molH_2O}$

Therefore, the resulting mass of water is:

$m_{H_2O}=11.3gH_2O$

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4 0

2 years ago

At 2000°C, the equilibrium constant for the reaction below is Kc = 4.10 ´ 10–4 . If 0.600 moles of NO is placed in a 1.0-L react

erastova [34]

Answer:

At equilibrium, the concentration of $N_{2 (g)}$ is going to be 0.30M

Explanation:

We first need the reaction.

With the information given we can assume that is:

$N_{2 (g)}$ + $O_{2 (g)}$ ⇄ 2 $NO_{(g)}$

If there is placed 0.600 moles of NO in a 1.0-L vessel, we have a initial concentration of 0.60 M NO; and no $N_{2 (g)}$ nor $O_{2 (g)}$ present. Immediately, $N_{2 (g)}$ and $O_{2 (g)}$ are going to be produced until equilibrium is reached.

By the ICE (initial, change, equilibrium) analysis:

I: [ $N_{2 (g)}$ ]=0 ; [ $O_{2 (g)}$ ]= 0 ; [ $NO_{(g)}$ ]=0.60M

C: [ $N_{2 (g)}$ ]=+x ; [ $O_{2 (g)}$ ]= +x ; [ $NO_{(g)}$ ]=-2x

E: [ $N_{2 (g)}$ ]=0+x ; [ $O_{2 (g)}$ ]= 0+x ; [ $NO_{(g)}$ ]=0.60-2x

Now we can use the constant information:

$K_{c}=\frac{[products]^{stoichiometric coefficient} }{[reactants]^{stoichiometric coefficient} }$