What’s the oxidation number of k2so4

What mass of aluminum at 144 Celsius will increase the temperature of 100 g of water by from 10 C to 15 C?

Andrej [43]

Answer:

$m_{Al} = 18.019\,g$

Explanation:

There is a heating process between an aluminium piece and water, which is described by the First Law of Thermodynamics:

$Q_{Al} = - Q_{w}$

$m_{Al}\cdot c_{Al}\cdot (T_{o,Al}-T) = m_{w}\cdot c_{w}\cdot (T-T_{o,w})$

The mass of the aluminium piece is therefore cleared and known variables are substituted:

$m_{Al} = \frac{m_{w}\cdot c_{w}\cdot (T-T_{o,w})}{c_{Al}\cdot (T_{o,Al}-T)}$

$m_{Al} = \frac{(100\,g)\cdot \left(4.184\,\frac{J}{g\cdot^{\circ}C} \right)\cdot (15^{\circ}C-10^{\circ}C)}{\left(0.900\,\frac{J}{g\cdot ^{\circ}C} \right)\cdot (144^{\circ}C-15^{\circ}C)}$

$m_{Al} = 18.019\,g$

5 0

3 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:

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