How much heat is required to convert 0.3 kilograms of ice at 0°C to water at the same temperature?

oksian1 [2.3K]

Hydrogen is a covalent bond. (A bond where one or more pairs of electrons are shared by two atoms)

7 0

3 years ago

1. The solubility of AgNO3 at 20°C is 222.0g AgNO3/100g H2O. What mass of AgNO3 can be dissolved in 250 g of water at 20°C? Reca

neonofarm [45]

Answer :

(1) The mass of silver nitrate is, 555 g

(2) The solubility of the gas will be, 0.433 g/L

Solution for Part 1 :

From the given data we conclude that

In 100 gram of water, the amount of silver nitrate = 222 g

In 250 gram of water, the amount of silver nitrate = $\frac{222}{100}\times 250=555g$

Therefore, the mass of silver nitrate is, 555 g

Solution for Part 2 :

Formula used : $S_1P_1=S_2P_2$ (at constant temperature)

where,

$S_1$ = initial solubility of methane gas = 0.026 g/L

$S_2$ = final solubility of methane gas

$P_1$ = initial pressure of methane gas = 1 atm

$P_2$ = final pressure of methane gas = 0.06 atm

Now put all the given values in the above formula, we get the solubility of methane gas.

$(0.026g/L)\times (1atm)=S_2\times (0.06atm)$

$S_2=0.433g/L$

Therefore, the solubility of the gas will be, 0.433 g/L

5 0

3 years ago

The combustion reaction of magnesium results in what compound?

Tasya [4]

C. MgO because mg+o= MgO

7 0

3 years ago

Read 2 more answers

What super power would you want to have and why? I will give Brainly to the best answer

Darya [45]

Answer:

If I could have a super power it would be invisibility. Sometimes you wish you weren't there and if you were hiding from someone then they couldn't find you.

3 0

3 years ago

Use the given data at 500 K to calculate ΔG°for the reaction

Anton [14]

Answer : The value of $\Delta G^o$ for the reaction is -959.1 kJ

Explanation :

The given balanced chemical reaction is,

$2H_2S(g)+3O_2(g)\rightarrow 2H_2O(g)+2SO_2(g)$

First we have to calculate the enthalpy of reaction $(\Delta H^o)$ .

$\Delta H^o=H_f_{product}-H_f_{reactant}$

$\Delta H^o=[n_{H_2O}\times \Delta H_f^0_{(H_2O)}+n_{SO_2}\times \Delta H_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta H_f^0_{(H_2S)}+n_{O_2}\times \Delta H_f^0_{(O_2)}]$

where,

$\Delta H^o$ = enthalpy of reaction = ?

n = number of moles

$\Delta H_f^0$ = standard enthalpy of formation

Now put all the given values in this expression, we get:

$\Delta H^o=[2mole\times (-242kJ/mol)+2mole\times (-296.8kJ/mol)}]-[2mole\times (-21kJ/mol)+3mole\times (0kJ/mol)]$

$\Delta H^o=-1035.6kJ=-1035600J$

conversion used : (1 kJ = 1000 J)

Now we have to calculate the entropy of reaction $(\Delta S^o)$ .

$\Delta S^o=S_f_{product}-S_f_{reactant}$

$\Delta S^o=[n_{H_2O}\times \Delta S_f^0_{(H_2O)}+n_{SO_2}\times \Delta S_f^0_{(SO_2)}]-[n_{H_2S}\times \Delta S_f^0_{(H_2S)}+n_{O_2}\times \Delta S_f^0_{(O_2)}]$