The mass of an atom may be found by adding the

Water’s boiling point would be close to

aleksandrvk [35]

Waters boiling point is 100°C and 212°F

4 0

3 years ago

I NEED HELP PLEASE!!

VikaD [51]

Answer:

The amount of heat absorbed is <u>5.183889 kJ</u> .

Explanation:

In conversion of water to ice it rejects some heat while in conversion of ice to water it absorbs heat which is called latent heat which is given as 6.02 kJ/mol.

The amount of ice given is 15.5 g.

Converting it to moles as the latent heat is given in per moles:

$\frac{given\\weight \\ (in\\grams)}{molecular\\weight\\(in\\grams)}$

Molecular mass of Hydrogen (H) and Oxygen (O) is 1 u and 16 u respectively.

Molecular mass of water is 18 g ( $H_{2} O$ ⇒ 2*1+16=18 ).

mole = 15.5/18 ≈ 0.8611 moles

Therefore the amount of heat absorbed by 15.5 g of ice ( 0.8611 moles) = <em>Latent heat * moles </em>

Heat absorbed = 6.02*0.8611

= 6.02*(15.5/18)

≈ 5.183889 kJ

5 0

3 years ago

For the equilibrium PCl5(g) PCl3(g) + Cl2(g), Kc = 2.0 × 101 at 240°C. If pure PCl5 is placed in a 1.00-L container and allowed

Cloud [144]

Answer:

the equilibrium concentration of [PCl₅] is 3.64*10⁻³ M

Explanation:

for the reaction

PCl₅(g) → PCl₃(g) + Cl₂(g)

where

Kc= [PCl₃]*[Cl₂]/[PCl₅] = 2.0*10¹ M = 20 M

and [A] denote concentrations of A

if initially the mixture is pure PCl₅ , then it will dissociate according to the reaction and since always one mole of PCl₃(g) is generated with one mole of Cl₂(g) , the total number of moles of both at the end is the same → they have the same concentration → [PCl₃(g)] = [Cl₂]=0.27 M

therefore

Kc= [PCl₃]*[Cl₂]/[PCl₅] = 0.27 M* 0.27 M /[PCl₅] = 20 M

[PCl₅] = 0.27 M* 0.27 M / 20 M = 3.64*10⁻³ M

[PCl₅] = 3.64*10⁻³ M

the equilibrium concentration of [PCl₅] is 3.64*10⁻³ M

6 0

3 years ago

Ga3+ and Br1- is what formula?

Sladkaya [172]

$\text{GaBr}_3$

8 0

2 years ago

Balance the following half reaction in basic conditions. Then, indicate the coefficients for H2O and OH– for the balanced half r

Ugo [173]

Answer:

The ballance half reactions are:

Mg²⁺ + 2e⁻ → Mg

6OH⁻ + Si → SiO₃²⁻ + 4e⁻ + 3 H₂O

Coefficients for H2O and OH– are 3 for H₂O (in products side) and 6 for OH⁻ (in reactants side)

Explanation:

Si (s) + Mg(OH)₂ (s) → Mg (s) + SiO₃²⁻ (aq)

Let's see the oxidations number.

As any element in ground state, we know that oxidation state is 0, so Si in reactants and Mg in products, have 0.

Mg in reactants, acts with +2, so the oxidation number has decreased.

This is the reduction, so it has gained electrons.

Si in reactants acts with 0 so in products we find it with +4. The oxidation number increased it, so this is oxidation. The element has lost electrons.

Let's take a look to half reactions:

Mg²⁺ + 2e⁻ → Mg

Si → SiO₃²⁻ + 4e⁻

In basic medium, we have to add water, as the same amount of oxygen we have, IN THE SAME SIDE. We have 3 oxygens in products, so we add 3 H₂O and in the opposite site we can add OH⁻, to balance the hydrogen. The half reaciton will be:

6OH⁻ + Si → SiO₃²⁻ + 4e⁻ + 3 H₂O

If we want to ballance the main reaction we have to multiply (x2) the half reaction of oxidation. So the electrons can be ballanced.

2Mg²⁺ + 4e⁻ → 2Mg

Now, that they are ballanced we can sum the half reactions:

2Mg²⁺ + 4e⁻ → 2Mg

6OH⁻ + Si → SiO₃²⁻ + 4e⁻ + 3 H₂O

2Mg²⁺ + 4e⁻ + 6OH⁻ + Si → 2Mg + SiO₃²⁻ + 4e⁻ + 3 H₂O

7 0

3 years ago