All categories

2 years ago

Can someone help me please.

Chemistry

1 answer:

Vika [28.1K]2 years ago

4 0

Answer:

ccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccccc

Explanation:

ccccccccccccccccccccccccccc

You might be interested in

The molar heat of fusion for water is 6.01 kJ/mol. How much energy must be added to a 75.0-g block of ice at 0°C to change it to

andreyandreev [35.5K]

Answer is: 25,06 kJ of energy must be added to a 75 g block of ice.
ΔHfusion(H₂O) = 6,01 kJ/mol.
T(H₂O) = 0°C.
m(H₂O) = 75 g.
n(H₂O) = m(H₂O) ÷ M(H₂O).
n(H₂O) = 75 g ÷ 18 g/mol.
n(H₂O) = 4,17 mol.
Q = ΔHfusion(H₂O) · n(H₂O)
Q = 6,01 kJ/mol · 4,17 mol
Q = 25,06 kJ.

7 0

2 years ago

Read 2 more answers

What are the concentrations of h3o+ and oh− in tomatoes that have a ph of 4.10?

IRINA_888 [86]

Answer : The concentration of $H_3O^+$ and $OH^-$ are $7.94\times 10^{-5}$ and $1.258\times 10^{-10}$ respectively.

Solution : Given,

pH = 4.10

pH : pH is defined as the negative logarithm of hydronium ion concentration.

Formula used : $pH=-log[H_3O^+]$

First we have to calculate the hydronium ion concentration by using pH formula.

$4.10=-log[H_3O^+]$

$[H_3O^+]=antilog(-4.10)$

$[H_3O^+]=7.94\times 10^{-5}$

Now we have to calculate the pOH.

As we know, $pH+pOH=14$

$4.10+pOH=14$

$pOH=9.9$

Now we have to calculate the hydroxide ion concentration.

$pOH=-log[OH^-]$

$9.9=-log[OH^-]$

$[OH^-]=antilog(-9.9)$

$[OH^-]=1.258\times 10^{-10}$

Therefore, the concentration of $H_3O^+$ and $OH^-$ are $7.94\times 10^{-5}$ and $1.258\times 10^{-10}$ respectively.

5 0

2 years ago

If you move 60 meter in 10 seconds, what is your speed

Taya2010 [7]

the answer was S=d/t = 60/10 = 6m/s, I'm sorry for getting it wrong the first time I answered ^^

5 0

2 years ago

Read 2 more answers

Liquid alcohol will change to vapor when its temperature reaches the

Nina [5.8K]

C. boiling point
boiling point is the temperature at which all liquids change from liquid to gas

4 0

2 years ago

Be sure to answer all parts. ΔH o f of hydrogen chloride [HCl(g)] is −92.3 kJ/mol. Given the following data, determine the ident

goldenfox [79]

Answer:

NH₄Cl(s) → NH₃(g) + HCl(g)

ΔH°rxn 74,89 kJ/mol

Explanation:

The change in enthalpy of formation (ΔHf) is defined as the change in enthalpy in the formation of a substance from its constituent elements. For HCl(g):

(1) ¹/₂H₂(g) + ¹/₂ Cl₂ → HCl(g) ΔH = -92,3 kJ/mol

It is possible to sum ΔH of different reactions to obtain ΔH of a global reaction (Hess's law).

For the reactions:

(2) N₂(g) + 4H₂(g) + Cl₂(g) → 2NH₄Cl(s) ΔH°rxn = −630.78 kJ/mol

(3) N₂(g) + 3H₂(g) → 2NH₃(g) ΔH°rxn = −296.4 kJ/mol

The sum of -(2) + (3) gives:

-(2) 2NH₄Cl(s) → N₂(g) + 4H₂(g) + Cl₂(g) ΔH°rxn = +630.78 kJ/mol

(3) N₂(g) + 3H₂(g) → 2NH₃(g) ΔH°rxn = −296.4 kJ/mol

-(2) + (3) 2NH₄Cl(s) → 2NH₃(g) + H₂(g) + Cl₂(g)

ΔH°rxn = +630.78 kJ/mol −296.4 kJ/mol = +334,38 kJ/mol

Now, the sum of -(2) + (3) + 2×(1)

-(2) + (3) 2NH₄Cl(s) → 2NH₃(g) + H₂(g) + Cl₂(g) ΔH°rxn = +334,38 kJ/mol

2×(1) H₂(g) + Cl₂(g)→ 2HCl(g) ΔH = 2×-92,3 kJ/mol

-(2) + (3) + 2×(1) 2NH₄Cl(s) → 2NH₃(g) + 2HCl(g)

ΔH°rxn = +334,38 kJ/mol + 2×-92,3 kJ/mol = 149,78 kJ/mol

The reaction of:

NH₄Cl(s) → NH₃(g) + HCl(g)

Has ΔH°rxn = 149,78kJ/mol / 2 = 74,89 kJ/mol

I hope it helps!

4 0

3 years ago