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Alinara [238K]
3 years ago
5

What would be the coefficient of the NaOH molecule in the balanced equation?

Chemistry
1 answer:
Scilla [17]3 years ago
6 0
2NaOH=Na2O +H2O
The coefficient of NaOH is 2
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What is the pH of a 0.028M solution (pH= -log(M))<br><br> A) 3.56<br> B) 2.88<br> C) 1.55<br> D) 1
kupik [55]

Answer:

1.55

Explanation:

-log(M)=pH

- Hope that helps! Please let me know if you need further explanation.

4 0
3 years ago
Calculate the specific heat of a substance if a 35g sample absorbs 48 j as the temperature is raised from 293 k to 313 k
bulgar [2K]

Answer:

c = 0.07 j/g.k

Explanation:

Given data:

Mass of sample = 35 g

Heat absorbed = 48 j

Initial temperature = 293 K

Final temperature = 313 K

Specific heat of substance = ?

Solution:

Specific heat capacity:

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Formula:

Q = m.c. ΔT

Q = amount of heat absorbed or released

m = mass of given substance

c = specific heat capacity of substance

ΔT = change in temperature

ΔT = Final temperature - initial temperature

ΔT = 313 k - 293 K

ΔT = 20 k

Now we will put the values in formula.

48 j = 35 g × c× 20 k

48 j = 700 g.k ×c

c = 48 j/700 g.k

c = 0.07 j/g.k

8 0
3 years ago
An ultraviolet ray of light has a wavelength of 1.5x10^-8 m. What is the ray's<br> frequency?
frutty [35]

Answer:

2 E16 Hz    or     2 * 10^16 Hz

Explanation:

The formula to determine frequency is f = c / λ.

f = frequency

c = speed of light

λ = wavelength

f = 3E8 / 1.5E-8

f = 2E16

This makes sense because UV light exists roughly

between 8E14 Hz and 3E16 Hz ----- 2E16 Hz falls in that range

8 0
3 years ago
Given the following data:N2(g) + O2(g)→ 2NO(g), ΔH=+180.7kJ2NO(g) + O2(g)→ 2NO2(g), ΔH=−113.1kJ2N2O(g) → 2N2(g) + O2(g), ΔH=−163
statuscvo [17]

Answer:

ΔH = +155.6 kJ

Explanation:

The Hess' Law states that the enthalpy of the overall reaction is the sum of the enthalpy of the step reactions. To do the addition of the reaction, we first must reorganize them, to disappear with the intermediaries (substances that are not presented in the overall reaction).

If the reaction is inverted, the signal of the enthalpy changes, and if its multiplied by a constant, the enthalpy must be multiplied by the same constant. Thus:

N₂(g) + O₂(g) → 2NO(g) ΔH = +180.7 kJ

2NO(g) + O₂(g) → 2NO₂(g) ΔH = -113.1 kJ

2N₂O(g) → 2N₂(g) + O₂(g) ΔH = -163.2 kJ

The intermediares are N₂ and O₂, thus, reorganizing the reactions:

N₂(g) + O₂(g) → 2NO(g) ΔH = +180.7 kJ

NO₂(g) → NO(g) + (1/2)O₂(g) ΔH = +56.55 kJ (inverted and multiplied by 1/2)

N₂O(g) → N₂(g) + (1/2)O₂(g) ΔH = -81.6 kJ (multiplied by 1/2)

------------------------------------------------------------------------------------

N₂O(g) + NO₂(g) → 3NO(g)

ΔH = +180.7 + 56.55 - 81.6

ΔH = +155.6 kJ

5 0
3 years ago
An aqueous CsCl solution is 8.00 wt% CsCl and has a density of 1.0643 g/mL at 20°C. What is the boiling point of this solution?
umka2103 [35]

<u>Answer:</u> The boiling point of solution is 100.53

<u>Explanation:</u>

We are given:

8.00 wt % of CsCl

This means that 8.00 grams of CsCl is present in 100 grams of solution

Mass of solvent = (100 - 8) g = 92 grams

The equation used to calculate elevation in boiling point follows:

\Delta T_b=\text{Boiling point of solution}-\text{Boiling point of pure solution}

To calculate the elevation in boiling point, we use the equation:

\Delta T_b=iK_bm

Or,

\text{Boiling point of solution}-\text{Boiling point of pure solution}=i\times K_b\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ (in grams)}}

where,

Boiling point of pure solution = 100°C

i = Vant hoff factor = 2 (For CsCl)

K_b = molal boiling point elevation constant = 0.51°C/m

m_{solute} = Given mass of solute (CsCl) = 8.00 g

M_{solute} = Molar mass of solute (CsCl) = 168.4  g/mol

W_{solvent} = Mass of solvent (water) = 92 g

Putting values in above equation, we get:

\text{Boiling point of solution}-100=2\times 0.51^oC/m\times \frac{8.00\times 1000}{168.4g/mol\times 92}\\\\\text{Boiling point of solution}=100.53^oC

Hence, the boiling point of solution is 100.53

6 0
3 years ago
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