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Veronika [31]
4 years ago
14

4.1 moles of sodium carbonate to molecules of sodium carbonate.​

Chemistry
1 answer:
docker41 [41]4 years ago
3 0
<h3>Answer:</h3>

2.47 × 10^24 molecules

<h3>Explanation:</h3>

One mole of a compound contains molecules equivalent to the Avogadro's number, 6.022 × 10^23.

That is, 1 mole of a compound =  6.022 × 10^23 molecules

Therefore,

1 mole of Na₂CO₃ = 6.022 × 10^23 molecules

Thus, we can calculate the number of molecules in 4.1 moles of Na₂CO₃

we get,

 = 4.1 moles × 6.022 × 10^23 molecules

 = 2.47 × 10^24 molecules

Hence, 4.1 moles of Na₂CO₃ contains 2.47 × 10^24 molecules

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vova2212 [387]

Answer:

The balanced chemical equation is Pb(NO3)2 + 2KI produces PbI2 + 2K(NO)3. In chemistry, this is called a double replacement reaction. Lead nitrate and potassium iodide, which are the reactants, are powders that react to form a yellow-colored lead iodide and a colorless potassium nitrate.

Explanation:

3 0
3 years ago
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From which category of elements would you choose to make a container that would not shatter if dropped? Explain your answer plea
Eduardwww [97]
Container must be made up of non metallic elements which can typically covalently  bond and must have lone pairs of electrons  dative bonds in order to allow further stability.
6 0
3 years ago
Determine the amount of heat (in kj) required to vaporize 1.55 kg of water at its boiling point. for water, δhvap
Usimov [2.4K]
Heat required to vaporize 1 mol of water from water at 100C to steam at 100C = 40.7 kJ 
<span>1 mol of water weighs =  18.015g
</span>1.55 kg = <span>1550/18.015 mol = 86.03 mol 

</span><span>Heat required to vaporize :
</span>= 86.03 mol  x <span>40.7 kJ 
</span>
= 3501.421 kJ
5 0
3 years ago
In a certain city, electricity costs $0.17 per kW·h. What is the annual cost for electricity to power a lamp-post for 5.50 hours
Anon25 [30]

Answer:

(a) = $34.123

(b) = $8.532

(c) Additional cost of fluorescent bulb is justified

Explanation:

Cost of electricity = $0.17 per kW·h

(a) For a 100 watt bulb which is the same as 100/1000 or 0.1 kW, the cost per hour =

0.1 × 0.17 = $0.017/h

and for 5.5 hours = 0.017×5.5 = $0.0935

The annual cost, which is 365 days, we have

Annual cost = $0.0935 × 365 = $34.123

(b) For the energy efficient 25-watt bulb, we have

25/1000 = 0.025kW

Power cost per annum =

0.025kW×$0.17 per kW·h×5.5×365 = $8.532

(c) Total cost of incandescent bulb = $0.89 total cost of using the incandescent bulb is $34.123 + $0.89 = $35.02

Total cost of using the energy efficient fluorescent bulb is about $3.49

Total cost of using the energy efficient bulb = $8.532 + $3.49 = $12.02

Total cost of incandescent bulb = $35.02 while total cost of energy efficient bulb is = $12.02

$12.02 <$35.02

Additional cost of fluorescent bulb is justified

8 0
3 years ago
Determine the ph of a 0.18 m h2co3 solution. carbonic acid is a diprotic acid whose ka1 = 4.3 × 10-7 and ka2 = 5.6 × 10-11. dete
oee [108]
Answer is: ph value is 3.56.
Chemical reaction 1: H₂CO₃(aq) ⇄ HCO₃⁻(aq) + H⁺(aq); Ka₁ = 4,3·10⁻⁷.
Chemical reaction 2: HCO₃⁻(aq) ⇄ CO₃²⁻(aq) + H⁺(aq); Ka₂ = 5,6·10⁻¹¹.
c(H₂CO₃) = 0,18 M.
[HCO₃⁻] = [H⁺<span>] = x.
</span>[H₂CO₃] = 0,18 M - x.
Ka₁ = [HCO₃⁻] · [H⁺] / [H₂CO₃].
4,3·10⁻⁷ = x² / (0,18 M -x).
Solve quadratic equation: x = [H⁺] =0,000293 M.
pH = -log[H⁺] = -log(0,000293 M).
pH = 3,56; second Ka do not contributes pH value a lot.


5 0
3 years ago
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