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elixir [45]
3 years ago
10

How many kilograms of water at 15oC can be heated to 95oC by burning 143 grams of methane, CH4, assuming that 100% of the heat i

s used to heat the water. The heat of combustion of methane is 891 kJ per mole of methane. (The specific heat of water is 4.184 J/goC.)
Chemistry
1 answer:
mezya [45]3 years ago
8 0

Answer:

23.7 kg is the mass of H₂O that can be heated

Explanation:

First of all, we need to determine the heat used.

891 kJ per mol of methane is released in the combustion, but we don't have 1 mol, we have 143 g

We convert the mass to moles → 143 g . 1mol / 16 g = 8.93 moles

Now, a rule of three:

1 mol of CH₄ release a heat of 891 kJ

8.93 moles of CH₄ will release (8.93 . 891 kJ) / 1 = 7956.63 kJ

If we convert to J → 7956.63 kJ . 1000 J / kJ = 7956630 Joules

Now we determined the heat released, we can apply the calorimetry formula

Q = m . C . (Final T° - Initial T°)

7956630 J = mass . 4.184 J/g°C ( 95°C - 15°C)

7956630 J / 4.184 J/g°C . 80°C = mass

23771 g = mass

If we convert from g to kg → 23771 g . 1kg / 1000 g = 23.7 kg

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What is the product of reduction of ethyl 4-oxobutanoate with sodium borohydride in ethanol at room temperature for 30 minutes?
zheka24 [161]

The product of reduction of ethyl 4-oxobutanoate with sodium borohydride in ethanol at room temperature for 30 minutes is ethyl 4- hydroxybutanoate .

Sodium borohydride is a relatively selective reducing agent  Ethanolic solutions of Sodium borohydride reduces aldehyde , and ketone , in the presence of acid chloride , ester , epoxide , lactones , acids , nitriles , nitro groups.

The sodium borohydride does not reduce ester group because sodium borohydride is not strong enough and the electrophilicity at carbony carbon of ester is not more as compare toaldehyde , and ketone

The product of reduction of ethyl 4-oxobutanoate with sodium borohydride in ethanol at room temperature for 30 minutes is ethyl 4- hydroxybutanoate .

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4 0
1 year ago
What was the biggest challenge you faced in getting to where you are today and how did you overcome it?
Alchen [17]
It’s all based on you but yours could be anything hard you went through and how you overcame it. Also say how it made you the person you are today
4 0
3 years ago
If you react 156.0 g of calcium chloride with an excess NaOH, how much sodium chloride should you get?
laiz [17]

Answer:

164.3g of NaCl

Explanation:

Based on the chemical equation:

CaCl2 + 2NaOH → 2NaCl + Ca(OH)2

<em>where 1 mole of CaCl2 reacts with 2 moles of NaOH</em>

To solve this question we must convert the mass of CaCl2 to moles. Using the chemical equation we can find the moles of NaCl and its mass:

<em>Moles CaCl2 -Molar mass: 110.98g/mol-</em>

156.0g CaCl₂ * (1mol / 110.98g) = 1.4057 moles CaCl2

<em>Moles NaCl:</em>

1.4057 moles CaCl2 * (2mol NaCl / 1mol CaCl2) = 2.811 moles NaCl

<em>Mass NaCl -Molar mass: 58.44g/mol-</em>

2.811 moles NaCl * (58.44g / mol) = 164.3g of NaCl

7 0
2 years ago
How many equivalents are present in 10 g of Ca2+? <br> 1 <br> 0.5 <br> 1.5 <br> 2
Lynna [10]

0.5

Explanation:

Given parameters:

Mass of Ca²⁺ = 10g

unknown:

Equivalent weight = ?

Solution:

Equivalent weight that is the amount of electrons which a substance gains or loses per mole.

Ca²⁺ has +3 charge

It lost 2e⁻;

therefore;

  In 1 mole of  Ca²⁺, we have 2 equivalent weight

1 mol  Ca²⁺ = 2eq. wts. 

1 mol Ca x (40 g / 1 mol ) x (1 mol / 2 eq.wts.) = 20.0 g = 1 eq.wt. 

Therefore;

10.0 g  Ca²⁺ x (1 eq.wt. / 20.0 g) = 0.5 eq.wts.

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8 0
3 years ago
4.80 x 10^25 formula units pf calcium iodide (Cal2) will have what mass?
maw [93]

Answer:

mass CaI2 = 23.424 Kg

Explanation:

From the periodic table we obtain for CaI2:

⇒ molecular mass CaI2: 40.078  + ((2)(126.90)) = 293.878 g/mol

∴ mol CaI2 = (4.80 E25 units )×(mol/6.022 E23 units) = 79.708  mol CaI2

⇒ mass CaI2 = (79.708 mol CaI2)×(293.878 g/mol) = 23424.43 g

⇒ mass CaI2 = 23.424 Kg

7 0
3 years ago
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