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4 years ago

1. Convert 1365.4 joules into kilocalories.

Chemistry

1 answer:

Hunter-Best [27]4 years ago

3 0

Answer:

1365.4 J = 0.326 kcal

Explanation:

To solve this question, we use the unitary method.

1 Joule = 4184 kcal

According to the above equation, one joule is equivalent to that of 4.184 cal or 4184 kcal.

Joule(J) and kilocalories(kcal) are both units of energy.

$1365.4J=\frac{1365.4}{4184}kcal=0.326kcal$

Therefore, the answer is 0.326 kcal.

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Two carbon atoms cannot be linked to each other by more than three covalent bonds why

Leto [7]

Answer:since,there is a single bond between the two carbon atoms and both share their one atom therefore for completing its shell it need to combine with three atoms of carbon or other element . therefore it cannot be linked to more than 3 covalent bonds since its shell will be completed to become stable .

Explanation:

6 0

3 years ago

Given that the freezing point depression constant for water is 1.86°c kg/mol, calculate the change in freezing point for a 0.907

melamori03 [73]

Answer : The correct answer for change in freezing point = 1.69 ° C

Freezing point depression :

It is defined as depression in freezing point of solvent when volatile or non volatile solute is added .

SO when any solute is added freezing point of solution is less than freezing point of pure solvent . This depression in freezing point is directly proportional to molal concentration of solute .

It can be expressed as :

ΔTf = Freezing point of pure solvent - freezing point of solution = i* kf * m

Where : ΔTf = change in freezing point (°C)

i = Von't Hoff factor

kf =molal freezing point depression constant of solvent. $\frac{^0 C}{m}$

m = molality of solute (m or $\frac{mol}{Kg}$ )

Given : kf = 1.86 $\frac{^0 C*Kg}{mol}$

m = 0.907 $\frac{mol}{Kg}$ )

Von't Hoff factor for non volatile solute is always = 1 .Since the sugar is non volatile solute , so i = 1

Plugging value in expression :

ΔTf = 1* 1.86 $\frac{^0 C*Kg}{mol}$ * 0.907 $\frac{mol}{Kg}$ )

ΔTf = 1.69 ° C

Hence change in freezing point = 1.69 °C

5 0

3 years ago

11. As early as 1938. the use of NaOH was suggested as a means of removing CO2 from the cabin of a spacecraft according to the f

vagabundo [1.1K]

For a (unbalanced) reaction: NaOH +CO2-Na2CO3 + H2O, the moles of NaOH and moles of each product are formed are mathematically given as

a) Moles of NaOH =44.05

b) Moles of Na2CO3=21.0

Moles of H2O= 21.0

<h3>What is the moles of NaOH and what moles of each product are formed?</h3>

Generally, the equation for the Chemical reaction is mathematically given as

2 NaOH(aq)+ CO2(g)------> Na2CO3(aq)+ H2O(l)

Therefore

Moles of CO2= 925/44

Moles of CO2=21.0

Hence

Moles of NaOH = 2 x Moles of CO2

Moles of NaOH = 2x925/44

Moles of NaOH =44.05

In conclusion

Moles of Na2CO3 925/44

Moles of Na2CO3=21.0

And

Moles of H2O= 925/44

Moles of H2O= 21.0