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Semmy [17]
4 years ago
14

When a candle burns it produced 41,300 Joules per 1 gram. Use dimensional analysis to convert this to Calories per pound.

Chemistry
1 answer:
insens350 [35]4 years ago
3 0

Answer:

4477381.7 calories/pound

Explanation:

It is given that,

When a candle burns it produced 41,300 Joules per 1 gram.

We need to convert it into calories per pound.

We know that,

1 cal = 4.184 J

⇒ 1 J = (1/4.184) cal

1 pound = 453.592 grams

⇒1 g = (1/453.592) pounds

Now,

41300\ \text{Joules per gram}=\dfrac{41300\ \text{Joules}}{\text{gram}}\\\\=41300\times \dfrac{\dfrac{1}{4.184}\ \text{calories}}{\dfrac{1}{453.592}\ \text{gram}}\\\\=4477381.7\ \text{calories/pound}

Hence, 41,300 Joules/gram = 4477381.7 calories/pound.

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Answer - Ozone layer helps keep the harmful ultra violet rays hitting and keeping the atmospheric pressure in.

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6 0
3 years ago
Please help . also need the name of their binary ionic compound . ​
madreJ [45]

Answer:

See below!

Explanation:

For the chemical formula, you need to have enough of each atom so that the charge is zero.

Aluminum has a +3 charge, and fluorine has a -1 charge.  Since the charge has to be zero, you need three fluorines, giving you AlF₃.

Barium has a +2 charge, and oxygen has a -2 charge.  Since the charges are equal in magnitude but opposite in sign, you only need one of each atom giving you BaO.

The name of the ionic compound will be the metal and then the nonmetal.  When putting the nonmetal in, change the ending to "-ide".  For example "chlorine" would be "chloride.

CaCl₂  ==>  calcium chloride

Ga₂S₃  ==>  gallium sulfide

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BaO  ==>  barium oxide

6 0
3 years ago
Calculate ΔrG∘ at 298 K for the following reactions.CO(g)+H2O(g)→H2(g)+CO2(g)2-Predict the effect on ΔrG∘ of lowering the temper
KonstantinChe [14]

Answer:

1) ΔG°r(298 K) = - 28.619 KJ/mol

2) ΔG°r will decrease with decreasing temperature

Explanation:

  • CO(g) + H2O(g) → H2(g) + CO2(g)

1) ΔG°r = ∑νiΔG°f,i

⇒ ΔG°r(298 K) = ΔG°CO2(g) + ΔG°H2(g) - ΔG°H2O(g) - ΔG°CO(g)

from literature, T = 298 K:

∴ ΔG°CO2(g) = - 394.359 KJ/mol

∴ ΔG°CO(g) = - 137.152 KJ/mol

∴ ΔG°H2(g) = 0 KJ/mol........pure substance

∴ ΔG°H2O(g) = - 228.588 KJ/mol

⇒ ΔG°r(298 K) = - 394.359 KJ/mol + 0 KJ/mol - ( - 228.588 KJ/mol ) - ( - 137.152 KJ7mol )

⇒ ΔG°r(298 K) = - 28.619 KJ/mol

2) K = e∧(-ΔG°/RT)

∴ R = 8.314 E-3 KJ/K.mol

∴ T = 298 K

⇒ K = e∧(-28.619/(8.314 E-3)(298) = 9.624 E-6

⇒ ΔG°r = - RTLnK

If T (↓) ⇒ ΔG°r (↓)

assuming T = 200 K

⇒ ΔG°r(200 K) = - (8.314 E-3)(200)Ln(9.624E-3)

⇒ ΔG°r (200K) = - 19.207 KJ/mol < ΔG°r(298 K) = - 28.619 KJ/mol

6 0
3 years ago
Which atoms in the pair has the larger atomic radius ? Help please !!
Ne4ueva [31]

A

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2. Calcium, Ca.

3. Gallium, Ga.

4. Carbon, C.

5. Bromine, Br.

6. Barium, Ba.

7. Silicon, Si.

Explanation:

4 0
2 years ago
Read 2 more answers
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Ilya [14]

Answer:

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