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horrorfan [7]
2 years ago
6

If an endothermic process absorbs 256j how many kilocalories are absorbed

Chemistry
2 answers:
Finger [1]2 years ago
8 0
There are several units for expressing energy, Most common are joules and kilocalories.

The conversion factor of joule to kilocalories is 
 1 kilo calorie  = 4184 J
∴ x kilo calorie = 256 J

x = \frac{256}{4184} = 0.0611 kilo calories

Thus, 256 J = 0.0611 kilocalories
astra-53 [7]2 years ago
5 0

Answer: 0.0612 kcalories

Explanation:

Endothermic processes are those processes in which heat is absorbed and exothermic processes are those processes in which heat is released.

Standard units for heat measurement are Joules and calories wherein

1 joule = 0.239 calorie.

As 1 Joule is equivalent to 0.239 calorie

256 Joules are equivalent to=\frac{0.239}{1}\times 256=61.2calories

Now 1 cal = 0.001 kcalories

Therefore 61.2 calories=\frac{0.001}{1}\times 61.2=0.0612kcal

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An atom has the electron arrangement 2,8. What will it do to gain a full
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Alex dragged a log across the yard in 30 the log weighted 400n she did 900 j of work how much power did I she have
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The power used by Alex to drag the log across the yard is determined as 2,656 W.

<h3>Mass of the log</h3>

The mass of the log is calculated as follows;

W = mg

m = W/g

m = (400)/9.8

m = 40.82 kg

<h3>Velocity of the log</h3>

K.E = ¹/₂mv²

v² = 2K.E/m

v² = (2 x 900)/(40.82)

v² = 44.096

v = 6.64 m/s

<h3>Power used by Alex</h3>

P = Fv

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5 0
2 years ago
3/4(x-2)=x-1 please answer me ​
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Answer:

-2

Explanation:

3/2 (x-2) = x-1

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3 0
3 years ago
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The elements from this section of the periodic table all belong to the same
krok68 [10]

Answer:

Option C = same period.

Explanation:

All these elements belongs to second period of periodic table. This period consist of eight elements lithium, beryllium, boron, carbon, nitrogen, oxygen, fluorine and neon.

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Li₃ = [He] 2s¹

Electronic configuration of beryllium:

Be₄ = [He] 2s²

Electronic configuration of boron:

B₅ = [He] 2s² 2p¹

Electronic configuration of carbon:

C₆ = [He] 2s² 2p²

Electronic configuration of nitrogen:

N₇ = [He] 2s² 2p³

Electronic configuration of oxygen:

O₈ = [He] 2s² 2p⁴

Electronic configuration of fluorine:

F₉ = [He] 2s² 2p⁵

Electronic configuration of neon:

Ne₁₀ = [He] 2s² 2p⁶

All these elements present in same period having same electronic shell.

However their families, valance electrons and group are different. Boron have three valance electrons and belongs to group 3A. Carbon belongs to group 4A and have 4 valance electrons. Nitrogen belongs to group 5A and have five valance electrons. Oxygen belongs to group 6A and have six valance electrons. Fluorine belongs to group 7A and have seven valance electrons.

7 0
2 years ago
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