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Juli2301 [7.4K]
2 years ago
12

Convert 7.68cal/sec to Kcal/min

Chemistry
2 answers:
svp [43]2 years ago
8 0
First, you must convert 7.68cal/sec to cal/min. To do so, multiply 7.68x60(seconds, thus making a minute). 7.68x60=460.8 

Next, we must convert 460.8 cal to kcal. To do so, divide 460.8 by 1000 (460.8/1000). The result is .4608 kcal/min
sammy [17]2 years ago
6 0

Answer:

7.68 cal/sec = 0.461 kcal/min

Explanation:

Measurement = 7.68 cal/sec

Unit conversions:

1 cal (calorie) = 10⁻³ kcal (kilocalories)

1 min = 60 sec

Step 1: convert cal to kcal

7.68 cal/sec= \frac{7.68\ cal * 10^{-3} kcal}{1\ cal} *\frac{1}{sec} = 7.68*10^{-3} kcal/sec

Step 2: convert sec to min

7.68*10^{-3} kcal/sec=\frac{7.68*10^{-3} kcal}{sec} *\frac{60\ sec}{1\ min} = 0.461 kcal/min

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A pure substance or a homogeneous mixture consists of a single phase. A heterogeneous mixture consists of two or more phases. When oil and water are combined, they do not mix evenly, but instead form two separate layers.

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Which milligram quantity contains a total of four significant figures
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4.8g of calcium is added to 3.6g of water. The following reaction occurs
notka56 [123]
Q1)
the number of moles can be calculated as follows
number of moles = mass present / molar mass
number of moles is the amount of substance.
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q2)
next we are asked to calculate the number of moles of water present 
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2 years ago
When the reaction CO2(g) + H2(g) ⇄ H2O(g) + CO(g) is at equilibrium at 1800◦C, the equilibrium concentrations are found to be [C
UNO [17]

Answer:

The new molar concentration of CO at equilibrium will be  :[CO]=1.16 M.

Explanation:

Equilibrium concentration of all reactant and product:

[CO_2] = 0.24 M, [H_2] = 0.24 M, [H_2O] = 0.48 M, [CO] = 0.48 M

Equilibrium constant of the reaction :

K=\frac{[H_2O][CO]}{[CO_2][H_2]}=\frac{0.48 M\times 0.48 M}{0.24 M\times 0.24 M}

K = 4

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Concentration at eq'm:

0.24 M          0.24 M                 0.48 M            0.48 M

After addition of 0.34 moles per liter of CO_2 and H_2 are added.

(0.24+0.34) M    (0.24+0.34) M  (0.48+x)M         (0.48+x)M

Equilibrium constant of the reaction after addition of more carbon dioxide and water:

K=4=\frac{(0.48+x)M\times (0.48+x)M}{(0.24+0.34)\times (0.24+0.34) M}

4=\frac{(0.48+x)^2}{(0.24+0.34)^2}

Solving for x: x = 0.68

The new molar concentration of CO at equilibrium will be:

[CO]= (0.48+x)M = (0.48+0.68 )M = 1.16 M

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