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Rom4ik [11]
2 years ago
13

Si 4,8 moles de gas

Chemistry
1 answer:
Marat540 [252]2 years ago
7 0

Answer:

I don't speck or understand spanish

Explanation:

I can understand Si that means yes but that all

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PLEASE ANSWER
PolarNik [594]
I think its A.
if one force cannot overcome the other, the object remains stationary.
4 0
3 years ago
Calculate the percent errorin a length measurementof 4.45cm if the correct value is 4.06
FinnZ [79.3K]

Answer:

9.6 %

Explanation:

<u>Step 1: How to define  percent error ? </u>

⇒ % error is the difference between a measured value and the known or accepted value

⇒Percent error is calculated using the following formula:

⇒%error  =  | Experimental value-theoretical value/theoretical value |  x100%

⇔ this can be written as well as : error = (| Experimental value/ theoretical value | - | Theoretical value / Theoretical value | ) x100%

<u>Step 2: Calculate % error</u>

In this case, this means :

%error = ( |(4.45 cm - 4.06cm ) / 4.06cm | ) x100%

%error = 0.096 x100%

%error =9.6 %

3 0
3 years ago
A compound has a molar mass of 90. grams per mole and the empirical formula CH2O. What is the molecular formula of this compound
Elanso [62]
The molecular formula of this compound is C3H603 XD
8 0
3 years ago
How do you explode stuff
Svetradugi [14.3K]
Well. Depends. 
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You should try baking soda and vinegar, almost everyone knows about this.
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Hope this helps :D
8 0
3 years ago
Read 2 more answers
It is required to prepare 1250 kg of a solution composed of 12 wt.% ethanol and 88 wt.% water. Two solutions are available, the
bekas [8.4K]

Answer:

437.5 kg of first solution and 812.5 kg of second solution should be mixed to get desired solution.

Explanation:

Let the mass of the first solution be x and second solution be y.

Amount solution required = 1250 kg

x + y = 1250 kg....[1]

Percentage of ethanol in required solution = 12% of 1250 kg

Percentage of ethanol in solution-1 = 5% of x

Percentage of ethanol in required solution = 25% of y

5% of x +  25% of y =12% of 1250 kg

\frac{5}{100}\times x+\frac{25}{100}y=\frac{12}{100}\times 1250 kg

x + 5y = 3000 kg...[2]

Solving [1] and [2] we :

x = 437.5 kg   , y =  812.5 kg

437.5 kg of first solution and 812.5 kg of second solution should be mixed to get desired solution.

8 0
3 years ago
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