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Pavel [41]
3 years ago
9

PLS PLS HELP this is science

Chemistry
2 answers:
Ne4ueva [31]3 years ago
8 0

Answer:

its minerals :)

Explanation:

i hope its right

alsooo heyyyy

ladessa [460]3 years ago
6 0

Answer:

minerals!!!!! i think anyway

Explanation:

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Answer:

<u><em>Pentane </em></u>

Explanation:

since we have in here CH3-CH2-CH2-CH2-CH3 5 Carbon atoms and 12 Hydrogen making it C_{5} H_{12}

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3 years ago
What kind of change results in a new substance being produced
tatiyna
Chemical change creates a new substance
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Complete combustion of 8.10 g of a hydrocarbon produced 25.9 g of CO2 and 9.27 g of H2O. What is the empirical formula for the h
balu736 [363]

CxHy     +  O2    -->    x CO2     +    y/2  H2O

 

Find the moles of CO2 :     18.9g  /  44 g/mol   =    .430 mol CO2   = .430 mol of C in compound

Find the moles of H2O:      5.79g / 18 g/mol     =     .322 mol H2O   = .166 mol of H in compound

 

Find the mass of C and H in the compound:

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                              .166mol  x 1g   = .166g H   

 

When you add these up they indicate a mass of 5.33 g for the compound, not 5.80g as you stated in the problem.

Therefore it is likely that either the mass of the CO2 or the mass of H20 produced is incorrect (most likely a typo).

In any event, to find the formula, you would take the moles of C and H and convert to a whole number ratio (this is usually done by dividing both of them by the smaller value).

8 0
4 years ago
What volume (in L) of carbon dioxide will be produced from the reaction of 13.6 L of nitrogen monoxide?
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A. The reaction will proceed forward forming more CH4

B. The reaction will proceed forward forming more CH4

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D. Lowering the volume makes the gas particles to be more close together thereby enhancing their collisions leading to reaction. Therefore the reaction will proceed forward forming more CH4

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F. The following will favour CH4 at equilibrium

i. Catalyst to the reaction mixture,

ii. Both adding more H2 to the reaction mixture and lowering the volume of the reaction mixture

iii. Adding more C to the reaction mixture.

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