Are plants abiotic or biotic

Chemistry

1 answer:

UNO [17]3 years ago

6 0

Hey there!

The answer is biotic! Biotic are all living factors in an ecosystem.

Hope this helped!

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What is the excess reactant for the following reaction given we have 3.4 moles of Ca(NO3)2 and 2.4 moles of Li3PO4? Reaction: 3C

Yuki888 [10]

<u>Answer:</u> The excess reactant for the given reaction is $Li_3PO_4$

<u>Explanation:</u>

Limiting reagent is defined as the reagent which is present in less amount and it limits the formation of products.

Excess reagent is defined as the reagent which is present in large amount.

For the given chemical reaction:

$3Ca(NO_3)_2+2Li_3PO_4\rightarrow 6LiNO_3+Ca_3(PO_4)_2$

We are given:

Moles of calcium nitrate = 3.4 mol

Moles of lithium phosphate = 2.4 mol

By stoichiometry of the reaction:

3 moles of calcium nitrate reacts with 2 moles of lithium phosphate

So, 3.4 moles of calcium nitrate will react with = $\frac{2}{3}\times 3.4=2.27mol$ of lithium phosphate

As, the given amount of lithium phosphate is more than the required amount. So, it is considered as an excess reagent.

Thus, calcium nitrate is considered as the limiting reagent because it limits the formation of products.

Hence, the excess reactant for the given reaction is $Li_3PO_4$

7 0

3 years ago

What volume of CH4(g), measured at 25oC and 745 Torr, must be burned in excess oxygen to release 1.00 x 106 kJ of heat to the su

anastassius [24]

Answer:

$V=27992L=28.00m^3$

Explanation:

Hello,

In this case, the combustion of methane is shown below:

$CH_4+2O_2\rightarrow CO_2+2H_2O$

And has a heat of combustion of −890.8 kJ/mol, for which the burnt moles are:

$n_{CH_4}=\frac{-1.00x10^6kJ}{-890.8kJ/mol}= 1122.6molCH_4$

Whereas is consider the total released heat to the surroundings (negative as it is exiting heat) and the aforementioned heat of combustion. Then, by using the ideal gas equation, we are able to compute the volume at 25 °C (298K) and 745 torr (0.98 atm) that must be measured:

$PV=nRT\\\\V=\frac{nRT}{P}=\frac{1122.6mol*0.082\frac{atm*L}{mol*K}*298K}{0.98atm}\\\\V=27992L=28.00m^3$