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3 years ago

8. H2 + O2-> __H2O

Chemistry

1 answer:

Rus_ich [418]3 years ago

8 0

Anser:

Explanation:

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What mass of NH3 can be made from 35.0g of N2?

egoroff_w [7]

Formation of ammonia by nitrogen and hydrogen is habers process wher 28g N2 results in formation of 34g NH3
so 35g N2 will form 34*35/28=42.5g NH3 where it given that reaction takes place in excess of H2
N2+3H2 gives 2NH3

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3 years ago

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What is a non-example for an Electric circuit?

Mademuasel [1]

Answer:

Impressionist paintings are a non-example of electric circuits.

Explanation:

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How many milliliters of 0.150 M NaOH are required to neutralize 85.0 mL of 0.300 M H2SO4 ? The balanced neutralization reaction

nekit [7.7K]

Answer : The volume of $NaOH$ required to neutralize is, 340 mL

Explanation :

To calculate the volume of base (NaOH), we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is $H_2SO_4$

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is NaOH.

We are given:

$n_1=2\\M_1=0.300M\\V_1=85.0mL\\n_2=1\\M_2=0.150M\\V_2=?$

Putting values in above equation, we get:

$2\times 0.300M\times 85.0mL=1\times 0.150M\times V_2\\\\V_2=340mL$

Hence, the volume of $NaOH$ required to neutralize is, 340 mL

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4 years ago

How many moles of HNO3 are present in 450 g of HNO3?

iVinArrow [24]

Since there are 63.01284 grams to one mole of HNO3, then 12.5 moles would be in 450 grams of it.

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3 years ago

When methanol, CH 3 OH , is burned in the presence of oxygen gas, O 2 , a large amount of heat energy is released. For this reas

luda_lava [24]

<u>Answer:</u> The mass of methanol that must be burned is 24.34 grams

<u>Explanation:</u>

We are given:

Amount of heat produced = 581 kJ

For the given chemical equation:

$CH_3OH(g)+\frac{3}{2}O_2(g)\rightarrow CO_2(g)+2H_2O(l);\Delta H=-764kJ$

By Stoichiometry of the reaction:

When 764 kJ of heat is produced, the amount of methanol reacted is 1 mole

So, when 581 kJ of heat will be produced, the amount of methanol reacted will be = $\frac{1}{764}\times 581=0.7605mol$

To calculate mass for given number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Moles of methanol = 0.7605 moles

Molar mass of methanol = 32 g/mol

Putting values in above equation, we get:

$0.7605mol=\frac{\text{Mass of methanol}}{32g/mol}\\\\\text{Mass of methanol}=(0.7605mol\times 32g/mol)=24.34g$

Hence, the mass of methanol that must be burned is 24.34 grams

4 0

3 years ago

8. __H2 + __O2-> __H2O

8. H2 + O2-> __H2O