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2 years ago

When 1.57 mol o2 reacts with h2 to form h2o, how many moles of h2 are consumed in the process?

Chemistry

2 answers:

Talja [164]2 years ago

7 0

The reaction showing the synthesis of water from its elements:

2H₂ + O₂ --> 2H₂O

One mole of oxygen combines with 2 moles of hydrogen to form 1 mole of H₂O

If 1 mole O₂ consumes 2 moles H₂

then 1.57 moles of O₂ consumes = 1.57 x 2 moles of H₂

= 3.14 moles of H₂

Therefore, 3.14 moles of H₂ are consumed in the process.

natita [175]2 years ago

3 0

Here we have to get the moles of hydrogen (H₂) consumed to form water (H₂O) from 1.57 moles of oxygen (O₂)

In this process 3.14 moles of H₂ will be consumed.

The balanced reaction between oxygen (O₂) and hydrogen (H₂); both of which are in gaseous state to form water, which is liquid in nature can be written as-

2H₂ (g) + O₂ (g) = 2H₂O (l).

Thus form the equation we can see that 1 mole of oxygen reacts with 2 moles of hydrogen to form 2 moles of water.

So, 1.57 moles of oxygen will consume (1.57×2) = 3.14 moles of hydrogen to form water.

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How many sigma and pi bonds in propionic bond

Ierofanga [76]

Propanoic acid formula is ch ch 2 so it has 8 bonds

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2 years ago

A balance measures mass to 0.001 g. If you determine the mass of an object that weighs about 30 g, would you record the mass as

solong [7]

Answer:

The mass is recorded as 32.075 g

Explanation:

"The first digit of uncertainty is taken as the last significant digit", this is the rule for significant figures in the analysis. The balance measures the mass up to three decimal places, so it makes the most sense to note the whole figure.

8 0

2 years ago

Read 2 more answers

A solution is prepared by dissolving 10.0 g of NaBr and 10.0 g of Na2SO4 in water to make a 100.0 mL solution. This solution is

Colt1911 [192]

Answer:

$M_{Na^+}=1.36M$

$M_{Br^-}=1.58M$

Explanation:

Hello,

At first, it turns out convenient to compute the total moles of sodium that will be dissolved into the solution by considering the added amounts of sodium bromide and sodium sulfate:

$n_{Na^+}=n_{Na^+,NaBr}+n_{Na^+,Na_2SO_4}\\n_{Na^+,NaBr}=10.0gNaBr*\frac{1molNaBr}{103gNaBr}*\frac{1molNa^+}{1molNaBr}=0.0971molNa^+\\n_{Na^+,Na_2SO_4}=10.0gNa_2SO_4*\frac{1molNa_2SO_4}{142gNa_2SO_4}*\frac{2molNa^+}{1molNa_2SO_4} =0.141molNa^+\\n_{Na^+}=0.0971molNa^++0.141molNa^+\\n_{Na^+}=0.238molNa^+$

Once we've got the moles we compute the final volume via:

$V=100.0mL+75.0mL=175.0mL*\frac{1L}{1000mL}=0.1750L$

Thus, the molarity of the sodium atoms turn out into:

$M_{Na^+}=\frac{0.238mol}{0.1750L} =1.36M$

Now, we perform the same procedure but now for the bromide ions:

$n_{Br^-}=n_{Br^-,NaBr}+n_{Br^-,AlBr_3}\\n_{Br^-,NaBr}=10.0gNaBr*\frac{1molNaBr}{103gNaBr}*\frac{1molBr^-}{1molNaBr}=0.0971molBr^-\\n_{Br^-,AlBr_3}=0.0750L*0.800\frac{molAlBr_3}{L} *\frac{3molBr^-}{1molAlBr_3}=0.180molBr^- \\n_{Br^-}=0.0971molBr^-+0.180molBr^-\\n_{Br^-}=0.277molBr^-$

Finally, its molarity results:

$M_{Br^-}=\frac{0.277molBr^-}{0.1750L}=1.58M$

Best regards.

7 0

1 year ago

Cacodyl, which has an intolerable garlicky odor and is used in the manufacture of cacodylic acid, a cotton herbicide, has a mass

Papessa [141]

Answer:

The molecular formula of cacodyl is C₄H₁₂As₂.

Explanation:

<u>Let's assume we have 1 mol of cacodyl</u>, in that case we'd have 209.96 g of cacodyl and the<u> following masses of its components</u>:

209.96 g * 22.88/100 = 48.04 g C

209.96 g * 5.76/100 = 12.09 g H

209.96 g * 71.36/100 = 149.83 g As

Now we convert those masses into moles:

48.04 g C ÷ 12 g/mol = 4.00 mol C

12.09 g H ÷ 1 g/mol = 12.09 mol H

149.83 g As ÷ 74.92 g/mol = 2.00 mol As

Those amounts of moles represent the amount of each component in 1 mol of cacodyl, thus, the molecular formula of cacodyl is C₄H₁₂As₂.

3 0

1 year ago

The final overall chemical equation is Upper Ca upper O (s) plus upper C upper O subscript 2 (g) right arrow upper C a upper C u

GenaCL600 [577]

Answer:

the enthalpy of the second intermediate equation is halved and has its sign changed.

Explanation:

Let us take a look at the first and second intermediate reactions as well as the overall reaction equation for the process under review;

First reaction;

Ca (s) + CO₂ (g) + ½O₂ (g) → CaCO₃ (s) ΔH₁ = -812.8 kJ

Second reaction;

2Ca (s) + O₂ (g) → 2CaO (s) ΔH₂ = -1269 kJ

Hence the overall equation is now;

CaO (s) + CO₂ (g) → CaCO₃ (s) ΔH = ?

According to the Hess law of constant heat summation, the enthalpy of the overall reaction is supposed to be obtained as a sum of the enthalpy of both reactions but this will not give the enthalpy of the overall reaction in this case. The enthalpy of the overall reaction is rather obtained by halving the enthalpy of the second intermediate reaction and reversing its sign before taking the sum as shown below;

Enthalpy of Intermediate reaction 1 + ½(- Enthalpy of Intermediate reaction 2) = Enthalpy of Overall reaction

7 0

1 year ago