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3 years ago

When very electronegative atoms , like oxygen , bond to atoms with lower electronegativity , like hydrogen , what's the result?

1 answer:

5 0

Electronegative atoms or atoms where electrons spend more of their time orbiting the atom's nucleus giving the atom a negative charge. When electronegative atoms such as oxygen bond to hydrogen they pull the hydrogen's electron to themselves becoming even more electronegative. This causes the hydrogen side to become positive. The molecule formed is a polar molecule. Molecules with positive ends and negative ends are called polar.

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Think about the clouds in the water cycle. Where does the water cycle begin?

Tanzania [10]

The water cycle has no starting point. But, we'll begin in the oceans, since that is where most of Earth's water exists

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3 years ago

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After severe dust storms in the 1930s, practices were put in place to reduce soil erosion. Which

timurjin [86]

Answer: Farmers were paid to practice soil-conserving techniques like crop rotation and terracing

Explanation:

read about it here: https://www.pbs.org/wgbh/americanexperience/features/dust-bowl-surviving-dust-bowl/

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3 years ago

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Where the oxygen comes from the air (21% O2 and 79% N2). If oxygen is fed from air in excess of the stoichiometric amount requir

guajiro [1.7K]

Answer:

$y_{O2} =4.3$ %

Explanation:

The ethanol combustion reaction is:

$C_{2}H_{5} OH+3O_{2}$ → $2CO_{2}+3H_{2}O$

If we had the amount (x moles) of ethanol, we would calculate the oxygen moles required:

$x*1.10(excess)*\frac{3 O_{2}moles }{etOHmole}$

Dividing the previous equation by x:

$1.10(excess)*\frac{3 O_{2}moles}{etOHmole}=3.30\frac{O_{2}moles}{etOHmole}$

We would need 3.30 oxygen moles per ethanol mole.

Then we apply the composition relation between O2 and N2 in the feed air:

$3.30(O_{2} moles)*\frac{0.79(N_{2} moles)}{0.21(O_{2} moles)}=121.414 (N_{2} moles )$

Then calculate the oxygen moles number leaving the reactor, considering that 0.85 ethanol moles react and the stoichiometry of the reaction:

$3.30(O_{2} moles)-0.85(etOHmoles)*\frac{3(O_{2} moles)}{1(etOHmoles)} =0.75O_{2} moles$

Calculate the number of moles of CO2 and water considering the same:

$0.85(etOHmoles)*\frac{3(H_{2}Omoles)}{1(etOHmoles)}=2.55(H_{2}Omoles)$

$0.85(etOHmoles)*\frac{2(CO_{2}moles)}{1(etOHmoles)}=1.7(CO_{2}moles)$

The total number of moles at the reactor output would be:

$N=1.7(CO2)+12.414(N2)+2.55(H2O)+0.75(O2)\\ N=17.414(Dry-air-moles)$

So, the oxygen mole fraction would be:

$y_{O_{2}}=\frac{0.75}{17.414}=0.0430=4.3$ %

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3 years ago

Significant figures 5.3316 + 6.87 + 37.48

Aleksandr-060686 [28]

Answer:

Explanation:

01: 5.3316+6.87+37.48

02: 12.2016+37.48

03: 49.6816

04: 49.68

Answer: 49.68 (Decimals: 2; Significant Figures: 4)

HOPE THIS HELPS..

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3 years ago

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What mass of lead (II) chloride is produced when 200.0 mL of a 0.250 M solution of sodium chloride is mixed with 200.0 mL of a 0

kow [346]

Answer:

Option d. 6.95 g

Explanation:

First of all, we state the reaction:

2NaCl + Pb(NO₃)₂ → PbCl₂ + 2NaNO₃

We determine the moles of each reactant, to state the limiting

Firstly we convert volume frm mL to L

0.200 L . 0.250M = 0.05 moles of NaCl

0.200L . 0.250M = 0.05 moles of Pb(NO₃)₂

Acording to stoichiometry we know that relation is 1:2, so the limiting reagent is the NaCl.

For 1 mol of Pb(NO₃)₂ I need 2 moles of NaCl

For 0.05 moles of Pb(NO₃)₂ I would need, the double → 0.1 moles

(We only have, 0.05 moles of NaCl)

Stoichiometry to the formed product is 2:1

From 2 moles of NaCl I produce 1 mol of PbCl₂

From 0.05 moles I would produce, the half → 0.025 moles

Let's convert the moles to mass → 0.025 mol . 278.1 g / 1mol = 6.95 g

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3 years ago