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3 years ago

1. What kind of intermolecular forces act between a chlorine monofluoride molecule and a hydrogen bromide molecule

Chemistry

1 answer:

Arturiano [62]3 years ago

4 0

Answer:

Dipole- Dipole force.

Explanation:

Here both Chlorine monofloride( ClF) and hydrogen bromide( HBr) both are highly polar molecules. Polar molecules have dipole intermolecular forces. Polar molecules are adhering to their neighbors atom and tend to have higher boiling points, than non-polar molecules. One polar molecule's positive end is drawn by another polar molecule's negative end and vice versa.

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(worth 10pts!!!!!!) What did Dmitri Mendeleev contribute to science? A. He arranged the first periodic table. B. He developed th

vesna_86 [32]

Answer:

A. He arranged the first periodic table.

Explanation:

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3 years ago

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If 0.0714 moles of N2 gas occupies 1.25 L space, how many moles of N2 have a volume of 25.0 L? Assume temperature and pressure s

vfiekz [6]

Answer:

1.428 moles

Explanation:

If 0.0714 moles of N2 gas occupies 1.25 L space,

how many moles of N2 have a volume of 25.0 L?

Assume temperature and pressure stayed constant.

we experience it 0.0714 moles: 1.25L space

x moles : 25L of space

to get the x moles, cross multiply

(0.0714 x 25)/1.25

1.785/1.25 = 1.428 moles

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3 years ago

Which is a sign of a chemical change?

Igoryamba

The correct answer for the question it is option

>>>>D<<<<

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3 years ago

3. What should you do after an earthquake when you are in a building?

balu736 [363]

Answer:

3. You should walk fast in moving out.

4. You should stay in the evacuation center. Avoid going back to your car, aftershocks may occur.

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2 years ago

Read 2 more answers

Calculate the mass of MgCO3 (84.31 g/mol) precipitated by mixing 10.0 mL of a 0.300 M Na2CO3 solution with 6.00 mL of 0.0400 M M

I am Lyosha [343]

Answer:

$m_{MgCO_3}=0.0202molMgCO_3$

Explanation:

Hello,

In this case, for this purpose we first have to write the undergoing chemical reaction:

$Na_2CO_3+Mg(NO_3)_2\rightarrow MgCO_3+2NaNO_3$

Thus, since the mole ratio between the reactants is 1:1, we next identify the limiting reactant by computing the available moles of sodium carbonate and those moles of the same reactant consumed by the magnesium nitrate considering the given solutions:

$n_{Na_2CO_3}=0.010L*0.300\frac{molNa_2CO_3}{1L}=0.003molNa_2CO_3 \\\\n_{Na_2CO_3}^{consumed}=0.006L*0.0400\frac{molMg(NO_3)_2}{1L}*\frac{1molNa_2CO_3}{1molMg(NO_3)_2} =0.00024molNa_2CO_3$

In such a way, since less moles are consumed, we can say that the sodium carbonate is excess whereas the magnesium nitrate is the limiting one, therefore, the yielded mass of magnesium carbonate turns out:

$m_{MgCO_3}=0.00024molMg(NO_3)_2*\frac{1molMgCO_3}{1molMg(NO_3)_2}*\frac{84.31gMgCO_3}{1molMgCO_3} \\\\m_{MgCO_3}=0.0202molMgCO_3$

Regards.

7 0

3 years ago