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storchak [24]
3 years ago
12

Which statement(s) correctly compare the masses of protons, neutrons, and electrons? Select two options.

Chemistry
1 answer:
Ludmilka [50]3 years ago
5 0

Answer:

6 Electrons r smaller than a proton or a neutron

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A sample of strontium bicarbonate weighs 5.0020 mg. How many oxygen atoms are in the sample? I have the answer, but don't unders
hodyreva [135]

There are 14.4 * 10^18 oxygen atoms in 5.0020 mg of  strontium bicarbonate.

Mass of strontium bicarbonate Sr(HCO3)2 = 5.0020 mg

Molar mass of strontium bicarbonate Sr(HCO3)2 = 209.6537 g/mol

Number of moles of strontium bicarbonate Sr(HCO3)2 = 5.0020 * 10^-3g/209.6537 g/mol

= 2.3858 * 10^-5 moles

Given that we have 6 oxygen atoms per molecule of Sr(HCO3)2, the total number of oxygen atoms in Sr(HCO3)2 becomes;

2.3858 * 10^-5 moles * 6.02 * 10^23 = 14.4 * 10^18 oxygen atoms

Learn more: brainly.com/question/9743981

5 0
3 years ago
I need this asap Molar mass of C6H12O6
Zolol [24]

Answer:

180.156 g/mol

Explanation:

Hopes help

4 0
3 years ago
On the basis of dipole moments and/or hydrogen bonding, explain in a qualitative way the differences in the boiling points of ac
Inga [223]
Go on google bro you can get more help there and hope you get what your looking for and good luck

7 0
3 years ago
What are three Possible overlaps that can occur during bond formation​
mart [117]

Answer:

The Sigma (σ) Bond

S-S Overlapping. In this kind of overlapping, one 's' orbital from each participating atom undergoes head-on overlapping along the internuclear axis. ...

S-P Overlapping. ...

P-P overlapping.

4 0
2 years ago
The specific heat of copper is 0.093 cal/g0C. Calculate the temperature change that occurs if 28 g of copper at 25 0C absorbs 58
Umnica [9.8K]

Answer:

22.27 °C = ΔT

Explanation:

Specific heat capacity:

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Formula:

Q = m × c × ΔT

Given data:

mass = 28 g

heat absorbed = 58 cal

specific heat of copper =  0.093 cal/g .°C

temperature change =ΔT= ?

Solution:

Q = m × c × ΔT

58 cal = 28 g × 0.093 cal /g.°C × ΔT

58 cal = 2.604 cal.°C × ΔT

58 cal / 2.604 cal .°C = ΔT

22.27 °C = ΔT

5 0
3 years ago
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