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Alisiya [41]
2 years ago
15

Cuántos electrones contiene el átomo de helio​

Chemistry
1 answer:
Deffense [45]2 years ago
4 0

Answer: 2:Dos

Explanation:

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Help pleased alot of points​
JulsSmile [24]

Answer:

Nitrogen non metal

Gain 3 electrons

Negative ion N3-

Barium metal

Lose electrons

Positive charge

Selinium non metal

Gain electron

Negative charge

Cesium metal

Lose electrons

Positive charge

Explanation:

4 0
2 years ago
The Earth's diameter at the equators is _____ its diameter at the poles. A:greater than B:equal to C:less than
Svetlanka [38]
I think the answer is greater than
7 0
2 years ago
A 2.500g sample of compound containing only Carbon and Hydrogen is found containing 2.002g of Carbon. Determine the empirical fo
Dima020 [189]

The empirical formula : CH₃

<h3>Further explanation</h3>

Given

2.5 g sample

2.002 g Carbon

Required

The empirical formula

Solution

Mass of Hydrogen :

= 2.5 - 2.002

= 0.498

Mol ratio C : H :

C : 2.002/12 = 0.167

H : 0.498/1 = 0.498

Divide by 0.167 :

C : H = 1 : 3

7 0
2 years ago
A 99.8 mL sample of a solution that is 12.0% KI by mass (d: 1.093 g/mL) is added to 96.7 mL of another solution that is 14.0% Pb
andre [41]

Answer:

m_{PbI_2}=18.2gPbI_2

Explanation:

Hello,

In this case, we write the reaction again:

Pb(NO_3)_2(aq) + 2 KI(aq)\rightarrow PbI_2(s) + 2 KNO_3(aq)

In such a way, the first thing we do is to compute the reacting moles of lead (II) nitrate and potassium iodide, by using the concentration, volumes, densities and molar masses, 331.2 g/mol and 166.0 g/mol respectively:

n_{Pb(NO_3)_2}=\frac{0.14gPb(NO_3)_2}{1g\ sln}*\frac{1molPb(NO_3)_2}{331.2gPb(NO_3)_2}  *\frac{1.134g\ sln}{1mL\ sln} *96.7mL\ sln\\\\n_{Pb(NO_3)_2}=0.04635molPb(NO_3)_2\\\\n_{KI}=\frac{0.12gKI}{1g\ sln}*\frac{1molKI}{166.0gKI}  *\frac{1.093g\ sln}{1mL\ sln} *99.8mL\ sln\\\\n_{KI}=0.07885molKI

Next, as lead (II) nitrate and potassium iodide are in a 1:2 molar ratio, 0.04635 mol of lead (II) nitrate will completely react with the following moles of potassium nitrate:

0.04635molPb(NO_3)_2*\frac{2molKI}{1molPb(NO_3)_2} =0.0927molKI

But we only have 0.07885 moles, for that reason KI is the limiting reactant, so we compute the yielded grams of lead (II) iodide, whose molar mass is 461.01 g/mol, by using their 2:1 molar ratio:

m_{PbI_2}=0.07885molKI*\frac{1molPbI_2}{2molKI} *\frac{461.01gPbI_2}{1molPbI_2} \\\\m_{PbI_2}=18.2gPbI_2

Best regards.

5 0
3 years ago
Read 2 more answers
why would it not be necessary to heat the air in the balloon as much if the air outside the balloon is colder?
RUDIKE [14]

Hot-air balloons float because the air caught inside the balloon is heated up by a burner, making it less dense than the air outside. As the burner heats the air, it expands and some of the air escapes; that's what makes it less dense

5 0
2 years ago
Read 2 more answers
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