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Lerok [7]
3 years ago
6

Which elements have atoms that are likely to form an anion?

Chemistry
1 answer:
KengaRu [80]3 years ago
5 0
Answer:

Phosphorus, chlorine and selenium.

Explanation:

If you look at a periodic table, you will notice that phosphorus, chlorine and selenium are all in groups 15-17. Elements in these groups are more likely to gain electrons then they are to lose them, thus making them more likely to form an anion.
This is because they want to fill their outer valence shell of electrons so that they can have a full octet and for groups 15-17 it’s easier to gain 1-3 electrons then it is to loose 5-7 electrons.
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A 100.0 g sample of sugar contains carbon, hydrogen, and oxygen. The sample contains 40.0 g of carbon and 53.3 g of oxygen. What
Lynna [10]

Answer:

I think this is it.

Explanation:

As all the mass of other molecules are given and we have to find the mass of hydrogen so total subtracted by carbon and oxygen.

6 0
3 years ago
8. Bring the balloon in contact with the wall. What happens to the charges in the wall?
Ne4ueva [31]

When the charged balloon is brought near the wall, it repels some of the negatively charged electrons in that part of the wall. Therefore, that part of the wall is left repelled.

<u>Explanation</u>:  

  • Balloons don't stick to walls. However, if you rub the balloon on an appropriate piece of material such as clothing or a wall, electrons are pulled from the other material to the balloon.
  • The balloon now as more electrons than normal and therefore has an overall negative charge. Two balloons like this will repel each other.
  • The other material now has an overall positive charge. Because opposite charges attract, the balloon will now appear to stick to the other material. If you didn't rub the balloon first, it's charge would be neutral and it wouldn't stick to the wall.
7 0
3 years ago
(HELP ME ASAP!)
slega [8]
A) Ca(OH)2 + CO2 —> CaCO3 + H2O

B) when Ca(OH)2 is reacted with CO2, the CaCO3 produced is a precipitate which turns the solution milky
8 0
3 years ago
You are given 10ml (M) 20 Naoh solution in a conical flask and asked to titrate with (M) 20 Hcl and (M) 20 H2so4 separately. cal
Solnce55 [7]

Answer:

n_{HCl}=0.2molHCl\\n_{H_2SO_4}=0.1molH_2SO_4

Explanation:

Hello!

In this case, since the reactions between NaOH and the acids are:

NaOH+HCl\rightarrow NaCl+H_2O\\\\2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O

Whereas we can see the 1:1 and 2:1 mole ratios between NaOH and HCl and H2SO4 respectively. In such a way, at the equivalence point we realize that:

n_{HCl}=n_{NaOH}=V_{NaOH}M_{NaOH}=0.01L*20mol/L=0.2molHCl\\\\2n_{H_2SO_4}=n_{NaOH}\\\\n_{H_2SO_4}=\frac{1}{2} V_{NaOH}M_{NaOH}=\frac{0.01L*20mol/L}{2} =0.1molH_2SO_4

Best regards!

8 0
2 years ago
Perform the calculations and determine the absolute and percent relative uncertainty. Express each answer with the correct numbe
Lana71 [14]

Answer:

Explanation:

Given the equation:

\implies \dfrac{[9.8(\pm0.3)-2.31(\pm 0.01)]}{8.5(\pm0.6)}

The absolute uncertainty in a measurement is the term used to describe the degree of inaccuracy.

The first step is to determine the algebraic value on the numerator.

Algebraic value = 9.8 - 231

= 7.49

The absolute uncertainty = \sqrt{(abs. uncertainty_{v_1})^2+(abs. uncertainty_{v_2})^2}

absolute uncertainty = \sqrt{(0.3)^3 + (0.01)^2}

= \sqrt{0.09 + 0.0001}

= 0.300167

∴

[9.8(±0.3) - 2.31(±0.01)] = 7.49(±0.300167)

The division process now is:

\implies \dfrac{[9.8(\pm0.3)-2.31(\pm 0.01)]}{8.5(\pm0.6)}= \dfrac{7.49 (\pm 0.300167)}{8.5 (\pm0.6)}

Relative uncertainty = \dfrac{(\pm 0.300167)}{7.49}\times 100  \ , \  \dfrac{(\pm 0.6) }{8.5} \times 100

Relative uncertainty = ±4.007565% ,  ±7.058824%

\text{Relative uncertainty} = \sqrt{(4.007565)^2+(7.058824)^2}

\text{Relative uncertainty} = \sqrt{16.06057723+49.82699626}

\text{Relative uncertainty} = \sqrt{65.88757349}

\text{Relative uncertainty} = 8.117116

≅ 8%

The algebraic value = \dfrac{7.49}{8.5}

= 0.881176

≅ 0.88

The percentage of the relative uncertainty =\dfrac{\text{Absolute uncertainty }}{\text{calculated value} }\times 100

By cross multiplying:

\text{Absolute uncertainty} (\%) = \dfrac{\text{relative uncertainty} \times \text{calculated value}}{100}

\text{Absolute uncertainty} (\%) = \dfrac{8.117116\times 0.881176}{100}

\text{Absolute uncertainty} (\%) = 0.0715260

\mathbf{\text{Absolute uncertainty} (\%) \simeq 0.07}

Finally:

\mathbf{\implies \dfrac{[9.8(\pm0.3)-2.31(\pm 0.01)]}{8.5(\pm0.6)}= 0.88 \pm (0.07) \pm 8\%}

8 0
3 years ago
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