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julia-pushkina [17]
3 years ago
14

As you move from left to right across the periodic table elements ?

Chemistry
1 answer:
Sidana [21]3 years ago
6 0

Explanation:

Quite a number of properties varies across a period. Some remains constant whereas others decreases.

As one moves from left to right;

  • The energy level remains the same.
  • The ionization energy increases progressively as a result of increasing nuclear charge.
  • Electron affinity increases from left to right.
  • Electronegativity increases.
  • Electropositivity decreases.

learn more:

Periodic table brainly.com/question/2014634

#learnwithBrainly

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Answer the following question using the reaction: AB ---> A+B
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Answer:

So whats the question?

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Which of the following sentences describes oxygen at room temperature? (check all that apply)
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At room temperature, O2 is in gaseous state.

a gas has no definite volume or definite shape. It occupies volume of container and attains shape of container only.

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It has no definite volume and takes the shape of its container.

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It has more energy than it would at a cooler temperature: the kinetic energy of gas molecules increases with increase in temperature. Thus the energy increases with temperature and decreases with decrease in temperature.


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How many grams of sodium are in<br> 1.000 mole of sodium?
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Explanation:

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If 40.0 mL of a calcium nitrate solution reacts with excess potassium carbonate to yield 0.524 grams of a precipitate, what is t
tankabanditka [31]

Answer : The molarity of calcium ion on the original solution is, 0.131 M

Explanation :

The balanced chemical reaction is:

Ca(NO_3)_2+K_2CO_3\rightarrow CaCO_3+3KNO_3

When calcium nitrate react with potassium carbonate to give calcium carbonate as a precipitate and potassium nitrate.

First we have to calculate the moles of CaCO_3

\text{Moles of }CaCO_3=\frac{\text{Mass of }CaCO_3}{\text{Molar mass of }CaCO_3}

Given:

Mass of CaCO_3 = 0.524 g

Molar mass of CaCO_3 = 100 g/mol

\text{Moles of }CaCO_3=\frac{0.524}{100g/mol}=0.00524mol

Now we have to calculate the concentration of CaCO_3

\text{Concentration of }CaCO_3=\frac{\text{Moles of }CaCO_3}{\text{Volume of solution}}=\frac{0.00524mol}{0.040L}=0.131M

Now we have to calculate the concentration of calcium ion.

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CaCO_3\rightarrow Ca^{2+}+CO_3^{2-}

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Concentration of calcium ion = Concentration of CaCO_3 = 0.131 M

Thus, the concentration or molarity of calcium ion on the original solution is, 0.131 M

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