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sergey [27]
3 years ago
7

Which of the following are likely to form a covakent bond

Chemistry
1 answer:
Dmitriy789 [7]3 years ago
5 0

Answer:

For the most part, non-metals (excluding Nobel gases) are the most likely to form covalent bonds. Pure covalent bonds are formed between atoms with the same electronegativity, ie. they are trying to hold on to the electrons in the bond with the same strength.

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Put the following elements into five pairs of elements that have similar chemical reactivity: F, Sr, P, Ca, O, Br, Rb, Sb, Li, S
kirill115 [55]

Explanation:

Given elements:

  F, Sr, P, Ca, O, Br, Rb, Sb, Li, S

Elements with the same chemical reactivity will belong to the same group on the periodic table. This implies that elements in the same column will have the same reactivity;

  Li and Rb are both alkali metals in group 1

  Ca and Sr are both alkali earth metals in group 2

  F and Br are halogens in group 7

  O and S are group 6 elements

 P and Sb are both in group 5 on the periodic table

So these groupings show elements with the same chemical properties.

7 0
2 years ago
What is the molarity of a NaOH solution if 28.2 mL of a 0.355 M H2SO4 solution is required to neutralize a 25.0-mL sample of the
Verdich [7]

Answer:

[NaOH} = 0.4 M

Explanation:

In a reaction of neutralization, we determine the equivalence point of the titration. In this case, we have a strong base and a strong acid.

(H₂SO₄, is considered strong, but the first deprotonation is weak)

2NaOH  +  H₂SO₄  →  Na₂SO₄  + 2H₂O

As we have 2 protons in the acid, we need 2 OH⁻ from the base to form 2 molecules of water.

In the equivalence point we know mmoles of base = mmoles of acid

Let's finish the excersise with the formula

25 mL . M NaOH = 28.2 mL  .  0.355M

M NaOH = (28.2 mL  .  0.355M) / 25 mL → 0.400

8 0
3 years ago
A strontium hydroxide solution is prepared by dissolving 10.60 gg of Sr(OH)2Sr(OH)2 in water to make 47.00 mLmL of solution.What
NeTakaya

Answer:

Approximately 1.854\; \rm mol\cdot L^{-1}.

Explanation:

Note that both figures in the question come with four significant figures. Therefore, the answer should also be rounded to four significant figures. Intermediate results should have more significant figures than that.

<h3>Formula mass of strontium hydroxide</h3>

Look up the relative atomic mass of \rm Sr, \rm O, and \rm H on a modern periodic table. Keep at least four significant figures in each of these atomic mass data.

  • \rm Sr: 87.62.
  • \rm O: 15.999.
  • \rm H: 1.008.

Calculate the formula mass of \rm Sr(OH)_2:

M\left(\rm Sr(OH)_2\right) = 87.62 + 2\times (15.999 + 1.008) = 121.634\; \rm g \cdot mol^{-1}.

<h3>Number of moles of strontium hydroxide in the solution</h3>

M\left(\rm Sr(OH)_2\right) =121.634\; \rm g \cdot mol^{-1} means that each mole of \rm Sr(OH)_2 formula units have a mass of 121.634\; \rm g.

The question states that there are 10.60\; \rm g of \rm Sr(OH)_2 in this solution.

How many moles of \rm Sr(OH)_2 formula units would that be?

\begin{aligned}n\left(\rm Sr(OH)_2\right) &= \frac{m\left(\rm Sr(OH)_2\right)}{M\left(\rm Sr(OH)_2\right)}\\ &= \frac{10.60\; \rm g}{121.634\; \rm g \cdot mol^{-1}} \approx 8.71467\times 10^{-2}\; \rm mol\end{aligned}.

<h3>Molarity of this strontium hydroxide solution</h3>

There are 8.71467\times 10^{-2}\; \rm mol of \rm Sr(OH)_2 formula units in this 47\; \rm mL solution. Convert the unit of volume to liter:

V = 47\; \rm mL = 0.047\; \rm L.

The molarity of a solution measures its molar concentration. For this solution:

\begin{aligned}c\left(\rm Sr(OH)_2\right) &= \frac{n\left(\rm Sr(OH)_2\right)}{V}\\ &= \frac{8.71467\times 10^{-2}\; \rm mol}{0.047\; \rm L} \approx 1.854\; \rm mol \cdot L^{-1}\end{aligned}.

(Rounded to four significant figures.)

5 0
3 years ago
What willl happen if we nuke the sun some how
iogann1982 [59]

Answer:

The sun'll likely absorb the radiation if it is close enough (Which it will never be)

Explanation:

4 0
3 years ago
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You can be electrocuted if you try to use water to put out a class_ fire ?
olganol [36]
I think class C. fire
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