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Neporo4naja [7]
3 years ago
9

Based on the periodic table why are be, bg, ca and sr in the same column/group/family?

Chemistry
1 answer:
Lesechka [4]3 years ago
7 0

Answer:

The elements are in the same column/group IIA.

See the explanation below, please.

Explanation:

The elements Calcium, Strontium, Beryllium, Magnesium, Barium and Radio, belong to the group of alkaline earth metals located in group IIA of the periodic table, they require 2 electrons to complete their octet (they have 2 valence electrons). reagents than alkali metals.

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Answer: There is no question.

Explanation:

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Write a sentence that descrices how to determine the number of moles of a compound in known mass of a
sattari [20]

Explanation:

<u>Moles is denoted by given mass divided by the molecular mass ,  </u>

Hence ,  

n = w / m

n = moles ,  

w = given mass ,  

m = molecular mass .

For example ,

For  a compound X ,

The given mass i.e. w = 20 g

and the molecular mass ,i.e. , m = 10 g / mol

Then the moles can easily be calculated by using the above formula ,

n = w / m

n = 20 g / 10 g/mol = 2 mol

Hence , answer = 2 mol.

6 0
3 years ago
0.50 g of hydrogen chloride (HCl) is dissolved in water to make 4.0 L of solution. What is the pH of the resulting hydrochloric
svlad2 [7]

Explanation:

Given the mass of HCl is ---- 0.50 g

The volume of solution is --- 4.0 L

To determine the pH of the resulting solution, follow the below-shown procedure:

1. Calculate the number of moles of HCl given by using the formula:

number of moles of a substance=\frac{given mass of the substance}{its molecular mass}

2. Calculate the molarity of HCl.

3. Calculate pH of the solution using the formula:

pH=-log[H^+]

Since HCl is a strong acid, it undergoes complete ionization when dissolved in water.

HCl(aq)->H^+(aq)+Cl^-(aq)

Thus, [HCl]=[H^+]

Calculation:

1. Number of moles of HCl given:

number of moles of a substance=\frac{given mass of the substance}{its molecular mass}\\=0.50g/36.5g/mol\\=0.0137mol

2. Concentration of HCl:

Molarity of HCl=\frac{number of moles of HCl}{its molar mass}\\=\frac{0.0137 mol}{4.0 L} \\= 0.003425 M

3. pH of the solution:

pH=-log[H^+]\\=-log(0.003425)\\=2.47

Hence, pH of the given solution is 2.47.

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3 years ago
What reagent could be used to separate Br- from CH3 CO2 when added to an aqueous solution containing both.
Serjik [45]
C is the answer for this!
6 0
3 years ago
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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
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