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aalyn [17]
3 years ago
8

How many chloride ions are in 2.6 moles of CaCl2?

Chemistry
1 answer:
Rama09 [41]3 years ago
8 0

Answer:

31.31× 10²³ number of Cl⁻ are present in 2.6 moles of CaCl₂ .

Explanation:

Given data:

Number of moles of CaCl₂ = 2.6 mol

Number of Cl₂ ions = ?

Solution:

CaCl₂  → Ca²⁺ + 2Cl⁻

The given problem will solve by using Avogadro number.

It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.

The number 6.022 × 10²³ is called Avogadro number.

In one mole of CaCl₂ there are two moles of chloride ions present.

In 2.6 mol:

2.6×2 = 5.2 moles

1 mole Cl⁻ =   6.022 × 10²³ number of Cl⁻ ions

5.2 mol ×  6.022 × 10²³ number of Cl⁻  / 1mol

31.31× 10²³ number of Cl⁻

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Calculate the percent by mass of 4.35g of Na I dissolved in 105g of water​
Ludmilka [50]

<u>We are given:</u>

Mass of Na added = 4.35 grams

Mass of water = 105 grams

<u>Mass Percent of Na:</u>

Total mass of the solution = mass of solute + mass of solvent

Total mass of the solution = 4.35 + 105 = 109.35 grams

Mass percent of solute = (mass of solute / mass of solution) * 100

Mass percent of Solute = (4.35 / 109.35) * 100

Mass percent = 3.978 %

4 0
2 years ago
A 200.0 mL solution of 0.40 M ammonium chloride was titrated with 0.80 M sodium hydroxide. What was the pH of the solution after
choli [55]

Answer:

9.25

Explanation:

Let first find the moles of NH_4Cl and NaOH

number of moles of NH_4Cl = 0.40  mol/L × 200 ×  10⁻³L

= 0.08 mole

number of moles of NaOH = 0.80  mol/L × 50 ×  10⁻³L

= 0.04 mole

The equation for the reaction is expressed as:

NH^+_{4(aq)} \ + OH^-_{(aq)} ------> NH_{3(g)} \ + H_2O_{(l)}

The ICE Table is shown below as follows:

                            NH^+_{4(aq)} \ + OH^-_{(aq)} ------> NH_{3(g)} \ + H_2O_{(l)}

Initial (M)              0.08            0.04                            0

Change (M)         - 0.04          -0.04                          + 0.04

Equilibrium (M)      0.04             0                                0.04

K_a*K_b = 10^{-14} \ at \ 25^0C

K_a = \frac{10^{-14}}{K_b}

K_a = \frac{10^{-14}}{1.76*10^{-5}}

K_a= 5.68*10^{10}

pK_a = - log \ (K_a)

pK_a = - log \ (5.68*10^{-10})

pK_a = 9.25

pH = pKa + \ log (\frac{HB}{HA} )   for buffer solutions

pH = pKa + \ log (\frac{moles \ of \ base }{ moles\ of \ acid} ) since they are in the same solution

pH = 9.25 + \ log (\frac{0.04 }{ 0.04} )

pH = 9.25

8 0
3 years ago
1. Which of the following nuclear equations is correctly balanced? (choices attached)
lawyer [7]

1. <em>Balancing nuclear equations </em>

Answer:

_{18}^{37}\text{Ar} + _{-1}^{0}\text{e} \rightarrow _{17}^{37}\text{Cl}

Explanation:

The main point to remember in balancing nuclear equations is that the <em>sums of the superscripts</em> (the mass numbers) and the <em>subscripts</em> (the nuclear charges) <em>must balance</em>.

Mass numbers: 37 + 0 = 37; balanced.

Charges: 18 + 1 = 17; balanced

B is <em>wrong</em>. Mass numbers not balanced. 6 +2(1) ≠ 4 + 3.

C is <em>wron</em>g. Mass numbers not balanced. 254 + 4 ≠ 258 + 2(1).

D is <em>wron</em>g. Mass numbers not balanced. 14 + 4 ≠ 17 + 2.

===============

2. <em>Amount remaining </em>

Answer:

D. 5.25 g

Explanation:

The half-life of Th-234 (24 da) is the time it takes for half the Th to decay.  

After one half-life, half (50 %) of the original amount will remain.  

After a second half-life, half of that amount (25 %) will remain, and so on.

We can construct a table as follows:  

  No. of                  Fraction       Amount  

<u>half-lives</u>   <u>t/(da</u>)  <u>remaining</u>  <u>remaining/g</u>  

      1             24            ½                21.0  

      2            48            ¼                10.5  

      3             72           ⅛                 5.25  

      4             96          ¹/₁₆                 2.62  

We see that 72 da is three half-lives, and the amount of Th-234 remaining is 5.25 g.  

===============

3.<em> Calculating the half-life </em>

Answer:

a. 2.6 min

Explanation:

The fraction of the original mass remaining is 1.0 g/4.0 g ≈ ¼.

We saw from the previous table that it takes two half-lives to decay to ¼ of the original amount.

2 half-lives = 5.2 min       Divide both sides by 2

  1 half-life = 5.2 min/2 = 2.6 min

7 0
3 years ago
The rate of heat transfer by radiation is best described by which statement?
bogdanovich [222]

requires a medium to propagate

Explanation:

due to radiation z a dispersion of light ray into different multiple

7 0
3 years ago
How do I do this? What are the answers to the 5 questions shown?
frozen [14]

Answer:

1. C₃H₆O₃

2. C₆H₁₂

3. C₆H₂₄O₆

4. C₆H₆

5. N₂O₄

Explanation:

1. Determination of the molecular formula.

Empirical formula => CH₂O

Mass of compound = 90 g

Molecular formula =?

Molecular formula = n × Empirical formula = mass of compound

[CH₂O]ₙ = 90

[12 + (2×1) + 16]n = 90

[12 + 2 + 16]n = 90

30n = 90

Divide both side by 30

n = 90/30

n = 3

Molecular formula = [CH₂O]ₙ

Molecular formula = [CH₂O]₃

Molecular formula = C₃H₆O₃

2. Determination of the molecular formula.

Empirical formula => CH₂

Mass of compound = 84 g

Molecular formula =?

Molecular formula = n × Empirical formula = mass of compound

[CH₂]ₙ = 84

[12 + (2×1)]n = 84

[12 + 2]n = 84

14n = 84

Divide both side by 14

n = 84/14

n = 6

Molecular formula = [CH₂]ₙ

Molecular formula = [CH₂]₆

Molecular formula = C₆H₁₂

3. Determination of the molecular formula.

Empirical formula => CH₄O

Mass of compound = 192 g

Molecular formula =?

Molecular formula = n × Empirical formula = mass of compound

[CH₄O]ₙ = 192

[12 + (4×1) + 16]n = 192

[12 + 4 + 16]n = 192

32n = 192

Divide both side by 32

n = 192/32

n = 6

Molecular formula = [CH₄O]ₙ

Molecular formula = [CH₄O]₆

Molecular formula = C₆H₂₄O₆

4. Determination of the molecular formula.

Empirical formula => CH

Mass of compound = 78 g

Molecular formula =?

Molecular formula = n × Empirical formula = mass of compound

[CH]ₙ = 78

[12 + 1]n = 78

13n = 78

Divide both side by 13

n = 78/13

n = 6

Molecular formula = [CH]ₙ

Molecular formula = [CH]₆

Molecular formula = C₆H₆

5. Determination of the molecular formula.

Empirical formula => NO₂

Mass of compound = 92 g

Molecular formula =?

Molecular formula = n × Empirical formula = mass of compound

[NO₂]ₙ = 92

[14 + (2×16)]n = 92

[14 + 32]n = 92

46n = 92

Divide both side by 46

n = 92/46

n = 2

Molecular formula = [NO₂]ₙ

Molecular formula = [NO₂]₂

Molecular formula = N₂O₄

6 0
3 years ago
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