All categories

3 years ago

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⁻

You might be interested in

Valproic acid, used to treat seizures and bipolar disorder, is composed of C, H, and O. A 0.165-g sample is combusted to produce

Andru [333]

Answer:

The empirical formula is C4H8O

Explanation:

Step 1: Data given

Valproic acid is composed of C, H, and O

Mass of the sample = 0.165 grams

Mass of water = 0.166 grams

Mass of CO2 = 0.403 grams

Molar mass of water ( H2O) = 18.02 g/mol

Molar mass of CO2 = 44.01 g/mol

Atomic mass of C = 12.01 g/mol

Atomic mass of O = 16.0 g/mol

Atomic mass H = 1.01 g/mol (H2 = 2.02 g/mol)

Step 2: The equation

CxHyOz + O2 → CO2 + H2O

Step 3: Calculate the number of carbon in the sample

The carbon comes from CO2

Mass C = (12.01 g/mol/44.01 g/mol) * 0.403 grams

Mass C = 0.110 grams

Step 4: Calculate mass of hydrogen in the sample

The hydrogen comes from H2O

Mass H = (2.02/18.02) * 0.166 grams

MAss H = 0.0186 grams

Step 5: Calculate mass of O

The mass of O in the sample = Mass of sample - mass of H - mass of C

The mass of O = 0.165 grams - 0.110 grams - 0.0186 grams

The mass of O = 0.0364 grams

Step 6: Calculate moles

Moles C = 0.110 grams / 12.01 g/mol = 0.00916 moles

Moles H = 0.0186 / 1.01 = 0.0184 moles

Moles O = 0.0364/16.0 = 0.00228 moles

Step 7: Calculate empirical formula

We divide by the smallest amount of moles

C: 0.00916/ 0.00228 = 4

H: 0.0184/0.00228 = 8

O: 0.00228/0.00228 = 1

The empirical formula is C4H8O

7 0

3 years ago

Consider a 125 mL buffer solution at 25°C that contains 0.500 mol of hypochlorous acid (HOCl) and 0.500 mol of sodium hypochlori

Helga [31]

Answer:

pH = 6.82

Explanation:

To solve this problem we can use the<em> Henderson-Hasselbach equation</em>:

pH = pKa + log $\frac{[NaOCl]}{[HOCl]}$

We're given all the required data to <u>calculate the original pH of the buffer before 0.341 mol of HCl are added</u>:

pKa = -log(Ka) = -log(2.9x10⁻⁸) = 7.54

[HOCl] = [NaOCl] = 0.500 mol / 0.125 L = 4 M

pH = 7.54 + log $\frac{4}{4}$

pH = 7.54

By adding HCl, w<em>e simultaneously </em><u><em>increase the number of HOCl</em></u><em> and </em><u><em>decrease NaOCl</em></u>:

pH = 7.54 + log $\frac{[NaOCl-HCl]}{[HOCl+HCl]}$

pH = 7.54 + log $\frac{(0.500mol-0.341mol)/0.125L}{(0.500mol+0.341mol)/0.125L}$

pH = 6.82

6 0

3 years ago

Find the % composition for each element in Zinc Chlorate

Archy [21]

Answer:

chlorine ~ 30%

zinc ~ 28%

oxygen ~ 41%

Explanation:

3 0

3 years ago

Si tenemos una solución de ácido láctico 0.025M con un pH = 2.75. ¿Cuál es la constante de equilibrio Ka?

rodikova [14]

Answer:

La constante de equilibrio Ka del ácido láctico es 1.38x10⁻⁴.

Explanation:

El ácido láctico es un ácido débil cuya reacción de disociación es la siguiente:

CH₃CHOHCOOH + H₂O ⇄ CH₃CHOHCOO⁻ + H₃O⁺ (1)

0.025M - x x x

La constante de acidez del ácido es:

$Ka = \frac{[CH_{3}CHOHCOOH^{-}][H_{3}O^{+}]}{[CH_{3}CHOHCOOH]}$

Sabemos que la concentración del ácido inicial es:

[CH₃CHOHCOOH] = 0.025 M

Y que a partir del pH podemos hallar [H₃O⁺]:

$pH = -log[H_{3}O^{+}]$

$[H_{3}O^{+}] = 10^{-pH} = 10^{-2.75} = 1.78 \cdo 10^{-3} M$

Debido a que el ácido se disocia en agua para producir los iones CH₃CHOHCOO⁻ y H₃O⁺ de igual manera (según la reacción (1)), tenemos:

[CH₃CHOHCOO⁻] = [H₃O⁺] = 1.78x10⁻³ M

Y por esa misma disociación, la concentración del ácido en el equilibrio es:

$[CH_{3}CHOHCOOH^{-}] = 0.025 M - 1.78 \cdo 10^{-3} M = 0.023 M$

Entonces, la constante de equilibrio Ka del ácido láctico es:

$Ka = \frac{[CH_{3}CHOHCOOH^{-}][H_{3}O^{+}]}{[CH_{3}CHOHCOOH]} = \frac{(1.78 \cdo 10^{-3})^{2}}{0.023} = 1.38 \cdot 10^{-4}$

Espero que te sea de utilidad!

3 0

3 years ago

Cacl2+na3po4 ca3po42 + nacl

Alex787 [66]

<h3>Answer:</h3>

3CaCl₂ + 2Na₃PO₄→ Ca₃(PO₄)₂ + 6NaCl

<h3>Explanation:</h3>

We are given the Equation;

CaCl₂ + Na₃PO₄→ Ca₃(PO₄)₂ + NaCl

Assuming the question requires us to balance the equation;

A balanced chemical equation is one that has equal number of atoms of each element on both sides of the equation.
Balancing chemical equations ensures that they obey the law of conservation of mass in chemical equations.
According to the law of conservation of mass in chemical equation, the mass of the reactants should always be equal to the mass of the products.
Balancing chemical equations involves putting appropriate coefficients on the reactants and products.

In this case;

To balance the equation we are going to put the coefficients 3, 2, 1, and 6.

Therefore; the balanced equation will be;

3CaCl₂ + 2Na₃PO₄→ Ca₃(PO₄)₂ + 6NaCl

5 0

3 years ago