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Alik [6]
3 years ago
5

How many neutrons are in a Cs-130 isotope?

Chemistry
1 answer:
den301095 [7]3 years ago
5 0
It has 78 neutrons in a cesium
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Identify each substance as an acid or a base and write a chemical equation showing how it is an acid or a base according to the
faust18 [17]

Answer:

(a) Acid

(b) Base

(c) Acid

(d) Base

Explanation:

According to the Arrhenius acid-base theory:

  • An acid is a substance that releases H⁺ in aqueous solution.
  • A base is a substance that releases OH⁻ in aqueous solution.

(a) H₂SO₄ is an acid according to the following equation:

H₂SO₄(aq) ⇒ 2 H⁺(aq) + SO₄²⁻(aq)

(b) Sr(OH)₂ is a base according to the following equation:

Sr(OH)₂(aq) ⇄ Sr²⁺(aq) + 2 OH⁻(aq)

(c) HBr is an acid according to the following equation:

HBr(aq) ⇒ H⁺(aq) + Br⁻(aq)

(d) NaOH is a base according to the following equation:

NaOH(aq) ⇒ Na⁺(aq) + OH⁻(aq)

6 0
3 years ago
Each of the following values was read on an instrument of measuring device. In each case the last digit was estimated. Tell what
Drupady [299]

Answer:

<h3>160 cm</h3>

Explanation:

6 0
2 years ago
Determine the empirical and molecular formula for chrysotile asbestos. Chrysotile has the following percent composition: 28.03%
nlexa [21]

<u>Answer:</u> The empirical and molecular formula of chrysotile is Mg_3Si_2H_3O_4 and Mg_6Si_4H_6O_{16}

<u>Explanation:</u>

We are given:

Percentage of Mg = 28.03 %

Percentage of Si = 21.60 %

Percentage of H = 1.16 %

Percentage of O = 49.21 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of Mg = 28.03 g

Mass of Si = 21.60 g

Mass of H = 1.16 g

Mass of O = 49.21 g

To formulate the empirical formula, we need to follow some steps:

  • <u>Step 1:</u> Converting the given masses into moles.

Moles of Magnesium = \frac{\text{Given mass of Magnesium}}{\text{Molar mass of Magnesium}}=\frac{28.03g}{24g/mole}=1.17moles

Moles of Silicon = \frac{\text{Given mass of Silicon}}{\text{Molar mass of Silicon}}=\frac{21.06g}{28g/mole}=0.752moles

Moles of Hydrogen = \frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{1.16g}{1g/mole}=1.16moles

Moles of Oxygen = \frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{49.21g}{16g/mole}=3.07moles

  • <u>Step 2:</u> Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.752 moles.

For Magnesium = \frac{1.17}{0.752}=1.5

For Silicon = \frac{0.752}{0.752}=1

For Hydrogen = \frac{1.16}{0.752}=1.5

For Oxygen = \frac{3.07}{0.485}=4.08\approx 4

To convert the mole ratios into whole numbers, we multiply individual mole ratios by 2

Mole ratio of Magnesium = (2 × 1.5) = 3

Mole ratio of Silicon = (2 × 1) = 2

Mole ratio of Hydrogen = (2 × 1.5) = 3

Mole ratio of Oxygen = (2 × 4) = 8

  • <u>Step 3:</u> Taking the mole ratio as their subscripts.

The ratio of Mg : Si : H : O = 3 : 2 : 3 : 8

The empirical formula for the given compound is Mg_3Si_2H_3O_8

For determining the molecular formula, we need to determine the valency which is multiplied by each element to get the molecular formula.

The equation used to calculate the valency is :

n=\frac{\text{Molecular mass}}{\text{Empirical mass}}

We are given:

Mass of molecular formula = 520.8 g/mol

Mass of empirical formula = [(24 × 3) + (28 × 2) + (1 × 3) + (16 × 8)] = 259 g/mol

Putting values in above equation, we get:

n=\frac{520.8g/mol}{259g/mol}=2

Multiplying this valency by the subscript of every element of empirical formula, we get:

Mg_{(3\times 2)}Si_{(2\times 2)}H_{(3\times 2)}O_{(8\times 2)}=Mg_6Si_4H_6O_{16}

Hence, the empirical and molecular formula of chrysotile is Mg_3Si_2H_3O_4 and Mg_6Si_4H_6O_{16}

5 0
3 years ago
What is the pH for a solution that has an OH ion
Basile [38]
Assuming that you mean 10^-4 M then this would be basic and would have a pH of 10.

pOH = -log[OH].
So pOH = 4
pH=14-pOH
pH = 10
3 0
2 years ago
How many moles of na2co3 are necessary to reach stoichiometric quantities with cacl2
lbvjy [14]

0.0102 moles Na₂CO₃ = 1.08g of Na₂CO₃ is necessary  to reach stoichiometric quantities with cacl2.

<h3>Explanation:</h3>

Based on the reaction

CaCl₂ + Na₂CO₃ → 2NaCl + CaCO₃

1 mole of CaCl₂ reacts per mole of Na₂CO₃

we have to calculate how many moles of CaCl2•2H2O are present in 1.50 g

  • We must calculate the moles of CaCl2•2H2O using its molar mass (147.0146g/mol) in order to answer this issue.
  • These moles, which are equal to moles of CaCl2 and moles of Na2CO3, are required to obtain stoichiometric amounts.
  • Then, we must use the molar mass of Na2CO3 (105.99g/mol) to determine the mass:

<h3>Moles CaCl₂.2H₂O:</h3>

1.50g * (1mol / 147.0146g) = 0.0102 moles CaCl₂.2H₂O = 0.0102moles CaCl₂

Moles Na₂CO₃:

0.0102 moles Na₂CO₃

Mass Na₂CO₃:

0.0102 moles * (105.99g / mol) = 1.08g of Na₂CO₃ are present

Therefore, we can conclude that 0.0102 moles Na₂CO₃  is necessary.to reach stoichiometric quantities with cacl2.

To learn more about stoichiometric quantities visit:

<h3>brainly.com/question/28174111</h3>

#SPJ4

7 0
2 years ago
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