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Lapatulllka [165]
3 years ago
13

Write a balanced chemical equation for the standard formation reaction of solid sodium orthosilicate

Chemistry
2 answers:
Mumz [18]3 years ago
8 0
Equation for the standard formation of solid sodium orthosilicate:
4 Na(s) + Si(s) + 2 O₂(g) → Na₄SiO₄(s)
Brums [2.3K]3 years ago
3 0

Answer : Si_{(s)} + 4NaOH_{(aq)}---> Na_{4}SiO_{4}_{(s)} + 2H_{2}_{(g)}

Explanation :

Silicon in presence of aqueous sodium hydroxide undergoes single displacement/replacement reaction and forms solid sodium orthosilicate which is a stable compound.

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What is the freezing point depression of a solution that contains 0.705?
Sveta_85 [38]
Colligative properties calculations are used for this type of problem. Calculations are as follows:<span>
</span>

<span>ΔT(freezing point)  = (Kf)m
ΔT(freezing point)  = 1.86 °C kg / mol (0.705)
ΔT(freezing point)  = 1.3113 °C

</span>

<span>
</span>

<span>Hope this answers the question. Have a nice day.</span>

6 0
3 years ago
MgS is the chemical formula for the ionic compound named
irina1246 [14]

Answer:

Magnesium Chloride

Explanation:

3 0
2 years ago
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Arturiano [62]
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6 0
2 years ago
You have 350 mL of 3.4 M hydrochloric acid (HCl). How many grams of HCl gas are dissolved? Bonus: what is the volume of the HCl
Crank

Answer:

1. 43.44g of HCl

2. 26.67 L of HCl

Explanation:

1) Molarity of a solution = number of moles (n) ÷ Volume (V)

According to the provided information in this question,

V = 350 mL = 350/1000 = 0.350L

Molarity = 3.4 M

Using Molarity = n/V

3.4 = n/0.350

n = 3.4 × 0.350

n = 1.19mol

Using the formula below to calculate the mass of HCl;

mole = mass/molar mass

Molar mass of HCl = 1 + 35.5 = 36.5g/mol

mole = mass/MM

mass = 1.19 mol × 36.5g/mol

mass = 43.44g of HCl

2) At STP, HCl has a pressure of 1atm, a temperature of 273K

V = ?

n = 1.19 mol

R = 0.0821 Latm/molK

Using PV = nRT

V = nRT/P

V = 1.19 × 0.0821 × 273/1

Volume = 26.67L

5 0
3 years ago
Neptunium-237 undergoes a series of α-particle and β-particle productions to end up as thallium-205. How many α particles and β
Fantom [35]
<h3>Answer:</h3>

8 alpha particles

4 beta particles

<h3>Explanation:</h3>

<u>We are given;</u>

  • Neptunium-237
  • Thallium-205
  • Neptunium-237 undergoes beta and alpha decay to form Thallium-205.

We are required to determine the number of beta and alpha particles produced to complete the decay series.

  • We need to know that when a radioisotope emits an alpha particle the mass number reduces by 4 while the atomic number decreases by 2.
  • When a beta particle is emitted the mass number of the radioisotope increases by 1 while the atomic number remains the same.

In this case;

Neptunium-237 has an atomic number 93, while,

Thallium-205 has an atomic number 81.

Therefore;

²³⁷₉₃Np → x⁴₂He + y⁰₋₁e + ²⁰⁵₈₁Ti

We can get x and y

237 = 4x + y(0) + 205

237-205 = 4x

4x = 32

 x = 8

On the other hand;

93 = 2x + (-y) + 81

but x = 8

93 = 16 -y + 81

y = 4

Therefore, the complete decay equation is;

²³⁷₉₃Np → 8⁴₂He + 4⁰₋₁e + ²⁰⁵₈₁Ti

Thus, Neptunium-237 emits 8 alpha particles and 4 beta particles to become Thallium-205.

5 0
3 years ago
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