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2 years ago

Write the first and second ionization equations for H2SeO3 please.

1 answer:

3 0

As a diprotic acid, the H₂SeO₃ can ionize by step. First step is H₂SeO₃ =(reversible reaction) H⁺ + HSeO₃⁻. And second step is HSeO₃⁻ =(reversible reaction) H⁺ +SeO₃ ²⁻.

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It's 32°F (0°C) outside under normal atmospheric conditions (1 atm) at a stunt

Korolek [52]

The volume of the balloon is approximately 2652 liters.

<h3>How to determine the volume occupied by the gas in a balloon </h3>

Let suppose that <em>flammable</em> hydrogen behaves ideally. GIven the molar mass ( $M$ ), in kilograms per kilomole, and mass of the gas ( $m$ ), in kilograms. The volume occupied by the gas ( $V$ ), in cubic centimeters, is found by the equation of state for <em>ideal</em> gases:

$V = \frac{m\cdot R_{u}\cdot T}{P\cdot M}$ (1)

Where:

$R_{u}$ - Ideal gas constant, in kilopascal-cubic meters per kilomole-Kelvin.
$T$ - Temperature, in Kelvin
$P$ - Pressure, in kilopascals

If we know that $m = 0.239\,kg$ , $R_{u} = 8.314\,\frac{kPa\cdot m^{3}}{kmol\cdot K}$ , $T = 273.15\,K$ , $P = 101.325\,kPa$ and $M = 2.02\,\frac{kg}{kmol}$ , then the volume of the balloon is:

$V = \frac{(0.239\,kg)\cdot \left(8.314\,\frac{kPa\cdot m^{3}}{kmol\cdot K} \right)\cdot (273.15\,K)}{\left(101.325\,kPa\right)\cdot \left(2.02\,\frac{kg}{kmol} \right)}$

$V = 2.652\,m^{3}$ ( $2652\,L$ )

The volume of the balloon is approximately 2652 liters. $\blacksquare$

To learn more on ideal gases, we kindly invite to check this verified question: brainly.com/question/8711877

4 0

2 years ago

What is the molarity (M) of the following solutions?

Dennis_Churaev [7]

Answer:

The molarity (M) of the following solutions are :

A. M = 0.88 M

B. M = 0.76 M

Explanation:

A. Molarity (M) of 19.2 g of Al(OH)3 dissolved in water to make 280 mL of solution.

Molar mass of Al(OH)3 = Mass of Al + 3(mass of O + mass of H)

= 27 + 3(16 + 1)

= 27 + 3(17) = 27 + 51

= 78 g/mole

$Al(OH)_3$ = 78 g/mole

Given mass= 19.2 g/mole

$Mole = \frac{Given\ mass}{Molar\ mass}$

$Mole = \frac{19.2}{78}$

Moles = 0.246

$Molarity = \frac{Moles\ of\ solute}{Volume\ of\ solution(L)}$

Volume = 280 mL = 0.280 L

$Molarity = \frac{0.246}{0.280)}$

Molarity = 0.879 M

Molarity = 0.88 M

B .The molarity (M) of a 2.6 L solution made with 235.9 g of KBr

Molar mass of KBr = 119 g/mole

Given mass = 235.9 g

$Mole = \frac{235.9}{119}$

Moles = 1.98

Volume = 2.6 L

$Molarity = \frac{Moles\ of\ solute}{Volume\ of\ solution(L)}$

$Molarity = \frac{1.98}{2.6)}$

Molarity = 0.762 M

Molarity = 0.76 M

4 0

3 years ago

When 12 moles of methanol (ch3oh) and 24 moles of oxygen gas react according to the chemical equation below, what is the limitin

Nataliya [291]

The balanced reaction equation for the reaction between CH₃OH and O₂ is
2CH₃OH(l) + 3O₂(g) → 2CO₂(g) + 4H₂O(l)
Initial moles 12 24
Reacted moles 12 18
Final moles - 6 12 24

The stoichiometric ratio between CH₃OH and O₂ is 2 : 3
Hence,
reacted moles of O₂ = reacted moles of CH₃OH x (3/2)
= 12 mol x 3 / 2
= 18 mol

All of CH₃OH moles react with O₂.

Hence, the limiting agent is CH₃OH.

Excess reagent is O₂.
Amount of moles of excess reagent left = 24 - 18 mol = 6 mol

5 0

3 years ago

Elements have been discovered with up to ____ energy levels.

qaws [65]

im not sure but i think

8 0

3 years ago

How many liters of CO2 gas can be produced at 30.0 °C and 1.50 atm from the reaction of 5.00 mol of C3H8 and an excess of O2 acc

lbvjy [14]

Answer:

249 L

Explanation:

Step 1: Write the balanced equation

C₃H₈(g) + 5 O₂(g) → 3 CO₂(g) + 4 H₂O(g)

Step 2: Calculate the moles of CO₂ produced from 5.00 moles of C₃H₈

The molar ratio of C₃H₈ to CO₂ is 1:3. The moles of CO₂ produced are 3/1 × 5.00 mol = 15.0 mol

Step 3: Convert "30.0°C" to Kelvin

We will use the following expression.

K = °C + 273.15

K = 30.0°C + 273.15 = 303.2 K

Step 4: Calculate the volume of carbon dioxide

We will use the ideal gas equation.

P × V = n × R × T

V = n × R × T/P

V = 15.0 mol × 0.0821 atm.L/mol.K × 303.2 K/1.50 atm

V = 249 L

5 0

3 years ago