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2 years ago

10

How do I balance this equation? ?KNO3 → ?K2O + ?N2(g)+ ?O2(g) and what type of reaction is it?

1 answer:

nirvana33 [79]2 years ago

3 0

Answer:

4KNO3 ==> 2K2O + 2N2 + 5O2

Explanation:

It's a decomposition, but not a simple one.

KNO3 ==> K2O + N2 + O2 I don't usually do this, but I think the easiest way to proceed is to balancing the K and N together. That will require a 2 in front of KNO3

4KNO3 ==> 2K2O + 2N2 + 5O2

Now you have (3*4) = 12 oxygens. Two are on the K2O. So the other 10 must be on the O2

That should do it.

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What is the name of the product isotope formed when Thorium-234 decays by beta decay?

defon

Answer:

protactinium-234,

Explanation:

This is what happens when thorium-234 releases a W- boson, which then decays to an electron and an electron antineutrino.

7 0

2 years ago

How much positive charge is in 0.7 kg of lithium? with each atom having 3 protons and 3 electrons. The elemental charge is 1.602

hichkok12 [17]

Explanation:

As a neutral lithium atom contains 3 protons and its elemental charge is given as $1.602 \times 10^{-19} C$ . Hence, we will calculate its number of moles as follows.

Moles = $\frac{mass}{\text{molar mass}}$

= $\frac{0.7 \times 1000 g}{7 g/mol}$

= 100 mol

According to mole concept, there are $6.023 \times 10^{23}$ atoms present in 1 mole. So, in 100 mol we will calculate the number of atoms as follows.

No. of atoms = $100 \times 6.023 \times 10^{23}$

= $6.023 \times 10^{25}$ atoms

Since, it is given that charge on 1 atom is as follows.

$3 \times 1.602 \times 10^{-19}C$

= $4.806 \times 10^{-19}C$

Therefore, charge present on $6.023 \times 10^{25}$ atoms will be calculated as follows.

$6.023 \times 10^{25} atoms \times 4.806 \times 10^{-19} C$

$28.95 \times 10^{6}C$

Thus, we can conclude that a positive charge of $28.95 \times 10^{6}C$ is in 0.7 kg of lithium.

3 0

2 years ago

An ionic bond is a bond between

OleMash [197]

between two oppositely charged ions

7 0

3 years ago

Read 2 more answers

What is the pressure of 1.50 moles of a gas in a 30.0L tank at a temperature of<br> 285K?

julia-pushkina [17]

The pressure of the gas : 1.1685 atm

<h3>Further explanation</h3>

In general, the gas equation can be written

$\large {\boxed {\bold {PV = nRT}}}$

where

P = pressure, atm

V = volume, liter

n = number of moles

R = gas constant = 0.08206 L.atm / mol K

T = temperature, Kelvin

n=moles=1.5

V=volumes = 30 L

T=temperature=285 K

The pressure :

$\tt P=\dfrac{nRT}{V}\\\\n=\dfrac{1.5\times 0.082\times 285}{30}\\\\P=1.1685~atm$

7 0

2 years ago

What volume, in milliliters, of a 0.997 M KOH solution is needed to neutralize 30.0 mL of 0.0400 M HCl?

deff fn [24]

Answer:

1.2 mL

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated:

HCl + KOH —> KCl + H₂O

From the balanced equation,

Mole ratio of the acid, HCl (nₐ) = 1

Mole ratio of base, KOH (n₆) = 1

Finally, we shall determine the volume of the base, KOH needed to neutralize the acid, HCl as follow:

Molarity of base, KOH (M₆) = 0.997 M

Volume of acid, HCl (Vₐ) = 30 mL

Molarity of acid, HCl (Mₐ) = 0.0400 M

Volume of base, KOH (V₆) =?

MₐVₐ / M₆V₆ = nₐ/n₆

0.04 × 30 / 0.997 × V₆ = 1/1

1.2 / 0.997 × V₆ = 1

Cross multiply

0.997 × V₆ = 1.2

Divide both side by 0.997

V₆ = 1.2 / 0.997

V₆ = 1.2 mL

Thus, the volume of the base, KOH needed to neutralize the acid is 1.2 mL.

7 0

3 years ago