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Georgia [21]
3 years ago
12

What is the balanced equation for KOH+CO2—>K2CO3+H2O

Chemistry
1 answer:
insens350 [35]3 years ago
3 0

The balanced equation :

2KOH+CO₂⇒K₂CO₃+H₂O

<h3>Further explanation</h3>

Equalization of chemical reaction equations can be done using variables. Steps in equalizing the reaction equation:

1. gives a coefficient on substances involved in the equation of reaction such as a, b, or c etc.

2. make an equation based on the similarity of the number of atoms where the number of atoms = coefficient × index between reactant and product

3. Select the coefficient of the substance with the most complex chemical formula equal to 1

Reaction

KOH+CO₂⇒K₂CO₃+H₂O

  • give coeffiecient

aKOH+bCO₂⇒K₂CO₃+cH₂O

  • make equation

K, left=a, right=2⇒a=2

H, left=a, right=2c⇒a=2c⇒2=2c⇒c=1

O, left=a+2b, right=3+c⇒a+2b=3+c⇒2+2b=3+1⇒2b=2⇒b=1

the equation becomes :

2KOH+CO₂⇒K₂CO₃+H₂O

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Determine the latitude and longitude of the location A
malfutka [58]

Answer:

Latitude: 30\° N

Longitude: 45\° W

Explanation:

Lets begin by explaining the meaning of latitude and longitude as geogrephical coordinates:

Latitude is the angular distance between the equatorial line, and a specific point on the Earth. It is measured in degrees and is represented according to the hemisphere in which the point is located, which can be north or south latitude.  

In this sense latitude 0\° refers to the equatorial line that divides the Earth in two hemispheres (North and South).

Longitude represents the specific east–west position of a point on the Earth's surface, being longitude 0\° the prime meridian or Greenwich meridian.

So, according to the figure, where the model of the Earth is divided by latitude lines separated by 10\° and the longitude lines separated by 15\°; we only have to count the lines from the equator to the line where the point A is, and count the lines fromo the Prime meridian to the line where point A is located.

Hence, point A location is:

Latitude: 30\° N

Longitude: 45\° W

5 0
3 years ago
a substance did not change its chemical nature in a reaction. which most likely describes the reaction
Mkey [24]
This substance most likely is an inert. It is a substance that is not chemically reactive. It does not change its chemical nature in a reaction. It does not <span>easily react with other chemicals. Most of the group 8 gases in the periodic table are classified as inert, due to their having full outer electron shells. </span>
5 0
3 years ago
Read 2 more answers
When an electron in an atom moves from a higher, less stable level, down to a lower, more stable level, a photon is
Wittaler [7]

Answer:

Emitted

Explanation:

8 0
3 years ago
The energy needed to remove an electron is called ionization energy.
salantis [7]
<h2>                             Emma here<em>!</em></h2>

Answer:

im pretty sure its decreases going across a row (left to right)

Explanation:

As you move from left to right across a period on the periodic table the size of an atom will decrease

<h3>Hope this helps<em>! </em>Have a nice day<em>!</em></h3>
7 0
3 years ago
In another experiment, a 0.150 M BF4^-(aq) solution is prepared by dissolving NaBF4(s) in distilled water. The BF4^-(aq) ions in
Ilia_Sergeevich [38]

Answer:

A) Forward rate = 1.1934 × 10^(-4) M/min

B) I disagree with the claim

Explanation:

A) We are told that [HF] reaches a constant value of 0.0174 M at equilibrium.

The reversible reaction given to us is;

BF4-(aq) +H20(l) → BF3OH-(aq) + HF(aq)

From this, we can see that the stoichiometric ratio is 1:1:1:1

Thus, concentration of [BF4-] is now;

[BF4-] = 0.150 - 0.0174

[BF4-] = 0.1326 M

From the rate law, we are told the forward rate is kf [BF4-].

We are given Kf = 9.00 × 10^(-4) /min

Thus;

Forward rate = 9.00 × 10^(-4) /min × (0.1326M)

Forward rate = 1.1934 × 10^(-4) M/min

(B) The student claims that the initial rate of the reverse reaction is equal to zero can't be true because at equilibrium, rates for the forward and reverse reactions are usually equal.

Thus, I disagree with the claim.

3 0
3 years ago
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