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3 years ago

What’s the answer?????

Chemistry

1 answer:

Virty [35]3 years ago

5 0

Answer:

kinetic energy

potential energy

kinetic energy

Explanation:

kinetic energy- energy associated with motion

potential energy- the energy that a piece of matter has because of its position or nature.

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What type of reaction occurs initially with sodium azide when airbags initially deploy.

Luden [163]

The chemical at the heart of the air bag reaction is called sodium azide, or NaN3. CRASHES trip sensors in cars that send an electric signal to an ignitor. The heat generated causes sodium azide to decompose into sodium metal and nitrogen gas, which inflates the car's air bags.

8 0

3 years ago

A photon of light possesses 5 x 10^-19 J of energy. Calculate its frequency

saveliy_v [14]

Answer:

The frequency of photon is 0.75×10¹⁵ s⁻¹.

Explanation:

Given data:

Energy of photon = 5×10⁻¹⁹ J

Frequency of photon = ?

Solution:

Formula;

E = hf

h = planck's constant = 6.63×10⁻³⁴ Js

5×10⁻¹⁹ J = 6.63×10⁻³⁴ Js ×f

f = 5×10⁻¹⁹ J / 6.63×10⁻³⁴ Js

f = 0.75×10¹⁵ s⁻¹

The frequency of photon is 0.75×10¹⁵ s⁻¹.

4 0

3 years ago

3.4 x 10-25 kg = ? microounces

Thepotemich [5.8K]

Answer: 1.2 x 10^-17 microounces

Explanation:

Ounce = 28.5G microounce = 28.5*10^-6g

3.4*10^-25 kg = 3.4*10^-22 g = (3.4/2.85)*10^(-22+5) = 1.2*10-17

8 0

3 years ago

2BrO3- + 5SnO22-+ H2O5SnO32- + Br2+ 2OH- In the above reaction, the oxidation state of tin changes from to . How many electrons

Julli [10]

Answer :

The oxidation state of tin changes from (+2) to (+4).

The number electrons transferred in the reaction are, 10 electron.

Explanation :

Redox reaction or Oxidation-reduction reaction : It is defined as the reaction in which the oxidation and reduction reaction takes place simultaneously.

Oxidation reaction : It is defined as the reaction in which a substance looses its electrons. In this, oxidation state of an element increases. Or we can say that in oxidation, the loss of electrons takes place.

Reduction reaction : It is defined as the reaction in which a substance gains electrons. In this, oxidation state of an element decreases. Or we can say that in reduction, the gain of electrons takes place.

Rules for the balanced chemical equation in basic solution are :

First we have to write into the two half-reactions.

Now balance the main atoms in the reaction.

Now balance the hydrogen and oxygen atoms on both the sides of the reaction.

If the oxygen atoms are not balanced on both the sides then adding water molecules at that side where the more number of oxygen are present.

If the hydrogen atoms are not balanced on both the sides then adding hydroxide ion $(H^+)$ at that side where the less number of hydrogen are present.

Now balance the charge.

The given chemical reaction is,

$2BrO_3^-+5SnO_2^{2-}+H_2O\rightarrow 5SnO_3^{2-}+Br_2+2OH^-$

The oxidation-reduction half reaction will be :

Oxidation : $SnO_2^{2-}\rightarrow SnO_3^{2-}$

Reduction : $BrO_3^-\rightarrow Br_2$

First balance the main element in the reaction.

Oxidation : $SnO_2^{2-}\rightarrow SnO_3^{2-}$

Reduction : $2BrO_3^-\rightarrow Br_2$

Now balance oxygen atom on both side.

Oxidation : $SnO_2^{2-}\rightarrow SnO_3^{2-}+H_2O$

Reduction : $2BrO_3^-+H_2O\rightarrow Br_2$

Now balance hydrogen atom on both side.

Oxidation : $SnO_2^{2-}+OH^-\rightarrow SnO_3^{2-}+H_2O$

Reduction : $2BrO_3^-+3H_2O\rightarrow Br_2+12OH^-$

Now balance the charge.

Oxidation : $SnO_2^{2-}+OH^-\rightarrow SnO_3^{2-}+H_2O+2e^-$

Reduction : $2BrO_3^-+3H_2O+10e^-\rightarrow Br_2+12OH^-$

The charges are not balance on both side of the reaction. We are multiplying oxidation reaction by 5 and then added both equation, we get the balanced redox reaction.

Oxidation : $5SnO_2^{2-}+5OH^-\rightarrow 5SnO_3^{2-}+5H_2O+10e^-$

Reduction : $2BrO_3^-+3H_2O+10e^-\rightarrow Br_2+12OH^-$

The balanced chemical equation in basic medium will be,

$2BrO_3^-+5SnO_2^{2-}+H_2O\rightarrow 5SnO_3^{2-}+Br_2+2OH^-$

The number electrons transferred in the reaction are, 10 electron.

7 0

3 years ago

Describe in detail how to prepare 500.0 mL of a 2.5 M calcium chloride solution. Calcium chloride is purchased as a solid. TO pi

antoniya [11.8K]

Answer:

In a volumetric flask of marking 500.0 mL add 138.75 grams of calcium chloride and add small amount of water to dissolve solute completely. After the solute gets completely soluble add more water up till the mark of 500 ml.

Explanation:

Concentration of calcium chloride = 2.5 M

Volume of the solution = 500.0 ml = 0.5000 L

Moles of calcium chloride = n

$c=\frac{n}{V(L)}$

n = moles of solute

c = concentration of solution

V = volume of the solution in L

$2.5 M=\frac{n}{0.5000 L}$

$n=2.5 M\times 0.5000 L = 1.2500 mol$

$n=\frac{\text{Mass of calcium chloride}}{\text{Molar mass of calcium chloride}}$

$1.2500 mol=\frac{\text{Mass of calcium chloride}}{111 g/mol}$

Mass of calcium chloride = 111 g/mol × 1.2500 mol = 138.75 g

In a volumetric flask of marking 500.0 mL add 138.75 grams of calcium chloride and add small amount of water to dissolve solute completely. After the the solute gets completely soluble add more water up till the mark of 500 ml.

4 0

3 years ago