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Marina CMI [18]
3 years ago
11

1. What two quantities must be known to calculate the density of a sample of matter?

Chemistry
2 answers:
aleksandr82 [10.1K]3 years ago
8 0
Remember the formula for density:
d =  \frac{m}{v}
where d is density, m is mass, and v is volume.

Therefore, B. mass and volume is needed to determine the density
Nikolay [14]3 years ago
3 0
mass and volume is the answer

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Given the partial equation:
Nikolay [14]

Answer : The balanced chemical equation in acidic medium will be,

IO_3^-(aq)+2Sn^{2+}(aq)+6H^+(aq)\rightarrow I^-(aq)+2Sn^{4+}(aq)+3H_2O(l)

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 acidic 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 less number of oxygen are present.

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

Now balance the charge.

The given chemical reaction is,

IO_3^-(aq)+Sn^{2+}(aq)\rightarrow I^-(aq)+Sn^{4+}(aq)

The oxidation-reduction half reaction will be :

Oxidation : Sn^{2+}\rightarrow Sn^{4+}

Reduction : IO_3^-\rightarrow I^-

First balance the main element in the reaction.

Oxidation : Sn^{2+}\rightarrow Sn^{4+}

Reduction : IO_3^-\rightarrow I^-

Now balance oxygen atom on both side.

Oxidation : Sn^{2+}\rightarrow Sn^{4+}

Reduction : IO_3^-\rightarrow I^-+3H_2O

Now balance hydrogen atom on both side.

Oxidation : Sn^{2+}\rightarrow Sn^{4+}

Reduction : IO_3^-+6H^+\rightarrow I^-+3H_2O

Now balance the charge.

Oxidation : Sn^{2+}\rightarrow Sn^{4+}+2e^-

Reduction : IO_3^-+6H^++4e^-\rightarrow I^-+3H_2O

The charges are not balanced on both side of the reaction. Thus, we are multiplying oxidation reaction by 2 and the adding both equation, we get the balanced redox reaction.

Oxidation : 2Sn^{2+}\rightarrow 2Sn^{4+}+4e^-

Reduction : IO_3^-+6H^++4e^-\rightarrow I^-+3H_2O

The balanced chemical equation in acidic medium will be,

IO_3^-(aq)+2Sn^{2+}(aq)+6H^+(aq)\rightarrow I^-(aq)+2Sn^{4+}(aq)+3H_2O(l)

5 0
3 years ago
Why do all humans have similar features? A. They have the same number and kinds of chromosomes. B. They have identical alleles.
Serggg [28]
I belive it is a we all have 46 chromosomes.I learned this in sience.
4 0
3 years ago
Read 2 more answers
10. When a 13.6 g sample of a compound containing only magnesium and oxygen is decomposed, 5.4 g of oxygen is obtained. What is
irinina [24]

Answer:5.4 g / 13.6 g *100

Explanation:Its is the correct answer

7 0
4 years ago
Read 2 more answers
Most substances become less dense as the temperature increases. Which substance is an exception to this rule
stealth61 [152]
The substance is water
6 0
4 years ago
Read 2 more answers
To what volume in millimeters must 50.0 mL of 18.0 M H2SO4 be diluted to obtain 4.35 M H2SO4?
poizon [28]
We know that to relate solutions of with the factors of molarity and volume, we can use the equation: M_{1}  V_{1} = M_{2}  V_{2}

**NOTE: The volume as indicated in this question is defined in L, not mL, so that conversion must be made. However it is 1000 mL = 1 L.

So now we can assign values to these variables. Let us say that the 18 M H_{2}  SO_{4} is the left side of the equation. Then we have:

(18 M)(0.050 L)=(4.35M) V_{2}

We can then solve for V_{2}:

V_{2}= \frac{(18M)(0.05L)}{4.35M} and V_{2} =0.21 L or 210 mL

We now know that the total amount of volume of the 4.35 M solution will be 210 mL. This is assuming that the entirety of the 50 mL of 18 M is used and the rest (160 mL) of water is then added.


7 0
4 years ago
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