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kogti [31]
2 years ago
15

If the volume occupied by the air in a bicycle pump is 525 cm3, and the pressure changes from 73.2 kPa to 122.5 k.Pa as the pist

on is pushed down, what is the new volume occupied by the air?
Chemistry
1 answer:
Musya8 [376]2 years ago
6 0

Answer:

V_2=313.71\ cm^3

Explanation:

Given that,

Initial volume, V_1=525\ cm^3

The pressure changes from 73.2 kPa to 122.5 k.Pa.

We need to find the new volume occupied by the air. Let it is V₂. It can be calculated using Boyle's law such that,

P_1V_1=P_2V_2\\\\V_2=\dfrac{P_1V_1}{P_2}\\\\V_2=\dfrac{73.2\times 525}{122.5}\\\\V_2=313.71\ cm^3

So, the new volume is 313.71\ cm^3.

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Part 2 please and explain please
vekshin1

1) 7.15 * 10^6

since the exponent of the scientific notation is positive, move 6 places to the right.

so, <em>7.15 * 10^6 = 7150000</em>

2) 3.03 * 10^-8

Since the exponent of the scientific notation is negative, move the decimal point 8 places to the left.

so,<em> 3.03 * 10^-8 = 0.0000000303</em>

<em />

3)<em> </em>4.9 * 10^-1

since the exponent of the scientific notation is negative, move 1 decimal place to the left.

so,<em> 4.9 * 10^-1 = 0.49</em>

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4) 2.886 * 10^5

since the exponent of the scientific notation is positive, move 5 decimal places to the right.

so,<em> 2.886 * 10^5 = 288600</em>

6 0
3 years ago
Need help asap!
Scilla [17]

Answer: 1.  C. polar covalent: electrons shared between silicon and sulfur but attracted more to the sulfur

2. B)  NH_3

3. B) Fluorine

Explanation:

1. A polar covalent bond is defined as the bond which is formed when there is a difference of electronegativities between the atoms.

Electronegativity difference = electronegativity of sulphur- electronegativity of silicon = 2.5 -1.8 = 0.7

Thus as electronegativity difference is less than 1.7 , the cond is polar covalent and as electronegativity of sulphur is more , the electrons will be more towards sulphur.

2. A molecular compound is usually composed of two or more nonmetal elements. Example: NH_3

Ionic compound is formed by the transfer of electrons from metals to non metals. Example: Mg_3N_2 , AlCl_3 and LiBr

3.  For formation of a neutral ionic compound, the charges on cation and anion must be balanced. The cation is formed by loss of electrons by metals and anions are formed by gain of electrons by non metals.

Here K is having an oxidation state of +1 and as the compound formed is KZ, the oxidation state of non metallic element Z should be -1. Thus the element Z is flourine which exists as diatomic gas F_2

4 0
2 years ago
To a 0.0001 m solution of mg(no3)2, naoh was added to a final concentration of 0.001m did a precipitate form?
Natalija [7]

I looked on a solubility chart to answer this question, and hydroxides are generally insoluble (with some exceptions of course). However, it says to consider Mg(OH)_{2} as an insoluble substance, though it may be moderately soluble.


The answer that you are most likely looking for is: Yes, a precipitate does form - this is due to the double placement reaction:


Mg(NO_{3})_{2}_{(aq)} + 2NaOH_{(aq)} → Mg(OH)_{2} {(s)} + 2NaNO_{3}_{(aq)}

8 0
3 years ago
Lactose, C12H22O11, is a naturally occurring sugar found in mammalian milk. A 0.335 M solution of lactose in water has a density
Dmitry_Shevchenko [17]

The molality of the solution is 0.00037 m.

<h3>What is concentration?</h3>

The term concentration refers to the amount of solute in a solution.

We have the following information;

Molarity = 0.335 M

Density =  1.0432 g/mL

Temperature = 20 o C

The molality of the solution is obtained from;

m = 0.335 M ×  1.0432 g/mL/ 1000(1.0432 g/mL) - 0.335 M (342 g/mol)

m = 0.344/1043.2 - 114.57

m =  0.344/928.63

m = 0.00037 m

Learn more about molality of solution: brainly.com/question/4580605

5 0
2 years ago
Complete the following single replacement reaction. If they don’t react, just write “NR”
Kipish [7]

Here we have to complete the given single replacement reactions.

The replacement reactions are-

1) Fe (s) + CuCl₂ (aq) → FeCl₂ (aq) + Cu (s)

2) Cu (s) + FeCl₂ (aq) → NA

3) K (s) + NiBr₂ (aq) → NA

4) Ni (s) + KBr (aq) → NiBr₂ (aq) + K (s)

5) Zn (s) + Ca(NO₃)₂ (aq) → Zn(NO₃)₂ (aq)  + Ca (s)

6) Ca (s) + Zn(NO₃)₂ (aq) → NA

The replacement reactions can be explained in light of the redox potential.

The standard reduction potential of the half cells involved in these reactions are:

Fe²⁺ + 2e → Fe (E° = -0.441V); Cu²⁺ + 2e → Cu (E° = 0.674V)

Ni²⁺ + 2e → Ni (E° = -0.23V); Zn²⁺ + 2e → Zn (E° = -0.763V)

We know the half cell reactions in which the standard reduction potentials are positive are allowed.

1) The reaction is possible as Cu²⁺/Cu and Fe/Fe²⁺ standard reduction potentials are positive.

2) The reaction is not possible as Cu/Cu²⁺ and Fe²⁺/Fe standard reduction potentials are negative.

3) The reaction is not possible as Ni²⁺/Ni standard reduction potential is negative.

4) The reaction is possible as Ni/Ni²⁺ standard reduction potential is positive.

5) The reaction is possible as Zn/Zn²⁺ standard reduction potential is positive.

6) The reaction is possible as Zn²⁺/Zn standard reduction potential is negative.

4 0
2 years ago
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