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yan [13]
3 years ago
5

What is the average velocity between 3.00 and 6.00 seconds?

Chemistry
2 answers:
diamong [38]3 years ago
8 0
1)
average velocity:
(3,9)
(6,36)
average velocity=(36-9)m/(6-3) s=27m/3s=9 m/s

Answer: 9 m/s, east. 

2)
(12, 144)
(15,225)
average velocity=(225-144)m/(15-12)s=81 m/3s=27 m/s

Answer: 27 m/s,east. 
ehidna [41]3 years ago
8 0

Answer :

(1) The correct option is, 9.00 m/s, east

(2) The correct option is, 27.00 m/s, east

Explanation :

Average velocity : It is defined as the change in the displacement with respect to the change in time.

Formula used :

V_{average}=\frac{\Delta d}{\Delta t}=\frac{d_{final}-d_{initial}}{t_{final}-t_{initial}}

where,

d = displacement

t = time

(1) Now we have to calculate the average velocity between 3.00 and 6.00 seconds.

V_{average}=\frac{36m-9m}{6s-3s}=9m/s

The average velocity between 3.00 and 6.00 seconds is, 9 m/s, east

(2) Now we have to calculate the average velocity between 12.00 and 15.00 seconds.

V_{average}=\frac{225m-144m}{15s-12s}=27m/s

The average velocity between 12.00 and 15.00 seconds is, 27 m/s, east

Therefore, (1) The correct option is, 9.00 m/s, east

(2) The correct option is, 27.00 m/s, east

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Suppose 0.795 g of sodium iodide is dissolved in 100. mL of a 39.0 m M aqueous solution of silver nitrate.
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Answer:

The final molarity of iodide anion is 0.053 M

Explanation:

<u>Step 1</u>: Data given

Mass of sodium iodide (NaI) = 0.795 grams

Volume of the solution = 100 mL = 0.1 L

Molarity of aqueous solution of silver nitrate (AgNO3) = 39 mM = 0.039M

The molecular mass of sodium iodide is 149.89 g/mol.

<u>Step 2:</u> The balanced equation

AgNO3(aq) + NaI(aq) → AgI(s) + NaNO3(aq)

<u>Step 3: </u>Calculate number of moles of sodium iodide

Moles NaI = mass NaI / Molar mass NaI

Moles NaI = 0.795 grams / 149.89 g/mol

Moles NaI = 0.0053 moles

For 1 mole AgNO3 consumed, we need 1 mole NaI to produce 1 mole AgI and 1 mole NaNO3

The sodium iodide will dissociate as followed:

NaI(aq) → Na+(aq) +  I-(aq)

<u>Step 4</u>: Calculate iodide ions

For 1 mole NaI, we have 1 mole of I-

For 0.0053 moles of NaI we'll have 0.0053 moles I-

<u>Step 5:</u> Calculate molarity of iodide ion

Molarity = moles I- / volume

Molarity I- = 0.0053 moles / 0.1 L

Molarity I- = 0.053 M

The final molarity of iodide anion is 0.053 M

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