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2 years ago

A ball was massed in a dish. The total mass was 15.6 grams. Treball was placed in a

Chemistry

1 answer:

iVinArrow [24]2 years ago

3 0

Answer:

14.57g

Explanation:

Given parameters:

Mass of dish + ball = 15.6g

Initial volume of water in the cylinder = 26.7mL

Final volume of water in the cylinder = 38.9mL

Mass of dish = ?

Unknown

Mass of the ball = ?

Solution;

Since the mass of ball and dish is 15.6g,

Mass of the ball =Mass of ball + dish - mass of the dish

Insert the parameters and solve;

Mass of the ball = 15.6g - 1.03g = 14.57g

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A plastic bottle is closed outdoors on a cold day when the temperature is −15.0°C and is later brought inside where the temperat

Lana71 [14]

Answer:

Pressure = 1.14 atm

Explanation:

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This question requires us to calculate the final pressure of the bottle after thermal equilibrium.

This is a direct application of pressure law which states that in a fixed mass of gas, the pressure of a given gas is directly proportional to its temperature, provided that volume remains constant.

Mathematically, what this implies is

P = kT k = P / T

P1 / T1 = P2 / T2 = P3 / T3 =........= Pn / Tn

P1 / T1 = P2 / T2

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T1 = -15°C = (-15 + 273.15)K = 258.15K

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P2 = (P1 × T2) / T1

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P2 = 1.14atm

The pressure of the gas after attaining equilibrium is 1.14atm

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3 years ago

Which two systems first respond to burning your finger?

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Answer:

B. is my guess

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A sample of 28 Mg decays initially at a rate of 53500 disintegrations per minute, but the decay rate falls to 10980 disintegrati

prisoha [69]

Answer : The half life of 28-Mg in hours is, 6.94

Explanation :

First we have to calculate the rate constant.

Expression for rate law for first order kinetics is given by:

$k=\frac{2.303}{t}\log\frac{a}{a-x}$

where,

k = rate constant

t = time passed by the sample = 48.0 hr

a = initial amount of the reactant disintegrate = 53500

a - x = amount left after decay process disintegrate = 53500 - 10980 = 42520

Now put all the given values in above equation, we get

$k=\frac{2.303}{48.0}\log\frac{53500}{42520}$

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Now we have to calculate the half-life.

$k=\frac{0.693}{t_{1/2}}$

$9.98\times 10^{-2}=\frac{0.693}{t_{1/2}}$

$t_{1/2}=6.94hr$

Therefore, the half life of 28-Mg in hours is, 6.94

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