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zvonat [6]
4 years ago
5

4. The volume of a sample of a gas at STP is 200.0 ml. If the pressure is increased to 4.00 atmospheres (temperature constant),

what is the new volume?
Chemistry
1 answer:
Elina [12.6K]4 years ago
6 0

Answer : The value of new volume is, 50.0 mL

Explanation :

Boyle's Law : It is defined as the pressure of the gas is inversely proportional to the volume of the gas at constant temperature and number of moles.

P\propto \frac{1}{V}

or,

P_1V_1=P_2V_2

where,

P_1 = initial pressure at STP = 1 atm

P_2 = final pressure =  4.00  atm

V_1 = initial volume at STP = 200.0 mL

V_2 = final volume = ?

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

1atm\times 200.0mL=4.00atm\times V_2

V_2=50.0mL

Therefore, the value of new volume is, 50.0 mL

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A 10.50 gram sample of a compound is decomposed to yield 3.40 g Na, 2.37 g S, and 4.73 g O. What is the mass percentage of each
9966 [12]

Answer:

Na = 32.4% , % S = 22.6% and %O = 45.0%

Explanation:

% Na = 3.4/10.5. × 100%

= 32.4%

%S = 2.37/10.5 × 100%

= 22.6%

% O= 4.73/10.5 × 100%

= 45.0%xplanation:

6 0
3 years ago
What items are true about a block of ice at -10°C as you continue to apply heat
zhenek [66]

1. its temperature will rise continuously until it melts

I don't believe that any of the other answers are correct because it can not stay at a certain temperature if it is melting

5 0
3 years ago
Suppose the half-life is 9.0 s for a first order reaction and the reactant concentration is 0.0741 M 50.7 s after the reaction s
bazaltina [42]

<u>Answer:</u> The time taken by the reaction is 84.5 seconds

<u>Explanation:</u>

The equation used to calculate half life for first order kinetics:

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

where,

t_{1/2} = half-life of the reaction = 9.0 s

k = rate constant = ?

Putting values in above equation, we get:

k=\frac{0.693}{9}=0.077s^{-1}

Rate law expression for first order kinetics is given by the equation:

k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}     ......(1)

where,

k = rate constant  = 0.077s^{-1}

t = time taken for decay process = 50.7 sec

[A_o] = initial amount of the reactant = ?

[A] = amount left after decay process =  0.0741 M

Putting values in equation 1, we get:

0.077=\frac{2.303}{50.7}\log\frac{[A_o]}{0.0741}

[A_o]=3.67M

Now, calculating the time taken by using equation 1:

[A]=0.0055M

k=0.077s^{-1}

[A_o]=3.67M

Putting values in equation 1, we get:

0.077=\frac{2.303}{t}\log\frac{3.67}{0.0055}\\\\t=84.5s

Hence, the time taken by the reaction is 84.5 seconds

6 0
3 years ago
The​ half-life of a certain tranquilizer in the bloodstream is 5050 hours. how long will it take for the drug to decay to 8686​%
Novosadov [1.4K]
Using the exponential decay model; we calculate "k"
We know that "A" is half of A0
A = A0 e^(k× 5050)
A/A0 = e^(5050k)
0.5 = e^(5055k)
In (0.5) = 5055k 
-0.69315 = 5055k
 k = -0.0001371
To calculate how long it will take to decay to 86% of the original mass
0.86 = e^(-0.0001371t)
In (0.86) = -0.0001371t
-0.150823 = -0.0001371 t
 t = 1100 hours

4 0
3 years ago
You are going to standardize your sodium hydroxide by titrating with potassium hydrogen phthalate. As an example, you dissolve 0
Kay [80]

Answer:

0.13 M

Explanation:

The reaction equation is;

NaOH(aq) + KHC8H4O4(aq) ------> KNaC8H4O4(aq) + H2O(l)

Molar mass of KHP = 204.22 g/mol

Amount of KHP= mass/ molar mass = 0.3365 g/204.22 g/mol = 1.65 × 10^-3 moles

n= CV

Where;

C= concentration

V= volume in dm^3

n= number of moles

C= n/V = 1.65 × 10^-3 moles × 1000/250 = 6.6 × 10^-3 M

If 1 mole of KHP reacts with 1 mole of NaOH

1.65 × 10^-3 moles of KHP will react with 1.65 × 10^-3 moles of NaOH

From

n= CV

We have that only 12.44 ml of NaOH reacted

C= n/V = 1.65 × 10^-3 moles × 1000/12.44

C= 0.13 M

At the equivalence point, the KHP solution turned light pink.

4 0
3 years ago
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