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wolverine [178]
3 years ago
11

A sample of oxygen gas at a pressure of 690 mmHg and a temperature of 32°C occupies a volume of 14.2 L if the gas is cooled at c

onstant pressure to a temperature of -1°C the volume of the gas sample will be
Chemistry
1 answer:
Diano4ka-milaya [45]3 years ago
5 0

Answer:

12.66 L

Explanation:

- Used combined gas law formula

- Change C to K

- Rearrange formula to fit your values

- If you would like further explanation or want to learn how to do these please let me know. Hope this helped!

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Iodine-131 is a beta emitter used as a tracer in radio immunoassays in biological systems. It follows first order kinetics. The
allsm [11]

<u>Answer:</u> The amount of Iodine-131 remain after 39 days is 0.278 grams

<u>Explanation:</u>

The equation used to calculate rate constant from given half life for first order kinetics:

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

where,

t_{1/2} = half life of the reaction = 8.04 days

Putting values in above equation, we get:

k=\frac{0.693}{8.04days}=0.0862days^{-1}

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

k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}

where,

k = rate constant = 0.0862days^{-1}

t = time taken for decay process = 39 days

[A_o] = initial amount of the sample = 8.0 grams

[A] = amount left after decay process = ?

Putting values in above equation, we get:

0.0862=\frac{2.303}{39}\log\frac{8.0}{[A]}

[A]=0.278g

Hence, the amount of Iodine-131 remain after 39 days is 0.278 grams

3 0
3 years ago
Calculate the ph of a 0.40 m solution of sodium benzoate (nac6h5coo) given that the ka of benzoic acid (c6h5cooh) is 6.50 x 10-5
asambeis [7]
Hello!

The dissociation reaction for Benzoic Acid is the following:

C₆H₅COOH + H₂O ⇄ C₆H₅COO⁻ + H₃O⁺

The Ka expression is the following and we clear for the concentration of H₃O⁺(X) assuming that the dissociation is little so we can rule it out in the denominator of the equation:

Ka= \frac{[C_6H_5COO^{-}]*[H_3O^{+}] }{[C_6H_5COOH]}=\frac{X*X }{0,40 -X}  (assume: 0,40-X\approx0,40)\\  \\ X= \sqrt{0,40*6,50*10^{-5} }=0.00510M \\  \\ pH=-log([H_3O^{+}]=2,29

So, the pH of this Benzoic Acid solution is 2,29

Have a nice day!


3 0
3 years ago
In the treatment of domestic water supplies, chlorine is added to the water to form chloric(I) acid,HClO.
Naya [18.7K]
Chlorine goes from Cl₂ to ClO⁻.

Oxidation state of Cl in Cl₂ is zero whereas in ClO⁻ it is +1. Hence change in oxidation state of Chlorine is 1.

Calculation of oxidation number of Cl in ClO⁻ or HClO⁻ :
Let x be oxidation number of Cl in ClO⁻ the. Now since the net charge on ClO⁻ is -1, sum of oxidation number of all must be equal to -1. 


Therefore,
x + (-2) = -1 .....[ oxidation number of O is -2]
∴ x = 2-1 = +1
Therefore oxidation number of Cl in ClO⁻ is +1

8 0
4 years ago
Please help asap 18 points if help and right answer please ​
barxatty [35]

Answer:

Lithium = 8%

Bromine = 92%

Explanation:

To calculate percent composition, you must:

- Calculate the molar mass.

- Divide the subtotal for each element's mass by the molar mass.

- Convert to a percentage

With that being said, given LiBr (Lithium Bromide), calculate the molar mass:

Lithium has an atomic weight of 7, and there is one Lithium atoms in LiBr. Bromine has an atomic weight of 80, and there is one Bromine atom in LiBr:

1(7)\\1(80)

Add the two products:

80+7

=87

The molar mass of LiBr is about 87 grams.

With that information, divide the subtotal of Lithium by the molar mass, and then multiply by 100 to convert to a percentage:

\frac{7}{87} × 100=

8.0459%

(Round to nearest percentage):

8%

Therefore, the percent composition of Lithium in the compound Lithium Bromide is about 8%.

Now, divide the subtotal of Bromine by the molar mass, and then multiply by 100 to convert to a percentage:

\frac{80}{87} × 100=

91.9540

(Round):

92%

Therefore, the percent composition of Bromine in the compound Lithium Bromide is about 92%.

6 0
3 years ago
25. In the formula for a molecular compound, which atom generally comes first? You may went
vovikov84 [41]

Answer:

For organic compounds, carbon and hydrogen are listed as the first elements in the molecular formula, and they are followed by the remaining elements in alphabetical order. For example, for butane, the molecular formula is C 4 H 10.

Explanation:

3 0
3 years ago
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