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vagabundo [1.1K]
2 years ago
11

A 6.00 L container of N_2 has a temperature of 273 K. Calculate the volume if the temperature is doubled.

Chemistry
1 answer:
vfiekz [6]2 years ago
4 0
V1/T1 =V2/T2 (using charles law)

V1=6.00
V2=?
T1=273
T2=273

Making V2 the subject of formula the equation then becomes

V2= V1xT2/T1

6.00x263/273=6.0L
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Answer:

Layer A is the youngest

It is at the top, it has all of the dinosaur bones which got buried after earth got hit with an asteroid and they went extinct. Layer D started it and all of these years dirt, soil, and fossils have been gathering making each layer before older and older.

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Explanation:

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Calculate the destiny of the object that has a mass of 453 g and occupies the volume of 225 cm3
Elan Coil [88]

density = 2.013g/cm3

Explanation:

density = mass/volume

density = 453g/225cm3

density = 2.013g/cm3

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What happens to the energy needed to remove an electron as the atomic number increases across a period?
lora16 [44]

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In general, the energy needed to remove an electron (the ionization energy) increases as the atomic number increases across a period.

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Each additional proton increases the attraction between the nucleus and the electron.

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4 years ago
What is the pH when 10.0 mL of 0.20 M potassium hydroxide is added to 30.0 mL of 0.10 M cinnamic acid, HC9H7O2 (Ka = 3.6 × 10–5)
astra-53 [7]

Answer:-

Solution:- As is clear from the given Ka value, Cinnamic acid is a weak acid. let's calculate the moles of acid and KOH added to it from their given molarities and mL.

For KOH,  10.0mL(\frac{1L}{1000mL})(\frac{0.20mol}{1L})

= 0.002 mol

For Cinnamic acid,  30.0mL(\frac{1L}{1000mL})(\frac{0.10mol}{1L})

= 0.003 mol

Acid and base react as:

HC_9H_7O_2(aq)+KOH(aq)\rightleftharpoons KC_9H_7O_2(aq)+H_2O(l)

The reaction takes place in 1:1 mol ratio. Since the moles of acid are in excess, the acid is still remaining when all the kOH is used.

0.002 moles of KOH react with 0.002 moles of Cinnamic acid to form 0.002 moles of potassium cinnamate. Excess moles of Cinnamic acid = 0.003 - 0.002 = 0.001

As the solution have weak acid and it's salt(or we could say conjugate base), it is a buffer solution and the pH of the buffer solution could easily be calculated using Handerson equation:

pH=pKa+log(\frac{base}{acid})

pKa could be caluted from given Ka value using the formula:

pKa = - log Ka

pKa=-log3.6*10^-^5

pKa = 4.44

let's plug in the values in Handerson equation and calculate the pH:

pH=4.44+log(\frac{0.002}{0.001})

pH = 4.44+0.30

pH = 4.74

So, the first choice is correct, pH is 4.74.

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.........................................

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