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3 years ago

Does anyone know how to do this??

Chemistry

1 answer:

mihalych1998 [28]3 years ago

4 0

How to do what? There's no problem here. If you send a picture I can do my best to help you

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In a problem, you are given two pressures and one temperature at constant volume and amount of gas. You are asked to find a seco

laiz [17]

I don't know who it was named for but the law is

P1/T1 = P2/T2

Make sure the pressure units are the same (atmospheres or kPa usually) and that the temperature is in Degrees Kelvin which is derived from Celsius degrees.

Try Charles' Law for the name.

3 0

4 years ago

Read 2 more answers

Plz answer for brainliest trolls will get deleted

Roman55 [17]

Answer:

A; C6H12O6 + 6O2

Explanation:

You can easily differentiate between the reactants and the products, as the reactants are on the left, while the products are on the right.

3 0

3 years ago

Which of the following equalities is not correct?

Kryger [21]

1cm^3 = 1L would be the correct answer. One cubic centimeter equals .001 liter, so this equality above is not correct.

Please let me know if you have any questions! :)

8 0

3 years ago

Read 2 more answers

An unknown compound has the following chemical formula:

Softa [21]

Answer:

Formula of the compound = $P_2O_5$

Explanation:

Given,

No. of mole of O = 8.20 mol

No. of mole of P = 3.30 mol

Chemical formula = $P_2O_x$

Ratio of P and O = $\frac{2}{x}$

$\frac{2}{x} =\frac{3.3}{8.20} \\x \times 3.3 = 8.20 \times 2\\x = \frac{8.20 \times 2}{3.3} \\x=4.9969\\ = 5$

formula of the compound = $P_2O_5$

5 0

3 years ago

A 25.0-mL sample of 0.150 M hydrazoic acid, HN3, is titrated with a 0.150 M NaOH solution. What is the pH after 13.3 mL of base

tamaranim1 [39]

Answer:

pH ≅ 4.80

Explanation:

Given that:

the volume of HN₃ = 25 mL = 0.025 L

Molarity of HN₃ = 0.150 M

number of moles of HN₃ = 0.025 × 0.150

number of moles of HN₃ = 0.00375 mol

Molarity of NaOH = 0.150 M

the volume of NaOH = 13.3 mL = 0.0133

number of moles of NaOH = 0.0133× 0.150

number of moles of NaOH = 0.001995 mol

The chemical equation for the reaction of this process can be written as:

$HN_3 + OH- ---> N^-_{3} + H_2O$

1 mole of hydrazoic acid react with 1 mole of hydroxide to give nitride ion and water

thus the new number of moles of HN₃ = 0.00375 - 0.001995 = 0.001755 mol

Total volume used in the reaction = 0.025 + 0.0133 = 0.0383 L

Concentration of $HN_3$ = $\dfrac{0.001755}{0.0383}$ = 0.0458 M

Concentration of $N^{-}_3$ = $\dfrac{ 0.001995 }{0.0383}$ = 0.0521 M

GIven that :

Ka = $1.9 x 10^{-5}$

Thus; it's pKa = 4.72

$pH =4.72 + log(\dfrac{ \ 0.0521}{0.0458})$

$pH =4.72 + log(1.1376)$

$pH =4.72 + 0.05598$

$pH =4.77598$

pH ≅ 4.80

3 0

3 years ago