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Galina-37 [17]
3 years ago
11

Help me 5-8 please and thank you

Chemistry
1 answer:
defon3 years ago
4 0
-3
explanation: if you subtract 5 from 8 you get 3 just put a negative sign and you have your answer :)
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Serine has pka1 = 2.21 and pka2 = 9.15. use the henderson-hasselbalch equation to calculate the ratio neutral form/protonated fo
malfutka [58]
Calculate the ratio by using Henderson-Hasselbalch equation:

pH = pKa + log [neutral form] / Protonated form

3.05 = 2.21 + log [neutral form] / [Protonated form]

3.05 - 2.21 = log [neutral form] / [Protonated form]

0.84 = log [neutral form] / [Protonated form]

[neutral form] / [protonated form] = anti log 0.84 = 6.91
8 0
3 years ago
Describe how elements are laid out on the periodic table of elements.
defon

Answer:

The chemical elements are arranged in order of increasing atomic number.

Explanation:

Hydrogen = 1

Helium = 2

Lithium = 3

8 0
3 years ago
Read 2 more answers
What is the density of CHCL3 vapor at 1.00atm and 298K?
Advocard [28]

Answer:

4.8 g/mL is the density of chloroform vapor at 1.00 atm and 298 K.

Explanation:

By ideal gas equation:

PV=nRT

Number of moles (n)

can be written as: n=\frac{m}{M}

where, m = given mass

M = molar mass

PV=\frac{m}{M}RT\\\\PM=\frac{m}{V}RT

where,

\frac{m}{V}=d which is known as density of the gas

The relation becomes:

PM=dRT    .....(1)

We are given:

M = molar mass of chloroform= 119.5 g/mol

R = Gas constant = 0.0821\text{ L atm }mol^{-1}K^{-1}

T = temperature of the gas = 298K

P = pressure of the gas = 1.00 atm

Putting values in equation 1, we get:

1.00atm\times 119.5g/mol=d\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 298K\\\\d=4.88g/L

4.8 g/mL is the density of chloroform vapor at 1.00 atm and 298 K.

7 0
3 years ago
Help!?
wolverine [178]
I think to measure the age of the rock 
6 0
3 years ago
How many moles are in 20 grams of argon​
Sladkaya [172]

Answer:

There are 0.5 mole in 20g of argon.

Explanation:

40 g of argon = 1mole

Then 20g of argon is,

→ 1/40 × 20

→ 0.5 mole

5 0
2 years ago
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