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2 years ago

Use the periodic table to answer the questions below.

Chemistry

2 answers:

Yanka [14]2 years ago

7 0

Answer:

where is the diagram shown

ExtremeBDS [4]2 years ago

4 0

Answer: 1s2, 2s2, 2p3

Explanation: Nitrogen is an element number 7 on the periodic table which means that it has 7 proton. And for an atom to maintain its electrical neutrality, the number of proton is equal to the number of electron. That is, equal number of positively and negatively charge.

So, 7 electrons of Nitrogen is

1s2, 2s2, 2p4

You might be interested in

During a physical change water can change it's (blank) but it will not change it's (blank).

Valentin [98]

Water can change it's STATE but it will not change it's COMPOUND.

8 0

3 years ago

A beaker with 155 mL of an acetic acid buffer with a pH of 5.000 is sitting on a benchtop. The total molarity of acid and conjug

OverLord2011 [107]

Answer:

ΔpH = 0.296

Explanation:

The equilibrium of acetic acid (CH₃COOH) in water is:

CH₃COOH ⇄ CH₃COO⁻ + H⁺

Henderson-Hasselbalch formula to find pH in a buffer is:

pH = pKa + log₁₀ [CH₃COO⁻] / [CH₃COOH]

Replacing with known values:

5.000 = 4.740 + log₁₀ [CH₃COO⁻] / [CH₃COOH]

0.260 = log₁₀ [CH₃COO⁻] / [CH₃COOH]

1.820 = [CH₃COO⁻] / [CH₃COOH] (1)

As total molarity of buffer is 0.100M:

[CH₃COO⁻] + [CH₃COOH] = 0.100M (2)

Replacing (2) in (1):

1.820 = 0.100M - [CH₃COOH] / [CH₃COOH]

1.820[CH₃COOH] = 0.100M - [CH₃COOH]

2.820[CH₃COOH] = 0.100M

[CH₃COOH] = 0.100M / 2.820

[CH₃COOH] = 0.035M

Thus: [CH₃COO⁻] = 0.100M - 0.035M = 0.065M

5.40 mL of a 0.490 M HCl are:

0.0054L × (0.490mol / L) = 2.646x10⁻³ moles HCl.

Moles of CH₃COO⁻ are: 0.155L × (0.065mol / L) = 0.0101 moles

HCl reacts with CH₃COO⁻ thus:

HCl + CH₃COO⁻ → CH₃COOH

After reaction, moles of CH₃COO⁻ are:

0.0101 moles - 2.646x10⁻³ moles = 7.429x10⁻³ moles of CH₃COO⁻

Moles of CH₃COOH before reaction are: 0.155L × (0.035mol / L) = 5.425x10⁻³ moles of CH₃COOH. As reaction produce 2.646x10⁻³ moles of CH₃COOH, final moles are:

5.425x10⁻³ moles + 2.646x10⁻³ moles = 8.071x10⁻³ moles of CH₃COOH. Replacing these values in Henderson-Hasselbalch formula:

pH = 4.740 + log₁₀ [7.429x10⁻³ moles] / [8.071x10⁻³ moles]

pH = 4.704

As initial pH was 5.000, change in pH is:

ΔpH = 5.000 - 4.740 = 0.296

4 0

2 years ago

the chemical compound C2F4 is used to make PTEE (Teflon). How manyC2F4 molecules are in 485 kilograms of this material?

monitta

Answer:

$molecules=2.92x10^{27}moleculesC_2F_4$

Explanation:

Hello,

In this case, we use the Avogadro's number to compute the molecules of C2F4 whose molar mass is 100 g/mol contained in a 485-kg sample as shown below:

$molecules=485kgC_2F_4*\frac{1000gC_2F_4}{1kgC_2F_4} *\frac{1molC_2F_4}{100gC_2F_4}*\frac{6.022x10^{23}molecules C_2F_4}{1molC_2F_4} \\\\molecules=2.92x10^{27}moleculesC_2F_4$