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3 years ago

What is the electron structure of sodium

Chemistry

2 answers:

yarga [219]3 years ago

8 0

Electron structure of sodium:
₁₁Na: 1s²2s²2p⁶3s¹

Elan Coil [88]3 years ago

7 0

Answer:

Na: 1s²2s²2p⁶3s¹

Explanation:

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Given a solution with a hydroxide ion concentration, oh=3.73×10^-5,what is the ph

Stells [14]

Answer:

pH = 9.73

Explanation:

pOH is the negative logarithm of the molar concentration of hydroxide ions and pH is the negative logarithm of the concentration of hydronium ions.

The mathematical expression to determine pOH is:

$pOH=-\log [OH^-]$

Where [OH⁻] means the molar concentration of hydroxide ions.

Substituting:

$pOH=-\log {(3.73\times 10^{-5})$

$pOH=4.43$

pH and pOH are related by the equation:

$pH+pOH=14$

From which:

$pH=14-pOH\\\\pH=14 - 4.43\\\\pH=9.57\leftarrow answer$

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3 years ago

Helium, neon, and argon are examples of

iogann1982 [59]

C) Noble gases
The six noble gases are helium (He), neon (Ne), argon (Ar), krypton (Kr), xenon (Xe), and radon (Rn). Their atomic numbers are, respectively, 2, 10, 18, 36, 54, and 86.

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3 years ago

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When blood passes through the lungs, carbon dioxide and water are formed from carbonic acid dissolved in the blood. The carbon d

NeTakaya

Answer is: decomposition.

Balanced chemical reaction: H₂CO₃ → CO₂ + H₂O.
H₂CO₃ is carbonic acid.
CO₂ is carbon (IV) oxide or carbon dioxide.
Chemical decomposition is the separation of a single chemical compound (in this example carbonic acid) into its two or more simpler compounds (in this example water and carbon dioxide).

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3 years ago

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What sample size (grams) of Na3PO4 (FW 164.00) known to be 50.00% pure should be used to consume exactly 40.00 mL of 0.1000 M HC

Mkey [24]

Answer:

0.109 g.

Explanation:

Equation of the reaction:

Na3PO4 + 3HCl --> 3NaCl + H3PO4

Number of moles of HCl = molar concentration × volume

= 0.1 × 0.04

= 0.004 mol.

By stoichiometry, 1 mole of Na3PO4 neutralises 3 moles of HCl. Therefore, number of moles of Na3PO4 = 0.004/3

= 0.0013 mol

Mass of Na3PO4 = molar mass × number of moles

= 0.0013 × 164

= 0.219 g

Since 50% of Na3PO4 was present in the sample. Let 100 g be the total mass of the substance

= 0.219 × 50 g/100 g

= 0.109 g.

3 0

3 years ago

mixture of N 2 And H2 Gases weighs 13.22 g and occupies a volume of 24.62 L at 300 K and 1.00 atm.Calculate the mass percent of

anygoal [31]

Answer: The mass percent of nitrogen gas and hydrogen gas is 91.41 % and 8.59 % respectively.

Explanation:

To calculate the number of moles, we use the equation given by ideal gas equation:

PV = nRT

where,

P = Pressure of the gaseous mixture = 1.00 atm

V = Volume of the gaseous mixture = 24.62 L

n = number of moles of the gaseous mixture = ?

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

T = Temperature of the gaseous mixture = 300 K

Putting values in above equation, we get:

$1.00atm\times 24.62L=n_{mix}\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 300K\\\\n_{mix}=\frac{1.00\times 24.62}{0.0821\times 300}=0.9996mol$

We are given:

Total mass of the mixture = 13.22 grams

Let the mass of nitrogen gas be 'x' grams and that of hydrogen gas be '(13.22 - x)' grams

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

For nitrogen gas:

Molar mass of nitrogen gas = 28 g/mol

$\text{Moles of nitrogen gas}=\frac{x}{28}mol$

For hydrogen gas:

Molar mass of hydrogen gas = 2 g/mol

$\text{Moles of hydrogen gas}=\frac{(13.22-x)}{2}mol$

Equating the moles of the individual gases to the moles of mixture:

$0.9996=\frac{x}{28}+\frac{(13.22-x)}{2}\\\\x=12.084g$

To calculate the mass percentage of substance in mixture we use the equation:

$\text{Mass percent of substance}=\frac{\text{Mass of substance}}{\text{Mass of mixture}}\times 100$

Mass of the mixture = 13.22 g

For nitrogen gas:

Mass of nitrogen gas = x = 12.084 g

Putting values in above equation, we get:

$\text{Mass percent of nitrogen gas}=\frac{12.084g}{13.22g}\times 100=91.41\%$

For hydrogen gas:

Mass of hydrogen gas = (13.22 - x) = (13.22 - 12.084) g = 1.136 g

Putting values in above equation, we get:

$\text{Mass percent of hydrogen gas}=\frac{1.136g}{13.22g}\times 100=8.59\%$

Hence, the mass percent of nitrogen gas and hydrogen gas is 91.41 % and 8.59 % respectively.

5 0

3 years ago