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3 years ago

7

What is atomic number of sodium

2 answers:

Sergeeva-Olga [200]3 years ago

6 0

Answer: Atomic number for sodium is 11

Symbol: Na

Melting point: 207.90°F (97.72°C)

The electronic configuration of sodium is 2, 8, 1

FromTheMoon [43]3 years ago

5 0

Answer:

Explanation:

Sodium is a chemical element with the symbol Na and atomic number 11. It is a soft, silvery-white, highly reactive metal. Sodium is an alkali metal, being in group 1 of the periodic table. Its only stable isotope is ²³Na. The free metal does not occur in nature, and must be prepared from compounds.

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You have 100 grams of potassium fluoride (KF) .

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Answer:

Option C = 1.72 mol

Explanation:

Given data:

Mass of KF = 100 g

Moles of KF = ?

Solution:

First of all we have to calculate the molar mass of KF.

Molar mass of KF = 39.0983 g/mol + 18.998403 g/mol

Molar mass of KF = 58. 0967 g/mol

Formula:

Number of moles = mass/molar mass

Number of moles = 100 g/ 58.0967 g/mol

Number of moles = 1.72 mol

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Answer:

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Explanation:

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Answer:

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3 years ago

Read 2 more answers

Assume that 50.0mL 50.0mL of 1.0MNaCl(aq) 1.0MNaCl(aq) and 50.0mL 50.0mL of 1.0M AgNO 3 (aq) 1.0MAgNO3(aq) were combined. Accord

S_A_V [24]

Answer:

The amount of precipitate formed would 7.175 grams of silver chloride.

Explanation:

$Moles (n)=Molarity(M)\times Volume (L)$

Moles of NaCl = n

Volume of NaCl solution = 50.0 mL = 0.050 L

Molarity of the hydrogen peroxide = 2.0 M

$n=2.0 M\times 0.050 L=0.100 mol$

Moles of silver nitarte = n'

Volume of silver nitrate solution = 50.0 mL = 0.050 L

Molarity of the silver nitrate = 1.0 M

$n'=1.0 M\times 0.050 L=0.050 mol$

$NaCl(aq)+AgNO_3(aq)\rightarrow AgCl(s)+NaNO_3(aq)$

According to reaction, 1 mole of of silver nitrate reacts with 1 mole of NaCl. Then 0.050 mole of silve nitrate will :

$\frac{1}{1}\times 0.050 mol=0.050 mol$ of NaCl

This means that silver nitrate is in limiting amount and NaCl is in excessive amount.

So, the amount of AgCl depends upon amount of silver nitrate.

According to reaction, 1 mole of silver nitrate gives 1 mole of AgCl.

Then 0.050 moles of silver nitrate will give;

$\frac{1}{1}\times 0.050 mol=0.050 mol$ of AgCl

Mass of 0.050 moles of AgCl ;

$0.050 mol\times 143.5 g/mol=7.175 g$

The amount of precipitate formed would 7.175 grams of silver chloride.

8 0

3 years ago

How many dekagrams are in a kilogram

Norma-Jean [14]

Answer:

1=100 1 Kilogram (kg) is equal to 100 dekagrams (dag). To convert kilograms to dekagrams, multiply the kilogram value by 100.

Explanation:

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1 year ago