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SashulF [63]
3 years ago
13

Please Help 3. Were there any metallic compounds that did not react with either the acid or the base? Write the type of metal, b

ased on your examination of the periodic table.
Chemistry
1 answer:
Norma-Jean [14]3 years ago
5 0

Answer:

1)KNO3 ( pottasium trioxonitrate (V)

2)Ca(NO3)2 (calcium trioxonitrate (V

Explanation:

1)KNO3 ( pottasium trioxonitrate (V)

2)Ca(NO3)2 (calcium trioxonitrate (V)

Thses two compounds are metallic compounds and does not react with either the acid or the base.

Write the type of metal, based on your examination of the periodic table?

1)KNO3 ( pottasium trioxonitrate (V) have a potassium element which belongs to group 2 of the Periodic table which is an alkaline metal and can react with water or steàm

2)Ca(NO3)2 (calcium trioxonitrate (V) has calcium metal element which is an alkaline earth metal in the Periodic table and they react with Hallogens

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What is the number of moles in 3.0 X 10^24 atoms of Carbon
evablogger [386]

Answer:the number of moles represented by 3.0 x 10^24 atoms of Ag is 0.500mol 0.500 m o l .

Explanation:

4 0
1 year ago
Answer the following for the reaction: 3AgNO3(aq)+Na3PO4(aq)→Ag3PO4(s)+3NaNO3(aq)
Brums [2.3K]

Answer:1) Volume of AgNO_3 required is 55.98 mL.

2) 0.62577 grams of Ag_3PO_4 is produced.

Explanation:

3AgNO_3(aq)+Na_3PO_4(aq)\rightarrow Ag_3PO_4(s)+3NaNO_3(aq)

1) Molarity of AgNO_3,M_1=0.225 M

Volume of AgNO_3.V_1=?

Molarity of Na_3PO_4,M_2=0.135 M

Volume of Na_3PO_4,V_2=31.1 mL=0.0311 L

Molarity=\frac{\text{number of moles}}{\text{volume of solution in liters}}

\text{number of moles }Na_3PO_4=M_2\times V_2=0.135 mol/L\times 0.0311 L=0.0041985 moles

According to reaction, 1 mole of Na_3PO_4 reacts with 3 mole of AgNO_3, then, 0.0041985 moles of Na_3PO_4 will react with:

\frac{3}{1}\times 0.0041985 moles of AgNO_3 that is 0.0125955 moles.

M_1=0.225 M=\frac{\text{number of moles of }AgNO_3}{V_1}

V_1=\frac{0.0125955 moles}{0.225 M}=0.05598 L=55.98 mL

Volume of AgNO_3 required is 55.98 mL.

2)

Molarity=0.195 M=\frac{\text{number of moles}}{\text{volume of solution in liters}}

Number of moles of AgNO_3=0.195\times 0.023 L=0.004485 moles

According to reaction, 3 moles of AgNO_3 gives 1 mole of Ag_3PO_4, then 0.004485 moles of AgNO_3 will give:\frac{1}{3}\times 0.004485 moles of Ag_3PO_4 that is 0.001495 moles.

Mass of Ag_3PO_4 =

Moles of Ag_3PO_4 × Molar Mass of Ag_3PO_4

= 0.001495 moles × 418.58 g/mol = 0.62577 g

0.62577 grams of Ag_3PO_4 is produced.

7 0
4 years ago
If 2.00 g of N2 occupies 1.25 L space, how many moles of N2
uranmaximum [27]

Answer:

n₂ =1.4 mol

Explanation:

Given data:

Mass of nitrogen = 2 g

Initial Volume occupy by nitrogen = 1.25 L

Final volume occupy by nitrogen = 25.0 L

Final number of moles = ?

Solution;

Formula:

V₁ / n₁ = V₂ / n₂

Number of moles of nitrogen:

Number of moles = mass/ molar mass

Number of moles = 2 g/ 28 g/mol

Number of moles = 0.07 mol

Now we will put the values in formula:

V₁ / n₁ = V₂ / n₂

n₂ = V₂× n₁ /V₁

n₂ = 25 L × 0.07 mol /  1.25 L

n₂ = 1.75 L. mol / 1.25 L

n₂ =1.4 mol

7 0
3 years ago
What is the volume occupied by 2.34 grams of CO2 gas at STP?​
krok68 [10]

Answer:

The volume occupied by 2.34 grams of CO2 gas at STP   is 1.18  L

8 0
3 years ago
Order frome largest to smallest <br>Allele,<br>Chromosome,<br>DNA, <br>nucleus of a cell​
svetlana [45]
Allele, dna, chromosome, nucleus of a cell
5 0
3 years ago
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