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Scorpion4ik [409]
3 years ago
15

Someone please help me please

Chemistry
1 answer:
mojhsa [17]3 years ago
3 0

Answer:

1. Solid

2. Melting Point

3. liquid

4. Boiling point

5. gas

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AlladinOne [14]

Formal charge can be calculated from the following formula

Formal charge = valency of central atom - (number of lone pair of electrons + number of covalent bonds)

a) for methylene:

Formal charge = 4 -( 2+ 2) = 0

b) For methyl free radical

Formal charge = 4- (3 +1) = 0

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3 years ago
Phosphorus trichloride gas and chlorine gas react to form phosphorus pentachloride gas: pcl3(g+cl2(g?pcl5(g. a 7.5-l gas vessel
Natasha2012 [34]
25655+6565++65+65+65+56+566+56+556+5+656+56+56+56+56+
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Which of the following is most likely to be covalent?<br> MgO <br> CO <br> CaO <br> BaO
Kisachek [45]

Answer:

CaO

Explanation:

CaO is the only compound that is a non-metal and a non-metal. The rest of the compounds are ionic, or metal and non-metal.

8 0
3 years ago
So I have an F in science, and everything is due today. Am I going to fail 9th grade because I didn't pass that class?
Elis [28]

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3 0
3 years ago
Read 2 more answers
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
3 years ago
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