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ollegr [7]
2 years ago
10

Al + AgNO3 -> Al(NO3)3 + Ag. As a balanced equation​

Chemistry
1 answer:
prisoha [69]2 years ago
4 0

Answer:

Al + 4AgNO3 >>Al(NO3)3+ 3Ag

Explanation:

the number of moles of No3 of the products is 3 therefore we have to balance the reactants by adding 3 before the "AgNO3" which also leades us to adding 3 mols to Ag on the products side

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kap26 [50]

Answer:

what you've asked so I guess the answer will be 10 days

8 0
2 years ago
what type of bond requires the give and take of electrons? a. ionic b. polar disalent c. polar covalent d. nonpolar covalent
Effectus [21]
Ionic bond involves electrostatic attraction between oppositely charged ions.
The ions are atoms that have gained 1 or more electrons and atoms that have lost 1 or more electrons.
Answer: The type of bond that requires the give and take of electrons is 
A ) ionic bond. 
3 0
3 years ago
Two processes are described below: Process A: rapid moving water strikes a rock on the bottom of a stream and breaks it into pie
worty [1.4K]

Answer:

both represent physical  weathering

Explanation:

7 0
3 years ago
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An indicator is what type of compound?
OleMash [197]
The letter (d.) strong base or acid would be the most appropriate answer to the question above. An indicator is a strong base or acid. Indicators are strong base and acid because through this you can determine if a compound is acid or base with its color.
3 0
3 years ago
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A 1.0 g sample of hydrogen reacts completely with 19.0 g of fluorine to form a compound of hydrogen and fluorine. a. What is the
Rashid [163]

Explanation:tr

a) Molar mass of HF = 20 g/mol

Atomic mass of hydrogen = 1 g/mol

Atomic mass of fluorine = 19 g/mol

Percentage of an element in a compound:

\frac{\text{Number of atoms of element}\times \text{Atomic mass of element}}{\text{Molar mass of compound }}\times 100

Percentage of fluorine:

\frac{1\times 19 g/mol}{20g/mol}\times 100=95\%

Percentage of hydrogen:

\frac{1\times 1g/mol}{20 g/mol}\times 100=5\%

b) Mass of hydrogen in 50 grams of HF sample.

Moles of HF = \frac{50 g}{20 g/mol}=2.5 mol

1 mole of HF has 1 mole of hydrogen atom.

Then 2.5 moles of HF will have:

1\times 2.5 mol=2.5 mol of hydrogen atom.

Mass of 2.5 moles of hydrogen atom:

1 g/mol × 2.5 mol = 2.5 g

2.5 grams of hydrogen would be present in a 50 g sample of this compound.

c) As we solved in part (a) that HF molecules has 5% of hydrogen by mass.

Then mass of hydrogen in 50 grams of HF compound we will have :

5% of 50 grams of HF = \frac{5}{100}\times 50 g=2.5 g

8 0
3 years ago
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