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Alja [10]
3 years ago
11

Silver nitrate can be produced by dissolving metallic silver in concentrated nitric acid, with an addition product being hydroge

n gas. Write the chemical formula for this reaction.
Chemistry
1 answer:
geniusboy [140]3 years ago
4 0

Answer:

2Ag(s) + 2HNO₃ (aq)  →  2AgNO₃(aq)  +  H₂(g)

Explanation:

Metallic silver = Ag (s)

Concentrated nitric acid = HNO₃ (aq)

Product being hydrogen gas = H₂(g)

2Ag(s) + 2HNO₃ (aq)  →  2AgNO₃(aq)  +  H₂(g)

Be careful AgNO₃ doesn't precipitate

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A 2.2 M solution is made by with 0.45 moles of a solute. What is the final volume of this solution?
Savatey [412]

Answer: The final volume of this solution is 0.204 L.

Explanation:

Given: Molarity of solution = 2.2 M

Moles of solute = 0.45 mol

Molarity is the number of moles of solute present divided by volume in liters.

Molarity = \frac{no. of moles}{Volume (in L)}

Substitute the values into above formula as follows.

Molarity = \frac{no. of moles}{Volume (in L)}\\2.2 M = \frac{0.45}{Volume}\\Volume = 0.204 L

Thus, we can conclude that the final volume of this solution is 0.204 L.

7 0
3 years ago
The reaction NA3PO4(aq) + 3 AgNO3(aq) → Ag3PO4() + 3NaNO 3( aq) is best classified as a(n)?
Natasha2012 [34]

Answer:

Double replacement reaction.

Explanation:

The Na and Ag atoms both (double) trade places (replacement) with each other.  

8 0
2 years ago
How many moles of neon occupy a volume of 14.3 l at stp? how many moles of neon occupy a volume of 14.3 l at stp? 1.57 moles 0.6
andriy [413]
The  number   of  neon  moles  that  occupy  a volume of 14.3 l  at STP is calculated as follows

At STP 1  mole = 22.4 liters

what about  14.3 liters

by cross  multiplication
= (1 mole x 14.3 l)/22.4 l =0.638  moles  of  neon


5 0
3 years ago
What is the mass of and object with a volume of 100 cm3 and a density of 10 g/cm3?
nadezda [96]

Answer:

1000 g

Explanation:

d = m/v

We are given d: 10g/cm3

and v: 100cm3

Plug them into the equation to get 10 = m/100

Then, cross multiply 10x100 to get mass which is: 1000g

4 0
3 years ago
The reaction of hydrogen and iodine to produce hydrogen iodide has a Kc of 54.3 at 703 K. Given the initial concentrations of H2
pentagon [3]

Answer:

[HI] = 0.7126 M

Explanation:

Step 1: Data given

Kc = 54.3

Temperature = 703 K

Initial concentration of H2 and I2 = 0.453 M

Step 2: the balanced equation

H2 + I2 ⇆ 2HI

Step 3: The initial concentration

[H2] = 0.453 M

[I2] = 0.453 M

[HI] = 0 M

Step 4: The concentration at equilibrium

[H2] = 0.453 - X

[I2] = 0.453 - X

[HI] = 2X

Step 5: Calculate Kc

Kc = [Hi]² / [H2][I2]

54.3 = 4x² / (0.453 - X(0.453-X)

X = 0.3563

[H2] = 0.453 - 0.3563 = 0.0967 M

[I2] = 0.453 - 0.3563 = 0.0967 M

[HI] = 2X = 2*0.3563 = 0.7126 M

3 0
3 years ago
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