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MrRissso [65]
3 years ago
6

What would you observe when zinc is added to a dilute hydrochloric acid? Write the chemical reaction that takes place?

Chemistry
1 answer:
STatiana [176]3 years ago
6 0

Answer:

1) Zn(s) + 2HCl(aq) --> ZnCl2(aq) + H2(g)

You would see the reaction forming bubbles vigorously. Hydrogen gas is produced in the process.

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The main group elements do not include which elements?
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Transition metals

Explanation:

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What is the standard cell potential of a hypothetical cell made of Ni/Ni2+ and Zn/Zn2+
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4 years ago
(1 point) How many mL of water is added to dilute a 10 mL of 0.4 M solution to a new concentration of 0.1 M?
LiRa [457]

Answer:

The volume of water added is 30 mL

Explanation:

Given data

  • Initial concentration (C₁): 0.4 M
  • Initial volume (V₁): 10 mL
  • Final concentration (C₂): 0.1 M
  • Final volume (V₂): ?

We can find V₂ using the dilution rule.

C₁ × V₁ = C₂ × V₂

0.4 M × 10 mL = 0.1 M × V₂

V₂ = 40 mL

The volume of water added is

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5 0
4 years ago
Read 2 more answers
Which sample of gas at STP has the same number of molecules as 6 liters of Cl2(g) at STP?
lara [203]

Answer:

The correct answer is option 2  (6 liters of N2)

Explanation:

The complete question

Which sample of gas at STP has the same number  of molecules as 6 liters of Cl2(g) at STP?

(1) 3 liters of O2(g)

(2) 6 liters of N2(g)

(3) 3 moles of O2(g)

(4) 6 moles of N2(g)

Step 1: Data given

Volume = 6 L

STP = 1 atm and 273 K

Step 2: Calculate moles of Cl2

p*V = n*R*T

n = (p*V)/ (R*T)

⇒with n = the number of moles Cl2 = TO BE DETERMINED

⇒with V = the volume of Cl2 = 6.0 L

⇒with p = the pressure of Cl2 = 1 atm

⇒with R = the gas constant = 0.08206 L*atm/mol*K

⇒ with T = the temperature = 273 K

n = (1.0 * 6.0 ) (0.08206 * 273)

n = 0.2678 moles Cl2

This means we need 0.2678 moles of a gas at STP

Option 3 has 3 moles of O2 ⇒ Not the same number of molecules

Option 4 has 6 moles of N2  ⇒ Not the same number of molecules

For 3 liters of O2 we'll have:

⇒with n = the number of moles O2 = TO BE DETERMINED

⇒with V = the volume of O2 = 3.0 L

⇒with p = the pressure of O2 = 1 atm

⇒with R = the gas constant = 0.08206 L*atm/mol*K

⇒ with T = the temperature = 273 K

n = (1.0 * 3.0 ) (0.08206 * 273)

n = 0,1339 moles ⇒ Not the same number of molecules

For 6 liters of N2 we'll have

⇒with n = the number of moles N2 = TO BE DETERMINED

⇒with V = the volume of N2 = 6.0 L

⇒with p = the pressure of N2 = 1 atm

⇒with R = the gas constant = 0.08206 L*atm/mol*K

⇒ with T = the temperature = 273 K

n = (1.0 * 6.0 ) (0.08206 * 273)

n = 0.2678 moles N2 ⇒ The same number of molecules

The correct answer is option 2  (6 liters of N2)

3 0
3 years ago
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