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3 years ago

2. What is the molarity of a HCl solution containing 1.2 grams of hydrochloric acid in 0.025 L of solution?

Chemistry

1 answer:

alisha [4.7K]3 years ago

3 0

Answer:

1.3M

Explanation:

Convert from grams to moles:

molar mass of HCl = 1.01g(molar mass of H) + 35.45g(molar mass of Cl) = 36.46g HCl

1.2g HCl (1mol HCl/36.46g HCl)

= .0329 mol HCl

Molarity = mol/L (important formula for concentration)

Plug your values in:

Molarity = .0329mol/.025L

1.317M - but you used two significant figures in the question, so:

1.3M

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barxatty [35]

Answer:

2 Moles

Explanation:

12 grams of carbon contains 1 mole

1 gram contains [1 mole/ 12 grams ] x 1 gram

24 grams contains [1 mole/ 12 grams ]* 24 gram = 2 moles.

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3 years ago

What are electrolytes

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An electrolyte is a substance that produces an electrically conducting solution when dissolved in a polar solvent, such as water. The dissolved electrolyte separates into cations and anions, which disperse uniformly through the solvent. Electrically, such a solution is neutral.

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3 years ago

For question #12, use the following picture: 12. What element is this? How do you know?

labwork [276]

Answer: The element shown in the image is Helium (He).

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Number of protons as shown in image = 2

Number of neutrons as shown in image = 2

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umka21 [38]

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3 years ago

When the following oxidation–reduction reaction in acidic solution is balanced, what is the lowest whole-number coefficient for

Svet_ta [14]

Answer:

c. 8, product side

Explanation:

In order to balance a redox reaction we use the ion-electron method, which has the following steps:

Step 1: identify oxidation and reduction half-reaction.

Oxidation: MnO₄⁻(aq) → Mn²⁺(aq)

Reduction: Br⁻(aq) → Br₂(l)

Step 2: perform the mass balance adding H⁺ and H₂O where necessary

8 H⁺(aq) + MnO₄⁻(aq) → Mn²⁺(aq) + 4 H₂O(l)

2 Br⁻(aq) → Br₂(l)

Step 3: perform the electrical balance adding electrons where necessary.

8 H⁺(aq) + MnO₄⁻(aq) + 5 e⁻ → Mn²⁺(aq) + 4 H₂O(l)

2 Br⁻(aq) → Br₂(l) + 2 e⁻

Step 4: multiply both half-reactions by numbers that secure that the number of electrons gained and lost are the same.

2 × (8 H⁺(aq) + MnO₄⁻(aq) + 5 e⁻ → Mn²⁺(aq) + 4 H₂O(l))

5 × (2 Br⁻(aq) → Br₂(l) + 2 e⁻)

Step 5: add both half-reactions side to side.

16 H⁺(aq) + 2 MnO₄⁻(aq) + 10 e⁻ + 10 Br⁻(aq) → 2 Mn²⁺(aq) + 8 H₂O(l) + 5 Br₂(l) + 10 e⁻

16 H⁺(aq) + 2 MnO₄⁻(aq) + 10 Br⁻(aq) → 2 Mn²⁺(aq) + 8 H₂O(l) + 5 Br₂(l)

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3 years ago