Answer:
The answer to your question is below
Explanation:
7.- Sublimation is when a solid turns into a gas without becoming a liquid.
8.- Solidification or freezing is when the temperature of a liquid diminishes
and it turns into a solid.
9.- Melting is the process when a solid change into a liquid because of the
temperature increases.
10.- Condensation is the change in the state of matter from gas phase to a liquid phase.
11.- Evaporation is the process when a liquid turns into a gas.
117 mL of 0.210 M K₂S solution
Explanation:
The question asks about the volume of 0.210 M K₂S (potassium sulfide) solution required to completely react with 175 mL of 0.140 M Co(NO₃)₂ (cobalt(II) nitrate).
We have the chemical reaction:
K₂S + Co(NO₃)₂ → CoS + 2 KNO₃
molar concentration = number of moles / volume (L)
number of moles = molar concentration × volume
number of moles of Co(NO₃)₂ = 0.140 × 175 = 24.5 mmoles
We see from the chemical reaction that 1 mmole of Co(NO₃)₂ is reacting with 1 mmole of K₂S, so 24.5 mmoles of Co(NO₃)₂ are reacting with 24.5 mmoles of K₂S.
volume = number of moles / molar concentration
volume of K₂S solution = 24.5 / 0.210 = 117 mL
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molar concentration
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The density, or volumetric mass<span> density, of a substance is its </span>mass per unit volume<span>. </span>
Answer:
This is on quizlet just search it :)
Explanation:
Answer:
1. V₁ = 2.0 mL
2. V₁ = 2.5 mL
Explanation:
<em>You are provided with a stock solution with a concentration of 1.0 × 10⁻⁵ M. You will be using this to make two standard solutions via serial dilution.</em>
To calculate the volume required (V₁) in each dilution we will use the dilution rule.
C₁ . V₁ = C₂ . V₂
where,
C are the concentrations
V are the volumes
1 refers to the initial state
2 refers to the final state
<em>1. Perform calculations to determine the volume of the 1.0 × 10⁻⁵ M stock solution needed to prepare 10.0 mL of a 2.0 × 10⁻⁶ M solution.</em>
C₁ . V₁ = C₂ . V₂
(1.0 × 10⁻⁵ M) . V₁ = (2.0 × 10⁻⁶ M) . 10.0 mL
V₁ = 2.0 mL
<em>2. Perform calculations to determine the volume of the 2.0 × 10⁻⁶ M solution needed to prepare 10.0 mL of a 5.0 × 10⁻⁷ M solution.</em>
C₁ . V₁ = C₂ . V₂
(2.0 × 10⁻⁶ M) . V₁ = (5.0 × 10⁻⁷ M) . 10.0 mL
V₁ = 2.5 mL