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Semenov [28]
3 years ago
9

What volume of a 2.5 M stock solution of acetic acid (HC¿HÀO¿) is required to prepare 100.0 milliliters of a 0.50 M acetic acid

solution?
Chemistry
1 answer:
Harman [31]3 years ago
5 0
M₁ x V₁ = M₂  x V₂

2.5 x V₁ = 0.50 x 100.0

2.5 x V₁= 50

V₁ = 50 / 2.5

V₁= 20 mL 

hope this helps!
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2. What ions are present in what ratio in a solution of aqueous calcium chloride?
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Answer:

\mathrm{Ca}^{2+} \text { and } \mathrm{Cl} \text { - ions are present in } 1: 2 \text { ratio in a solution of aqueous calcium chloride. }

Explanation:

Here in Calcium Chloride ionic bond is present in between calcium and chlorine atoms. As we know according to Octet rule calcium have two excess atoms and for matching nearest noble gas electronic configuration. It donate two electrons to gain more stability and form \mathrm{Ca}^{2+}, while chlorine is deficient from one electron to meet nearest noble gas electronic configuration therefore two chlorine atoms accept excess electron from calcium individually and form two\mathrm{Cl}^{-} ions.

\text { Equation is as follows: } \mathrm{Ca}^{2+}+2 \mathrm{Cl}^{-} \rightarrow \mathrm{CaCl}_{2}

Hence aqueous solution of calcium chloride breaks the ionic bond pairing in one \mathrm{Ca}^{2+}and two\mathrm{Cl}^{-}ions: \mathrm{CaCl}_{2} \longrightarrow \mathrm{H}_{2} \mathrm{O} \quad \mathrm{Ca}^{2+}(\mathrm{ag})+2 \mathrm{Cl}(\mathrm{ag})

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3 years ago
Which action most likely causes large rocks to become smaller pieces of rock?
algol13

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Explanation:

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According to Newton's third law of motion, for every action force, there is
Nonamiya [84]

Answer:

A. an equal and opposite reaction force

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This means that there is a natural reaction of force

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What is the significance of enzymes in maintaining homeostasis in living organisms ?
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The elementary reaction 2 H 2 O ( g ) − ⇀ ↽ − 2 H 2 ( g ) + O 2 ( g ) proceeds at a certain temperature until the partial pressu
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6.75 × 10⁻⁸is the value of the equilibrium constant at this temperature.

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The expression of Kp for the given chemical equation is:

K_p = \frac{[H_2]^2[O_2]}{H_2O}

= \frac{(0.00150^2)(0.00150)}{(0.0500)} \\= 6.75 * 10^-^8

6.75 × 10⁻⁸is the value of the equilibrium constant at this temperature

7 0
3 years ago
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