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3 years ago

2. In most reactions, energy is converted into what?

Chemistry

1 answer:

Lelu [443]3 years ago

8 0

Chemical energy

After a reaction, a chemical substance is formed

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Brrunno [24]

Answer:it's either the first or second one

Explanation:

the layers of the earth

4 0

3 years ago

How many outer planets are there?

Ivenika [448]

Answer:

four outer planetsThe Outer Planet the four outer planets and the Sun, with sizes to scale. From left to right, the outer planets are Jupiter, Saturn, Uranus, and Neptune. The gas giants are mostly made of hydrogen and helium. These are the same elements that make up most of the Sun.

Explanation:

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Ari

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3 years ago

An aqueous solution of perchloric acid is standardized by titration with a 0.191 M solution of barium hydroxide. If 13.8 mL of b

WARRIOR [948]

Answer:

Molarity of $HClO_{4}$ solution is 0.311 M

Explanation:

Neutralization reaction: $2HClO_{4}+Ba(OH)_{2}\rightarrow Ba(ClO_{4})_{2}+2H_{2}O$

According to balanced reaction, 1 mol of $Ba(OH)_{2}$ neutralizes 2 moles of $HClO_{4}$

Number of moles of $Ba(OH)_{2}$ in 13.8 mL of 0.191 M $Ba(OH)_{2}$ solution = $\frac{0.191}{1000}\times 13.8moles$ = $0.00264moles$

Let's assume molarity of $HClO_{4}$ solution is C (M) then-

Number of moles of $HClO_{4}$ in 17.0 mL of C (M) $HClO_{4}$ solution = $\frac{C}{1000}\times 17.0moles$ = $0.017Cmoles$

Hence, $0.017C=(2\times 0.00264)$

or, $C=0.311$

So, molarity of $HClO_{4}$ solution is 0.311 M

3 0

3 years ago

The final volume of buffer solution must be 100.00 mL and the final concentration of the weak acid must be 0.100 M. Based on thi

Grace [21]

Answer:

0.387 g

Explanation:

pH of the buffer = 1

V = Volume of solution = 100 mL

[HA] = Molarity of HA = 0.1 M

$K_a$ = Acid dissociation constant = $1.2\times 10^{-2}$

(assuming base as $Na_2SO_410H_2O$ )

Molar mass of base = 322.2 g/mol

pKa is given by

$pK_a=-\log K_a\\\Rightarrow pKa=-\log(1.2\times 10^{-2})\\\Rightarrow pK_a=1.92$

From the Henderson-Hasselbalch equation we get

$pH=pK_a+\log\dfrac{[A^-]}{[HA]}\\\Rightarrow pH-pK_a=\log\dfrac{[A^-]}{[HA]}\\\Rightarrow 10^{pH-pK_a}=\dfrac{[A^-]}{[HA]}\\\Rightarrow [A^-]=10^{pH-pK_a}[HA]\\\Rightarrow [A^-]=10^{1-1.92}\times0.1\\\Rightarrow [A^-]=0.01202\ \text{M}$