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3 years ago

C.) Is heating of cobalt (ii) chloride crystals a chemical change or a physical change (Imk)

1 answer:

8 0

Answer: Heating the hydrated forms of cobalt chloride reverses the reactions above, returning cobalt chloride to the blue, water-free, or anhydrous, state. Water is "liberated" in these reactions, known as dehydration reactions.

Explanation:

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0.378 mL + 42.3 mL - 1.5833 mL

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2 years ago

A 12.0g sample of a substance requires 50.9 J to raise its temperature by 6.00 c. What is it’s specific heat

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$\qquad\qquad\huge\underline{{\sf Answer}}$

Let's find the specific heat of sample using given formula :

$\qquad \tt \dashrightarrow \:Q = m \: s \triangle T$

where,

Q = heat transfer

m = mass

S = specific heat

$\tt \triangle T$ = change in temperature

$\qquad \tt \dashrightarrow \:50.9 = 12 \times \: s \times 6$

$\qquad \tt \dashrightarrow \:s = \dfrac{50.9}{72}$

$\qquad \tt \dashrightarrow \:s = 0.707 \:$

Therefore, specific heat of that substance is 0.707 J/g°C

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1 year ago

When 20.00 mL of an unknown monoprotic acid is titrated with 0.125 M NaOH, it takes 15.00 mL to reach the endpoint. What is the

ohaa [14]

Answer:

About 0.0940 M.

Explanation:

Recall that NaOH is a strong base, so it dissociates completely into Na⁺ and OH⁻ ions. Because the acid is monoprotic, we can represent it with HA. Thus, the reaction between HA and NaOH is:

$\displaystyle \text{HA}_\text{(aq)} + \text{OH}^-_\text{(aq)} \longrightarrow \text{H$_2$O}_\text{($\ell$)} + \text{A}^-_\text{(aq)}$

Using the fact that it took 15.00 mL of NaOH to reach the endpoint, determine the number of HA that was reacted with:

$\displaystyle \begin{aligned} 15.00\text{ mL} &\cdot \frac{0.125\text{ mol NaOH}}{1\text{ L}} \cdot \frac{1\text{ L}}{1000\text{ mL}} \\ \\ &\cdot \frac{1\text{ mol OH}^-}{1\text{ mol NaOH}} \cdot \frac{1\text{ mol HA}}{1\text{ mol OH}^-}\\ \\ & = 0.00188\text{ mol HA}\end{aligned}$

Therefore, the molarity of the original solution was:

$\displaystyle \left[ \text{HA}\right] = \frac{0.00188\text{ mol}}{20.00\text{ mL}} \cdot \frac{1000\text{ mL}}{1\text{ L}} = 0.0940\text{ M}$

In conclusion, the molarity of the unknown acid is about 0.0940 M.

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2 years ago

C.) Is heating of cobalt (ii) chloride crystals a chemical change or a physical change (Imk)​

C.) Is heating of cobalt (ii) chloride crystals a chemical change or a physical change (Imk)