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3 years ago

9

Which pair of elements would most likely bond to form a covalently bonded compound?

1 answer:

anastassius [24]3 years ago

3 0

Answer:

two nonmetal elements join together to form covalent compounds

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Pls help ASAP........ A student was roasting marshmallows over a fire. The student

Phantasy [73]

Answer:

A chemical change

Explanation:

The marshmallow turning brown and bubbling implies that a chemical change has taken place.

For chemical changes to occur, we observe any of the following:

a new kind of matter is formed.
it is always accompanied by energy changes
the process is not easily reversible
it involves a change in mass
requires considerable amount of energy.

ii. Two signs that shows a chemical change has taken place is that:

bubbles are being formed as it is roasted and it implies that new substances have been formed.
also, significant amount of heat energy is supplied for the roasting.

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2 years ago

2. What is the name of the following compound? *

alexgriva [62]

Answer: Cesium Nitrate

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3 years ago

How might the government make sure scientific research is done in an ethical<br> way?

antiseptic1488 [7]

Answer:

They could possibly make daily check ups to the research facility?

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3 years ago

Calculate the number of molecules of sulphur S8 in 8 gram of solid Sulphur

Luden [163]

Answer:

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Explanation:

test told me

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3 years ago

If a 95.27 mL sample of acetic acid (HC2H3O2) is titrated to the equivalence point with 79.06 mL of 0.113 M KOH, what is the pH

KATRIN_1 [288]

Answer:

8.73

Explanation:

The concentration of acetic acid can be determined as follows:

$M_1V_1 = M_2V_2\\(KOH) = (CH_3COOH)$

$M_{KOH}=0.113 M\\V_{KOH}=79.06 mL$

$V_{CH_3COOH}=95.27 \\\\M_{CH_3COOH)=?????$

$M_{CH3COOH} = \frac{M_{KOH}*V_{KOH}}{V_{CH_3COOH}}$

$M_{CH3COOH} = \frac{0.113*79.06}{95.27}$

$M_{CH3COOH} = 0.094 M$

Moles of $CH_3COOH$ = $95.27* 10^{-3}* 0.094$

=0.0090 moles

Moles of $KOH = 79.06*10^{-3}*0.113$

= 0.0090 moles

The equation for the reaction can be expressed as :

$CH_3COOH$ $+$ $KOH$ -----> $CH_3COO^{-}K^+$ $+$ $H_2O$

Concentration of $CH_3COO^{-$ ion = $\frac{0.0090}{Total volume (L)}$

= $\frac{0.0090}{(95.27+79.06)} *1000$

= 0.052 M

Hydrolysis of $CH_3COO^{-$ ion:

$CH_3COO^{-$ $+$ $H_2O$ -----> $CH_3COOH$ $+$ $OH^-$

$K = \frac{K_w}{K_a} = \frac{x*x}{0.052-x}$

⇒ $\frac{10^{-14}}{1.82*10^{-5}}= \frac{x*x}{0.052-x}$

= $0.5494*10^{-9}= \frac{x*x}{0.052-x}$

As K is so less, then x appears to be a very infinitesimal small number

0.052-x ≅ x

$0.5494*10^{-9}= \frac{x^2}{0.052}$

$x^2 = 0.5494*10^{-9}*0.052$

$x^2 = 0.286*10^{-10$

$x = \sqrt{0.286*10^{-10$

$x =0.535*10^{-5}M$

$[OH] = x =0.535*10^{-5}$

$pOH = -log[OH^-]$

$pOH = -log[0.535*10^{-5}]$

$pOH = 5.27$

pH = 14 - pOH

pH = 14 - 5.27

pH = 8.73

Hence, the pH of the titration mixture = 8.73

8 0

3 years ago