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3 years ago

What would you call the following chemical reaction, Zn(s) + H2O(g) – Zno (s) + H2(g)

Chemistry

1 answer:

egoroff_w [7]3 years ago

5 0

I'd say it's single replacement/displacement

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How many moles are equal to 83.4 L of O2?

ladessa [460]

Answer:

$3.72mol$

Explanation:

Hello,

In this case, we consider that at STP conditions (273 K and 1 atm) we know that the volume of 1 mole of a gas is 22.4 L, thereby, for 83.4 L, the resulting moles are:

$22.4L\rightarrow 1mol\\83.4L\rightarrow X\\X=\frac{83.4L*1mol}{22.4L}=3.72mol$

This is a case in which we apply the Avogadro's law which relates the volume and the moles as a directly proportional relationship.

Best regards.

8 0

3 years ago

A corpse discovered in the desert at 5:00 PM was found to have larvae from the blow fly species Phormia regina that had just dev

horrorfan [7]

Answer:

12.19 hours prior to discovery was the corpse placed in the desert.

Explanation:

Heat required for fly eggs to develop into first instar larvae = $Exposure\,time\times temperature$

From the given,

Exposure time = 16 hrs

Temperature = 27 C

$=16\times 27 = 427.2^{o}C$

The day time exposed hours is 10 hrs.-(7.00am-5.00pm)

$=10\times 37.8 = 378^{o}C$

So, the extra heat = 427.2-378= 49.2

The addition heat divided by average temperature.

Average temperature = 22.4 C

$=\frac{49.2}{22.4}= 2.19 hrs$

So, the total time exposed is night time hours to day time hours.

= 10+2.19 = 12.19 hrs.

Therefore, 12.19 hours prior to discovery was the corpse placed in the desert.

3 0

2 years ago

A solid is held in shape by strong forces.

n200080 [17]

Ok thanks for the valuble info.

8 0

3 years ago

Read 2 more answers

A student has an unknown sample. How can spectroscopy be used to identify the sample?

iogann1982 [59]

Answer: A. It can identify the elements in the sample.

Explanation: on edge

4 0

3 years ago

A buffer consists of 0.120 M HNO2 and 0.150 M NaNO2 at 25°C. pka of HNO2 is 3.40. a. What is the pH of the buffer? b. What is th

Mashcka [7]

Explanation:

It is known that $K_{a}$ of $HNO_{2}$ = $4.5 \times 10^{-4}$ .

(a) Relation between $K_{a}$ and $pK_{a}$ is as follows.

$pK_{a} = -log (K_{a})$

Putting the values into the above formula as follows.

$pK_{a} = -log (K_{a})$

= $-log(4.5 \times 10^{-4})$

= 3.347

Also, relation between pH and $pK_{a}$ is as follows.

pH = $pK_{a} + log\frac{[conjugate base]}{[acid]}$

= $3.347+ log \frac{0.15}{0.12}$

= 3.44

Therefore, pH of the buffer is 3.44.

(b) No. of moles of HCl added = $Molarity \times volume$

= $11.6 M \times 0.001 L$

= 0.0116 mol

In the given reaction, $NO^{-}_{2}$ will react with $H^{+}$ to form $HNO_{2}$

Hence, before the reaction:

No. of moles of $NO^{-}_{2}$ = $0.15 M \times 1.0 L$

= 0.15 mol

And, no. of moles of $HNO_{2}$ = $0.12 M \times 1.0 L$

= 0.12 mol

On the other hand, after the reaction :

No. of moles of $NO^{-}_{2}$ = moles present initially - moles added

= (0.15 - 0.0116) mol

= 0.1384 mol

Moles of $HNO_{2}$ = moles present initially + moles added

= (0.12 + 0.0116) mol

= 0.1316 mol

As, $K_{a}$ = $4.5 \times 10^{-4}$

$pK_{a} = -log (K_{a})$

= $-log(4.5 \times 10^{-4})$

= 3.347

Since, volume is both in numerator and denominator, we can use mol instead of concentration.

As, pH = $pK_{a} + log \frac{[conjugate base]}{[acid]}$

= 3.347+ log {0.1384/0.1316}

= 3.369

= 3.37 (approx)

Thus, we can conclude that pH after the addition of 1.00 mL of 11.6 M HCl to 1.00 L of the buffer solution is 3.37.

6 0

3 years ago