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3 years ago

As of January 2009, the USA has produced 60,000 metric tons of nuclear waste in 60 yesrs of operating 104 nuclear power plants.

Chemistry

2 answers:

goblinko [34]3 years ago

5 0

Answer: a. 0.8 tons p. month.

Explanation:

Given: Total nuclear waste = 60,000 metric tons

Time take = 60 years = 12 x 60 months [ 1 year = 12 months]

= 720 months

Total nuclear power plants = 104

Now , Average waste produced by each plant = $\dfrac{\text{Total waste}}{\text{Time x Number of plants}}$

$=\dfrac{60000}{720\times104}\approx0.8\text{ tons per month.}$

Hence, 0.8 tons p. month is produced by each plant.

So, option a. is correct.

Ipatiy [6.2K]3 years ago

3 0

Answer:

A on edge 2021

Explanation:

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3 years ago

46.6 grams of mercury II sulfate (HgSO4) reacts with an excess of sodium Chloride (NaCl). How many grams of mercury II chloride

slega [8]

Answer:

$m_{HgCl_2}=42.7gHgCl_2$

Explanation:

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In this case, the undergoing chemical reaction is:

$HgSO_4+2NaCl\rightarrow HgCl_2+Na_2SO_4$

In such a way, the mercury II sulfate (molar mass 296.65g/mol) is in a 1:1 molar ratio with the mercury II chloride (molar mass 271.52g/mol), for that reason the stoichiometry to find mass in grams of mercury II chloride turns out:

$m_{HgCl_2}=46.6gHgSO_4*\frac{1molHgSO_4}{296.65 gHgSO_4}*\frac{1molHgCl_2}{1molHgSO_4} *\frac{271.52gHgCl_2}{1molHgCl_2} \\\\m_{HgCl_2}=42.7gHgCl_2$

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3 years ago

What is the volume of the liquid in the this diagram?<br>Would it be 37 mL or 36.5 mL?

vivado [14]

The volume of the liquid in this diagram shown above would be equal to 36.5 mL.

<h3>What is a graduated cylinder?</h3>

A graduated cylinder is also known as measuring cylinder and it can be defined as a narrow, cylindrical piece of laboratory equipment with marked lines, which are used to measure the volume of a liquid.

In order to take a reading for the measurement of the volume of a liquid such as water, you should ensure that your eye level is even with the center of the meniscus.

In this scenario, the volume of the liquid in this diagram would be 36.5 mL because each of the small lines on the graduated cylinder measures 0.5 mL.

Read more on graduated cylinder here: brainly.com/question/24869562

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2 years ago

What is the most important use of an elements atomic number

Setler [38]

Answer:

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3 years ago

25.0cm3 of s saturated potassium hydroxide is neutralized by 35.0cm3 of hydrogen chloride acid of concentration 0.75 mol/dm3. Ca

Kamila [148]

Answer:

Concentration of the original $\rm KOH$ solution: approximately $1.05\; \rm mol \cdot dm^{-3}$ .

Explanation:

Notice that the concentration of the $\rm HCl$ solution is in the unit $\rm mol\cdot dm^{-3}$ . However, the unit of the two volumes is $\rm cm^{3}$ . Convert the unit of the two volumes to $\rm dm^{3}$ to match the unit of concentration.

$\begin{aligned} V(\mathrm{NaOH}) &= 25.0\; \rm cm^{3} \\ &= 25.0\; \rm cm^{3} \times \frac{1\; \rm dm^{3}}{1000\; \rm cm^{3}} \approx 0.0250\; \rm dm^{3} \end{aligned}$ .

$\begin{aligned} V(\mathrm{HCl}) &= 35.0\; \rm cm^{3} \\ &= 35.0\; \rm cm^{3} \times \frac{1\; \rm dm^{3}}{1000\; \rm cm^{3}} \approx 0.0350\; \rm dm^{3} \end{aligned}$ .

Calculate the number of moles of $\rm HCl$ molecules in that $0.0350\; \rm dm^{3}$ of $0.75\; \rm mol \cdot dm^{3}$ $\rm HCl\!$ solution:

$\begin{aligned}n(\mathrm{HCl}) &= c(\mathrm{HCl}) \cdot V(\mathrm{HCl}) \\ &= 0.00350\; \rm dm^{3} \times 0.75\; \rm mol \cdot dm^{3} \\ &\approx 0.02625\; \rm mol\end{aligned}$ .

$\rm HCl$ is a monoprotic acid. In other words, each $\rm HCl\!$ would release up to one proton $\rm H^{+}$ .

On the other hand, $\rm KOH$ is a monoprotic base. Each $\rm KOH\!$ formula unit would react with up to one $\rm H^{+}$ .

Hence, $\rm HCl$ molecules and $\rm KOH\!$ formula units would react at a one-to-one ratio.

${\rm HCl}\, (aq) + {\rm NaOH}\, (aq) \to {\rm NaCl}\, (aq) + {\rm H_2O}\, (l)$ .

Therefore, that $0.02625\; \rm mol$ of $\rm HCl$ molecules would neutralize exactly the same number of $\rm NaOH$ formula units. That is: $n(\mathrm{NaOH}) = 0.02625\; \rm mol$ .

Calculate the concentration of a $\rm NaOH$ solution where $V(\mathrm{NaOH}) = 0.0250\; \rm dm^{3}$ and $n(\mathrm{NaOH}) = 0.02625\; \rm mol$ :

$\begin{aligned}c(\mathrm{NaOH}) &= \frac{n(\mathrm{NaOH})}{V(\mathrm{NaOH})} \\ &= \frac{0.02625\; \rm mol}{0.0250\; \rm dm^{3}}\approx 1.05\; \rm mol \cdot dm^{-3}\end{aligned}$ .

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3 years ago