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Vilka [71]
3 years ago
15

Who’s is a difference between the focus of a nuclear engineer and the job of a nuclear scientist

Chemistry
1 answer:
yawa3891 [41]3 years ago
4 0

The primary difference between them is what they work with and what their goals are. Chemical engineers research and design production processes for things like food or gasoline while nuclear engineers focus on ways to harness and use radiation and nuclear energy.

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Which answer best describes what is happening in the following redox reaction?
kotegsom [21]

Answer:

you

Explanation:

not like that

7 0
3 years ago
What family contains the most<br> reactive nonmetals: halogen or alkali?
mart [117]
Halogen are the most reactive due to their electronic configuration
5 0
3 years ago
2.4 g of a solid is added to 91 g of water. What is the mass % concentration of the solution?
vladimir2022 [97]

Answer:

2.64%

Explanation:

mass percent = (grams of solute / grams of solution) x 100

mass percent = (2.4 / 91) × 100

mass percent = 2.64% to 3sf

8 0
1 year ago
When 70.4 g of benzamide (C7H7NO) are dissolved in 850. g of a certain mystery liquid X, the freezing point of the solution is 2
Arlecino [84]

Answer:

1.62

Explanation:

From the given information:

number of moles of benzamide  =\dfrac{70.4 \ g}{121.14 \ g/mol}

= 0.58 mole

The molality = \dfrac{mass \ of \ solute (i.e. \ benzamide )}{mass \ of \ solvent  }

= \dfrac{0.58 }{0.85 }

= 0.6837

Using the formula:

\mathbf {dT  = l   \times  k_f  \times m}

where;

dT = freezing point = 27

l = Van't Hoff factor = 1

kf = freezing constant of the solvent

∴

2.7 °C = 1 × kf ×  0.6837 m

kf = 2.7 °C/ 0.6837m

kf = 3.949 °C/m

number of moles of NH4Cl = \dfrac{70.4 \ g}{53.491 \  g /mol}

= 1.316 mol

The molality = \dfrac{1.316 \ mol}{0.85 \ kg}

= 1.5484

Thus;

the above kf value is used in determining the  Van't Hoff factor for  NH4Cl

i.e.

9.9 = l × 3.949 × 1.5484 m

l = \dfrac{9.9}{3.949 \times 1.5484 \ m}

l = 1.62

5 0
2 years ago
∆G° for the process benzene (l) benzene (g) is 3.7 Kj/mol at 60 °C , calculate the vapor pressure of benzene at 60 °C [R=0.0821
Alex787 [66]

Answer:

4) 0.26 atm

Explanation:

In the process:

Benzene(l) → Benzene(g)

ΔG° for this process is:

ΔG° = -RT ln Q

<em>Where Q = P(Benzene(g)) / P°benzene(l) P° = 1atm</em>

ΔG° = 3700J/mol = -8.314J/molK * (60°C + 273.15) ln P(benzene) / 1atm

1.336 = ln P(benzene) / 1atm

0.26atm = P(benzene)

Right answer is:

<h3>4) 0.26 atm </h3><h3 />
6 0
3 years ago
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