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andrey2020 [161]

3 years ago

10

A ____________ is a large body that moves around a star?

2 answers:

Travka [436]3 years ago

4 0

Answer:

A Planet

Explanation:

The earth for example, is a large body that orbits the sun, our local star

Makovka662 [10]3 years ago

4 0

Answer:

Planet

Explanation:

Just like Earth orbits the sun (a star).

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The town of Natrium, West Virginia, derives its name from the sodium produced in the electrolysis of molten sodium chloride (NaC

Alona [7]

Explanation:

The reaction equation will be as follows.

$Na^{+} + e^{-} \rightarrow Na(s)$

Hence, moles of Na = moles of electron used

Therefore, calculate the number of moles of sodium as follows.

No. of moles = $\frac{mass}{\text{molar mass}}$

= $\frac{4500 g}{23 g/mol}$ (as 1 kg = 1000 g)

= 195.65 mol

As, Q = $n \times F$ where F = Faraday's constant

= $195.65 mol \times 96500 C$

= $1.88 \times 10^{7}$ mol C

Relation between electrical energy and Q is as follows.

E = $Q \times V$

Hence, putting the given values into the above formula and then calculate the value of electricity as follows.

E = $Q \times V$

= $1.88 \times 10^{7} \times 5$

= $9.4 \times 10^{7} J$

As 1 J = $2.77 \times 10^{-7}$ kWh

Hence, $\frac{9.4 \times 10^{7}}{2.77 \times 10^{-7}}$ kWh

= 3.39 kWh

Thus, we can conclude that 3.39 kilowatt-hours of electricity is required in the given situation.

7 0

3 years ago

Read 2 more answers

A solution is made by mixing of water and of acetic acid . Calculate the mole fraction of water in this solution. Be sure your a

Molodets [167]

Answer:

The answer is "0.35".

Explanation:

please find the complete question in the attached file.

Mass of $CH_3C0_2 H$ $= 77 \ g \\\\$

Molar mass of $CH_3C0_2 H$ $= 60.05 \ \frac{g}{mol} \\\\$

No of moles in $n_{CH_3CO_2H}$ $= 77 \ g \times \frac{1 \ mol \ CH_3C0_2H}{60.05 \ g}$

$= 1.28 \ mol \ CH_3CO_2H$

Mass of water $(H_2O)$ $= 42 \ g$

The molar mass of water $(H_2O)$ $= 18.02 \ \frac{g}{mol}$

moles of water $n_{H_2O}$ :

$= 42 \ g \times \frac{1 mol H_2O}{18.02 \ g} \\\\= 2.33 \ mol \ H_2 O$

Molfraction of acetic acid:

$X CH_2CO_2H = \frac{n_{CH_3CO_2H}}{n_CH_3CO_2H +n_{H_2}}\\\\$

$=\frac{1.28 \ mol}{1.28 \ mo1 + 2.33 mol}\\\\ = 0.354\\\\= 0.35$

7 0

3 years ago

A. 6.07 x 10-15<br> How many significant figures are there in this equation

Harman [31]

Answer:

3

Explanation:

In scientific notation, the number of integers in the first part will always be the amount of significant figures.

For example,

1) 1.542505 x 10^17 =7 significant figures

1) 1.54 x 10^17 =3 significant figures

8 0

3 years ago

26.5 g of a solution with a density of 7.48 g/mL

Cloud [144]

Answer: Assuming the question: 3.54 ml (3 sig figs)

Explanation:

I don't see a question, but will assume it is "What volume is needed to obtain 26.5 grams?

If so:

(26.5 g)/(7.48 ml) = 3.54 ml (3 sig figs)

6 0

3 years ago

URGENTE!!!!!!!!!!!!!!

yawa3891 [41]

Answer:

English language please give me a

3 0

3 years ago