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3 years ago

10 points

Chemistry

1 answer:

Murljashka [212]3 years ago

7 0

Answer:

2. The metal would lose one electrons and the non metal would gain one electrons

Explanation:

An atom of a certain element reacts with the atoms of other elements in order to fullfill its outermost shell (called valence shell).

We notice the following:

- The elements in Group 1 (which are metals) have only 1 electron in their valence shell

- The elements in Group 17 (which are non-metals) have 1 vacancy (lack of electron) in their valence shell

This means that in order for both an atom of group 1 and an atom of group 17 to fullfill the valence shell, they have to:

- The atom in group 1 has to give away its only electron of the valence shell

- The atom in group 17 has to gain one electron in order to fullfill the shell

Therefore, the correct option is

2. The metal would lose one electrons and the non metal would gain one electrons

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An undergraduate weighed out 20grams of sodium hydroxide pellets. If Na =23, O = 16 andH = 1, What is the mole of this sodium hy

tatuchka [14]

The answer would be .5 mols because you take the total amount of grams, which is 20, and you had up the molar mass of sodium hydroxide, which would be 40. After you have this you would set this up as a stochiometry equation. With 1 mol on top you dived 20/40 to cancel out your grams. This leaves you with .5 mols

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3 years ago

20 N force moving a box to 30 meters distance. How much work id done? if it takes 10 sec to do this work, how much power is used

dalvyx [7]

Answer:

Work done = 600 J

Power used = 60 W

Explanation:

Given:

Force acting on the box is, $F=20\ N$

Displacement of the box is, $S=30\ m$

Time taken for the work, $t=10\ s$

Now, we know that, work is said to be done by a force only when there is displacement caused by the force in its direction.

Here, the force acting on the box causes a displacement of 30 m in its direction. So, work done is equal to the product of force and displacement caused.

Therefore, work done on the box is given as:

$Work=Force\times Displacement\\Work=F\times S\\Work=(20\ N)\times (30\ m)\\Work=600\ J$

Therefore, the work done is 600 J.

Now, we know that, power is given as work done per unit time.

So, power used is given as:

$Power=\frac{Work}{Time}\\\\Power=\frac{600\ J}{10\ s}\\\\Power=60\ W$

Therefore, the power used is 60 W.

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3 years ago

If a bag of peas weighs 454 grams, how many peas would be in the bag?

OverLord2011 [107]

In a bag of peas that weighs 454 grams, there are between 1261 and 4540 peas.

The average pea weighs between 0.1 and 0.36 grams.

If we take the lower value (0.1 g/pea), the number of peas in 454 g is:

$454 g \times \frac{1pea}{0.1 g} = 4540pea$

If we take the higher value (0.36 g/pea), the number of peas in 454 g is:

$454 g \times \frac{1pea}{0.36 g} \approx 1261pea$

In a bag of peas that weighs 454 grams, there are between 1261 and 4540 peas.

You can learn more about conversion factors here: brainly.com/question/1844638

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3 years ago

A gas used to extinguish fires is composed of 75 % CO2 and 25 % N2. It is stored in a 5 m3 tank at 300 kPa and 25 °C. What is th

tatyana61 [14]

Answer : The partial pressure of the $CO_2$ in the tank in psia is, 32.6 psia.

Explanation :

As we are given 75 % $CO_2$ and 25 % $N_2$ in terms of volume.

First we have to calculate the moles of $CO_2$ and $N_2$ .

$\text{Moles of }CO_2=\frac{\text{Volume of }CO_2}{\text{Volume at STP}}=\frac{75}{22.4}=3.35mole$

$\text{Moles of }N_2=\frac{\text{Volume of }N_2}{\text{Volume at STP}}=\frac{25}{22.4}=1.12mole$

Now we have to calculate the mole fraction of $CO_2$ .

$\text{Mole fraction of }CO_2=\frac{\text{Moles of }CO_2}{\text{Moles of }CO_2+\text{Moles of }N_2}$

$\text{Mole fraction of }CO_2=\frac{3.35}{3.35+1.12}=0.75$

Now we have to calculate the partial pressure of the $CO_2$ gas.

$\text{Partial pressure of }CO_2=\text{Mole fraction of }CO_2\times \text{Total pressure of gas}$

$\text{Partial pressure of }CO_2=0.75mole\times 300Kpa=225Kpa=225Kpa\times \frac{0.145\text{ psia}}{1Kpa}=32.625\text{ psia}$

conversion used : (1 Kpa = 0.145 psia)

Therefore, the partial pressure of the $CO_2$ in the tank in psia is, 32.6 psia.

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