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3 years ago

Which of the following elements has the lowest first ionizition energy ?

Chemistry

2 answers:

Hatshy [7]3 years ago

5 0

Answer:

I think Mg is correct I am not sure

Andrei [34K]3 years ago

4 0

Sodium (Na)

if u look at the ionization energy for each one its the lowest.

Li: 520 Kj/Mol

Mg:737.50

Na: 495.8

Al: 577.5

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What amount if force need to be applied to a 20 kg bowling ball to give it an acceleration of 5m/s

balandron [24]

Answer:

100N

Explanation:

Given parameters:

Mass of the bowling ball = 20kg

Acceleration = 5m/s²

Unknown:

Amount of force applied = ?

Solution:

To solve this problem, we apply newton's second law of motion.

Force = mass x acceleration

Now insert the parameters and solve;

Force = 20 x 5 = 100N

5 0

3 years ago

Which pair shares the same empirical formula?

ZanzabumX [31]

Answer:- 3. $CH_3$ and $C_2H_6$

Explanations:- An empirical formula is the simplest whole number ratio of atoms of each element present in the molecule/compound.

For example, the molecular formula of benzene is $C_6H_6$ . The ratio of C to H in it is 6:6 that could be simplified to 1:1. So, an empirical formula of benzene is CH.

In the first pair, the ratio of C to H in first molecule is 2:4 that could be simplified to 1:2 and the empirical formula is $CH_2$ . In second molecule the ratio of C to H is 6:6 and it could be simplified to 1:1. and the empirical formula is CH. Empirical formulas are different for both the molecules of first pair and so it is not the right choice.

In second pair, C to H ratio in first molecule is 1:2, so the empirical formula is $CH_2$ . The C to H ratio for second molecule is 1:4, so the empirical formula is $CH_4$ . Here also, the empirical formulas are not same and hence it is also not the right choice.

In third pair, C to H ratio in first molecule is 1:3, so the empirical formula is $CH_3$ . In second molecule the C to H ratio is 2:6 and it is simplified to 1:3. So, the empirical formula for this one is also $CH_3$ . Hence. this is the correct choice.

In fourth pair, first molecule empirical formula is CH. Second molecule has 2:4 that is 1:2 mole ratio of C to H and so its empirical formula is $CH_2$ . As the empirical formulas are different, it is not the right choice.

So, the only and only correct pair is the third one. 3. $CH_3$ and $C_2H_6$

4 0

3 years ago

Read 2 more answers

7. A square chunk of plastic has a length of 5 cm, width of 5 cm and height of 5 cm. It has a mass of 200 g. What is its density

BaLLatris [955]

Answer:

$d=1.6\ g/cm^3$

Explanation:

Given that,

The dimension of a square chunk of plastic is 5 cm × 5 cm × 5 cm

Mass, m = 200 g

We need to find its density

Density = mass/volume

It will form a cube. It volume = side³

So,

$d=\dfrac{200\ g}{5^3\ cm^3}\\\\d=1.6\ g/cm^3$

Hence density is $1.6\ g/cm^3$

5 0

4 years ago

Determine the theoretical yield of HCl if 60.0 g of BC13 and 37.5 g of H20 are reacted according to the following balanced react

Pie

Answer : The theoretical yield of HCl is, 56.1735 grams

Explanation : Given,

Mass of $BCl_3$ = 60 g

Mass of $H_2O$ = 37.5 g

Molar mass of $BCl_3$ = 117 g/mole

Molar mass of $H_2O$ = 18 g/mole

Molar mass of $HCl$ = 36.5 g/mole

First we have to calculate the moles of $BCl_3$ and $H_2O$ .

$\text{Moles of }BCl_3=\frac{\text{Mass of }BCl_3}{\text{Molar mass of }BCl_3}=\frac{60g}{117g/mole}=0.513moles$

$\text{Moles of }H_2O=\frac{\text{Mass of }H_2O}{\text{Molar mass of }H_2O}=\frac{37.5g}{18g/mole}=2.083moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$BCl_3(g)+3H_2O(l)\rightarrow H_3BO_3(s)+3HCl(g)$

From the balanced reaction we conclude that

As, 1 mole of $BCl_3$ react with 3 mole of $H_2O$

So, 0.513 moles of $BCl_3$ react with $3\times 0.513=1.539$ moles of $H_2O$

From this we conclude that, $H_2O$ is an excess reagent because the given moles are greater than the required moles and $BCl_3$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $HCl$ .

As, 1 mole of $BCl_3$ react to give 3 moles of $HCl$

So, 0.513 moles of $BCl_3$ react to give $3\times 0.513=1.539$ moles of $HCl$

Now we have to calculate the mass of $HCl$ .

$\text{Mass of }HCl=\text{Moles of }HCl\times \text{Molar mass of }HCl$

$\text{Mass of }HCl=(1.539mole)\times (36.5g/mole)=56.1735g$

Therefore, the theoretical yield of HCl is, 56.1735 grams

7 0

3 years ago

How does the temperature of absolute zero relate to the kinetic energy of a substance?

NeX [460]

At absolute zero the kinetic energy of the substance will be 0

7 0

3 years ago