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yulyashka [42]
3 years ago
15

Why do atoms of elements take part in chemical reaction?​

Chemistry
1 answer:
Marina CMI [18]3 years ago
4 0

Explanation:

During a chemical reaction, the atoms of the original substances gain, lose or share their electrons with those of the substances with which they are reacting. The reaction creates new substances made up of a new combination of atoms and a different configuration of electrons.

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If you find that a substance turns litmus paper to blue and contains hydrogen ions it is most likely what
Darya [45]
An acid, possibly a strong one
4 0
3 years ago
The Gateway Arch in St. Louis, MO is approximately 630 ft tall. How many U.S. dimes would be in a stack of the same
miskamm [114]

Answer:

142240

Explanation:

We are told in the question:

Height of Gateway Arch in St. Louis, MO = 630ft

We are asked, how many U.S. dimes would be in a stack of the same

height when 1 dime is 1.35 mm thick.

Step 1

Convert height in ft to mm

1 ft = 304.8 mm

630ft =

Cross Multiply

630ft × 304.8mm/1ft

= 192024 mm.

Step 2

To find how many US dimes would be in a stack of the same height

= Total thickness/ Thickness of 1 US dime

= 192024 mm/1.35mm

= 142240

Therefore, the number of dimes that would be in a stack of the same

height is 142240

6 0
3 years ago
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alexandr402 [8]

Answer:

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Explanation:

6 0
3 years ago
Equal moles of H2, N2, O2, and He are placed into separate containers at the same temperature. Assuming each gas behaves ideally
lbvjy [14]

Answer:

They would all exhibit the same pressure.

Explanation:

Since the same number of mole of each gas is placed in different containers, it means the gas will occupy the same volume.

Now, the gases were observed at the same temperature. This means they will all have the same pressure as their volume is the same.

Now we can further understand this by doing a simple calculation as follow:

Assumptions:

For H2:

Number of mole (n) = 1 mole

Volume (V) = 22.4L

Temperature (T) = 298K

Gas constant (R) = 0.0821 atm.L/Kmol

Pressure =..?

PV = nRT

Divide both side V

P = nRT /V

P = 1 x 0.0821 x 298 / 22.4

P = 1 atm

Therefore, H2 has a pressure of 1 atm.

For N2:

Number of mole (n) = 1 mole

Volume (V) = 22.4L

Temperature (T) = 298K

Gas constant (R) = 0.0821 atm.L/Kmol

Pressure =..?

PV = nRT

Divide both side V

P = nRT /V

P = 1 x 0.0821 x 298 / 22.4

P = 1 atm

Therefore, N2 has a pressure of 1 atm

For O2:

Number of mole (n) = 1 mole

Volume (V) = 22.4L

Temperature (T) = 298K

Gas constant (R) = 0.0821 atm.L/Kmol

Pressure =..?

PV = nRT

Divide both side V

P = nRT /V

P = 1 x 0.0821 x 298 / 22.4

P = 1 atm

Therefore, O2 has a pressure of 1 atm

For He:

Number of mole (n) = 1 mole

Volume (V) = 22.4L

Temperature (T) = 298K

Gas constant (R) = 0.0821 atm.L/Kmol

Pressure =..?

PV = nRT

Divide both side V

P = nRT /V

P = 1 x 0.0821 x 298 / 22.4

P = 1 atm

Therefore, He has a pressure of 1 atm.

From the above illustrations we can see that the gases have the same pressure since they have the same number of mole, volume and were observed at the same temperature.

4 0
3 years ago
A 5.00g of X, the product of organic synthesis is obtained in a 1.0 dm3 aqueous solution. Calculate the mass of X that can be ex
Anestetic [448]

Answer:

mass of X extracted from the aqueous solution by 50 cm³ of ethoxy ethane = 3.33 g

Explanation:

The partition coefficient of X between ethoxy ethane (ether) and water, K is given by the formula

K = concentration of X in ether/concentration of X in water

Partition coefficient, K(X) between ethoxy ethane and water = 40

Concentration of X in ether = mass(g)/volume(dm³)

Mass of X in ether = m g

Volume of ether = 50/1000 dm³ = 0.05 dm³

Concentration of X in ether = (m/0.05) g/dm³

Concentration of X in water = mass(g)/volume(dm³)

Mass of X in water left after extraction with ether = (5 - m) g

Volume of water = 1 dm³

Concentration of X in water = (5 - m/1) g/dm³

Using K = concentration of X in ether/concentration of X in water;

40 = (m/0.05)/(5 - m)

(m/0.05) = 40 × (5 - m)

(m/0.05) = 200 - 40m

m = 0.05 × (200 - 40m)

m = 10 - 2m

3m = 10

m = 10/3

m = 3.33 g of X

Therefore, mass of X extracted from the aqueous solution by 50 cm³ of ethoxy ethane = 3.33 g

8 0
3 years ago
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