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Romashka-Z-Leto [24]
2 years ago
11

Avogadro’s number is a constant that helps scientists count atoms and molecules. It is approximately equal to 602,200,000,000,00

0,000,000,000. How is this number expressed in scientific notation? A. B. C. 6.02/1023 D. 6.02 × 1023 E. 6.02 × 10-23
Chemistry
2 answers:
djverab [1.8K]2 years ago
8 0

Answer : The correct option is, (D) 6.022\times 10^{23}

Explanation :

Scientific notation : It is defined as the way or representation of expressing the number that are too big or too small that is written in the decimal form. That means always written in the power of 10 form.

For example : 700 is written as, 7\times 10^2

As we are given that the value of Avogadro’s number is, 602,200,000,000,000,000,000,000

This number is written in scientific notation as :

6.022\times 10^{23}

Therefore, the correct option is, (D) 6.022\times 10^{23}

kondaur [170]2 years ago
5 0

Avogrado's number = 6.022 × 10^23



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The elements chlorine and iodine have similar chemical properties because they _____. Group of answer choices have the same numb
worty [1.4K]

Answer:

Have the same number of electrons in their outer energy levels

Explanation:

Elements in the same group have similar chemical properties because they have the same number of valence electron(s) in their outermost shell.

Chlorine and Iodine have similar chemical properties because they have the same number of valence electron in their outermost shell. This can be seen from their electronic configuration as shown below:

Cl (17) => 1s² 2s²2p⁶ 3s²3p⁵

I (53) => [Kr] 4d¹⁰ 5s²5p⁵

From the above illustration:

Outer shell of Cl (3s²3p⁵) = 2 + 5 = 7 electrons

Outer shell of I (5s²5p⁵) = 2 + 5 = 7 electrons

Since they have the same number of valence electrons, therefore, they will have similar chemical properties.

7 0
1 year ago
Identify the functional groups attached to the benzene ring as either, being electron withdrawing, electron donating, or neither
Dominik [7]
-OH is elctron donating  -C=-N is electron withdrawing  -O-CO-CH3 is electron withdrawing  -N(CH3)2 is electron donating  -C(CH3)3 is electron donating  -CO-O-CH3 is electron withdrawing  -CH(CH3)2 is electron donating  -NO2 is electrong withdrawing  -CH2
8 0
2 years ago
A student in a chemistry laboratory has access to two acid solutions. The first one is 20% acid and the second solution is 45% a
Ksju [112]

Volume of each solution : 60 ml 20% and 40 ml 45%

<h3>Further explanation</h3>

Given

20% and 45% acid

100 ml of 30% acid

Required

Volume of each solution

Solution

Molarity from 2 solutions :

Vm Mm = V₁. M₁ + V₂. M₂  

m = mixed solution

V = volume

M = molarity

V₁ = x ml

V₂ = (100 - x) ml

Input the value :

100 . 0.3 = x . 0.2 + (100-x) . 0.45

30 = 0.2x+45-0.45x

0.25x=15

x= 60 ml

V₁ = 60 ml

V₂ = 100 - 60 = 40 ml

7 0
2 years ago
A 220.0 gram piece of copper is dropped into 500.0 grams of water 24.00 °C. If the final temperature of water is 42.00 °C, what
FrozenT [24]

Answer:

C. 481 °C.

Explanation:

  • At equilibrium:

The amount of heat absorbed by water = the amount of heat released by copper.

  • To find the amount of heat, we can use the relation:

<em>Q = m.c.ΔT,</em>

where, Q is the amount of energy.

m is the mass of substance.

c is the specific heat capacity.

ΔT is the difference between the initial and final temperature (ΔT = final T - initial T).

<em>∵ Q of copper = Q of water</em>

∴ - (m.c.ΔT) of copper = (m.c.ΔT) of water

m of copper = 220.0 g, c of copper = 0.39 J/g °C, ΔT of copper = final T - initial T = 42.00 °C - initial T.

m of water = 500.0 g, c of water = 4.18 J/g °C, ΔT of water = final T - initial T = 42.00 °C - 24.00 °C = 18.00 °C.

∴ - (220.0 g)( 0.39 J/g °C)(42.00 °C - Ti) = (500.0 g)(4.18 J/g °C)(18.00 °C)

∴ - (85.8)(42.00 °C - Ti) = 37620.

∴ (42.00 °C - Ti) = 37620/(- 85.8) = - 438.5.

∴ Ti = 42.00 °C + 438.5 = 480.5°C ≅ 481°C.

<em>So, the right choice is: C. 481 °C.</em>

4 0
2 years ago
A piece of iron metal is heated to 155 degrees C and placed into a calorimeter that contains 50.0 mL of water at 18.7 degrees C.
Korvikt [17]

Answer:

D = 28.2g

Explanation:

Initial temperature of metal (T1) = 155°C

Initial Temperature of calorimeter (T2) = 18.7°C

Final temperature of solution (T3) = 26.4°C

Specific heat capacity of water (C2) = 4.184J/g°C

Specific heat capacity of metal (C1) = 0.444J/g°C

Volume of water = 50.0mL

Assuming no heat loss

Heat energy lost by metal = heat energy gain by water + calorimeter

Heat energy (Q) = MC∇T

M = mass

C = specific heat capacity

∇T = change in temperature

Mass of metal = M1

Mass of water = M2

Density = mass / volume

Mass = density * volume

Density of water = 1g/mL

Mass(M2) = 1 * 50

Mass = 50g

Heat loss by the metal = heat gain by water + calorimeter

M1C1(T1 - T3) = M2C2(T3 - T2)

M1 * 0.444 * (155 - 26.4) = 50 * 4.184 * (26.4 - 18.7)

0.444M1 * 128.6 = 209.2 * 7.7

57.0984M1 = 1610.84

M1 = 1610.84 / 57.0984

M1 = 28.21g

The mass of the metal is 28.21g

3 0
1 year ago
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