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alekssr [168]
3 years ago
9

The mass numbers for two isotopes are unequal because they have different numbers of

Chemistry
2 answers:
Savatey [412]3 years ago
8 0

<u>Answer:</u> It is so because they have different number of neutrons.

<u>Explanation:</u>

Isotopes are defined as the chemical species which have same number of protons but differ in the number of neutrons.

They have same atomic number but different atomic mass.

<u>For Example:</u>  _6^{12}\textrm{C},_6^{13}\textrm{C}\text{ and }_6^{14}\textrm{C} are the three isotopes of carbon.

The number of protons in all the three isotopes is same which is 6 and the number of neutrons in the three isotopes are 6, 7 and 8 respectively.

Hence, it is so because they have different number of neutrons.

SSSSS [86.1K]3 years ago
5 0
The mass numbers for two isotopes are unequal because they have different numbers of NEUTRONS.

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2. A block of aluminum with a mass of 140g is cooled from 98.4oC to 62.2oC with a release of 1137J of heat. From these data, cal
NeTakaya
Q =  M * C *ΔT

Q / <span>ΔT  = M

</span>Δf - Δi =  98.4ºC - 62.2ºC = 36.2ºC
<span>
C = 1137 J / 140 * 36.2

C = 1137 / 5068

C = 0.224 J/gºC</span>
8 0
3 years ago
If the initial temperature of a movable cylinder was 50 degrees Celsius
slega [8]

Answer:

8.45 L

Explanation:

From the question given above, the following data were obtained:

Initial temperature (T₁) = 50 °C

Initial pressure (P₁) = 2 atm

Initial volume (V₁) = 5 L

Final temperature (T₂) = 0 °C

Final pressure (P₂) = 1 atm

Final volume (V₂) =?

Next, we shall convert celsius temperature to Kelvin temperature. This can be obtained as follow:

T(K) = T(°C) + 273

Initial temperature (T₁) = 50 °C

Initial temperature (T₁) = 50 °C + 273

Initial temperature (T₁) = 323 K

Final temperature (T₂) = 0 °C

Final temperature (T₂) = 0 °C + 273

Final temperature (T₂) = 273 K

Finally, we shall determine the new volume. This can be obtained as follow:

Initial temperature (T₁) = 323 K

Initial pressure (P₁) = 2 atm

Initial volume (V₁) = 5 L

Final temperature (T₂) = 273 k

Final pressure (P₂) = 1 atm

Final volume (V₂) =?

P₁V₁ / T₁ = P₂V₂ / T₂

2 × 5 / 323 = 1 × V₂ / 273

10 / 323 = V₂ / 273

Cross multiply

323 × V₂ = 10 × 273

323 × V₂ = 2730

Divide both side by 323

V₂ = 2730 / 323

V₂ = 8.45 L

Thus, the new volume is 8.45 L

5 0
3 years ago
Identify the compound with the smallest dipole moment in the gas phase. identify the compound with the smallest dipole moment in
stepladder [879]
Without solving for the dipole moment, we can easily determine which among the common gases has the smallest dipole moment just by determining the differences in their electronegativity. The greater the difference in the electronegativity, the higher is the value of the dipole moment. 

From the given above, there are obvious differences between the electronegativity between the atoms composing LiF, ClF, and HF. For Cl2, since this is the same molecule then, the difference in the electronegativity is zero.

Answer: Cl2. 
3 0
3 years ago
What are the four planets farthest away from the Sun categorized as?
Neko [114]

Answer:

They're called Jovian planets

6 0
3 years ago
Iron has a density of 7.87mL. How much mass would a 12.3mL sample contain?
algol [13]

The correct answer is 96.80 grams

Explanation:

The density of a substance shows the total mass the substance contains in 1 mL or 1 cm3 as density is calculated by using the formula D= M (mass) / V (volume). This implies, Iron contains 7.87 grams per milliliter. Moreover, this value and formula can be used to calculate the mass or volume of any other sample. The process to calculate the mass of an iron sample with a volume of 12.3 mL is shown below.

D = M ÷ V

7.87 mL = x ÷ 12.3 mL - x represents the missing value. Now find the value of x by solving the equation

x = 7.87 · 12.3  

x = 96.80

This means a sample of 12.3 mL contains a mass of 96.80 grams. Also, you can know this value is correct because if you divide the mass by the value the density is the same (96.80 grams ÷ 12.3 mL = 7.87 g/mL)

4 0
3 years ago
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