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3 years ago

107.854 how many significant

Chemistry

1 answer:

sattari [20]3 years ago

6 0

Answer:

Explanation:

look after DP theres 3 digits

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g 1.000 atm of oxygen gas, placed in a container having a pinhole opening in its side, leaks from the container 2.14 times faste

Ivan

The question is incomplete, the complete question is;

1.000 atm of Oxygen gas, placed in a container having apinhole opening in its side. leaks from the container 2.14 timesfaster thatn 1.000 atm of an unknown gas placed in this sameapparatus. Which of the following species could be theunknown gas?

A. CL2

B. SF6

C. Kr

D. UF6

E. Xe

Answer:

SF6

Explanation:

From Graham's law;

Let the rate of diffusion of oxygen be R1

Let the rate of diffusion of unknown base be R2

Let the molar mass of oxygen by M1

Let the molar mass of unknown gas be M2

Hence;

R1/R2 = √M2/M1

So;

2.14/1 = √M2/32

(2.14/1)^2 = M/32

M= (2.14/1)^2 × 32

M= 146.6

This is the molar mass of SF6 hence the answer above.

7 0

3 years ago

5.6 g of Iron reacts with excess fluorine gas.

s2008m [1.1K]

Answer: FeF3 Iron(III) fluoride

Explanation: 5.6 g = 1/10 mole Fe

11.3-5.6g F = 5.7 = 3/10 mole F

4 0

3 years ago

Was i correct? -w- please answer below if i was im totally sure i am

earnstyle [38]

Answer:

y e s good job

Explanation:

7 0

3 years ago

When an atom undergoes alpha decay

jekas [21]

Are you asking for a specific atom or just the general definition?

The general definition of alpha decay is the loss of two protons and two neutrons from a nucleus. (Also equal to a helium atom!) This decreases the total mass number by 4 and the atomic number by 2.

7 0

2 years ago

Suppose that a fictitious element, X , has two isotopes: 59 X ( 59.015 amu ) and 62 X ( 62.011 amu ) . The lighter isotope has a

Gemiola [76]

Answer: The average atomic mass of element X is 59.3 amu.

Explanation:

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

$\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i$ .....(1)

For isotope 1 (X-59) :

Mass of isotope 1 = 59.015 amu

Percentage abundance of isotope 1 = 91.7 %

Fractional abundance of isotope 1 = 0.917

For isotope 2 (X-62) :

Mass of isotope 2 = 62.011 amu

Percentage abundance of isotope 2 = (100 - 91.7) % = 8.3 %

Fractional abundance of isotope 2 = 0.083

Putting values in equation 1, we get:

$\text{Average atomic mass of X}=[(59.015\times 0.917)+(62.011\times 0.083)]$

$\text{Average atomic mass of X}=59.3amu$

Hence, the average atomic mass of element X is 59.3 amu.

8 0

3 years ago