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3 years ago

If the mass of 293nv is 293.15 amu and that of 295nv is 295.30 amu , what is the atomic weight of nv?

1 answer:

5 0

Honestly, I am quite confused with what Nv stands for because there is no element with that symbol. However, I still get the concept of finding the average molecular mass of an element. Let's just assume that nv stands for a specific type of element and it has two isotopes: nv-293 and nv-295. Isotopes have the same number of protons but differ in mass number (protons+neutrons).

To find the average atomic weight, just multiply the individual weights with the respective composition of the isotope. Since there are only two isotopes, they constitute 50% each. So, the average atomic weight is

(50%)(293.15 amu) + (50%)(295.30 amu) = 294.225 amu

Hence, the atomic weight of nv is 294.225 atomic mass units.

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Answer:

2.72 m/s

Explanation:

In 3 meters a person running 0.5 m/s accelerates 1.2 m/s².

It means,

Distance, s = 3 m

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Acceleration, a = 1.2 m/s²

We need to find the final velocity of the person. Using equation of motion to find it as follows :

$v^2-u^2=2as\\\\v^2=2as+u^2\\\\v^2=2\times 1.2\times 3+(0.5)^2\\\\v=2.72\ m/s$

So, the final velocity of the person is 2.72 m/s.

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3 years ago

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3 years ago

Read 2 more answers

An element has the following natural abundances and isotopic masses: 90.92% abundance with 19.99 amu, 0.26% abundance with 20.99

sashaice [31]

Answer: The average atomic mass of the given element is 20.169 amu.

Explanation:

Average atomic mass of an element is defined as the sum of masses of the isotopes 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)

We are given:

For isotope 1:

Mass of isotope 1 = 19.99 amu

Percentage abundance of isotope 1 = 90.92 %

Fractional abundance of isotope 1 = 0.9092

For isotope 2:

Mass of isotope 2 = 20.99 amu

Percentage abundance of isotope 2 = 0.26%

Fractional abundance of isotope 2 = 0.0026

For isotope 3:

Mass of isotope 3 = 21.99 amu

Percentage abundance of isotope 3 = 8.82%

Fractional abundance of isotope 3 = 0.0882

Putting values in equation 1, we get:

$\text{Average atomic mass}=[(19.99\times 0.9092)+(20.99\times 0.0026)+(21.99\times 0.0882)]$

$\text{Average atomic mass}=20.169amu$

Hence, the average atomic mass of the given element is 20.169 amu.

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3 years ago

The voltage across the terminals of a generator is 5.7 v when it supplies a current of 0.3 A. It becomes 5.1 V when I=0.9A. Find

snow_tiger [21]

Answer:

The emf of the generator is 6V

The internal resistance of the generator is 1 Ω

Explanation:

Given;

terminal voltage, V = 5.7 V, when the current, I = 0.3 A

terminal voltage, V = 5.1 V, when the current, I = 0.9 A

The emf of the generator is calculated as;

E = V + Ir

where;

E is the emf of the generator

r is the internal resistance

First case:

E = 5.7 + 0.3r -------- (1)

Second case:

E = 5.1 + 0.9r -------- (2)

Since the emf E, is constant in both equations, we will have the following;

5.1 + 0.9r = 5.7 + 0.3r

collect similar terms together;

0.9r - 0.3r = 5.7 - 5.1

0.6r = 0.6

r = 0.6/0.6

r = 1 Ω

Now, determine the emf of the generator;

E = V + Ir

E = 5.1 + 0.9x1

E = 5.1 + 0.9

E = 6 V

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