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3 years ago

How is the molar mass of an element different from the atomic mass of an element?

1 answer:

5 0

1.Molar mass is the mass of one mole per single element while atomic mass is the mass of an atom at rest or is the number of protons and neutrons.
2.Molar mass is measured in grams per mole while atomic mass is “unitless.”
3.Atomic mass is measured via mass spectrometry while molar mass is computed via atomic weight.

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The atomic masses of 151eu and 153eu are 150.919860 and 152.921243 amu, respectively. The average atomic mass of europium is 151

vitfil [10]

Answer:- The natural abundance of $^1^5^1_E_u$ is 0.478 or 47.8% and $^1^5^3_E_u$ is 0.522 or 52.2% .

Solution:- Average atomic mass of an element is calculated from the atomic masses of it's isotopes and their abundances using the formula:

Average atomic mass = mass of first isotope(abundance) + mass of second isotope(abundance)

We have been given with atomic masses for $^1^5^1_E_u$ and $^1^5^3_E_u$ as 150.919860 and 152.921243 amu, respectively. Average atomic mass of Eu is 151.964 amu.

Sum of natural abundances of isotopes of an element is always 1. If we assume the abundance of $^1^5^1_E_u$ as n then the abundance of $^1^5^3_E_u$ would be 1-n .

Let's plug in the values in the formula:

$151.964=150.919860(n)+152.921243(1-n)$

151.964=150.919860n+152.921243-152.921243n

on keeping similar terms on same side:

$151.964-152.921243=150.919860n-152.921243n$

$-0.957243=-2.001383n$

negative sign is on both sides so it is canceled:

$0.957243=2.001383n$

$n=\frac{0.957243}{2.001383}$

$n=0.478$

The abundance of $^1^5^1_E_u$ is 0.478 which is 47.8%.

The abundance of $^1^5^3_E_u$ is = $1-0.478$

= 0.522 which is 52.2%

Hence, the natural abundance of $^1^5^1_E_u$ is 0.478 or 47.8% and $^1^5^3_E_u$ is 0.522 or 52.2% .

3 0

3 years ago

Write the complete ionic equations, spectator ions and net ionic equation for the following.

gizmo_the_mogwai [7]

Answer:

Explanation:

1) ZnBr₂ (aq) + AgNO₃ (aq)

Chemical equation:

ZnBr₂ (aq) + AgNO₃ (aq) →Zn(NO₃)₂(aq) + AgBr(s)

Balanced chemical equation:

ZnBr₂ (aq) + 2AgNO₃ (aq) →Zn(NO₃)₂(aq) + 2AgBr(s)

Ionic equation:

Zn²⁺(aq) + Br₂²⁻ (aq) + 2Ag⁺ (aq)+ 2NO⁻₃ (aq) → Zn²⁺(aq) +(NO₃)₂²⁻(aq) + 2AgBr(s)

Net ionic equation:

Br₂²⁻ (aq) + 2Ag⁺ (aq) → 2AgBr(s)

The Zn²⁺((aq) and NO⁻₃ (aq) are spectator ions that's why these are not written in net ionic equation. The AgBr can not be splitted into ions because it is present in solid form.

Spectator ions:

These ions are same in both side of chemical reaction. These ions are cancel out. Their presence can not effect the equilibrium of reaction that's why these ions are omitted in net ionic equation.

2) HgCl₂ (aq) + KI (aq) →

Chemical equation:

HgCl₂ (aq) + KI (aq) → KCl + HgI₂

Balanced chemical equation:

HgCl₂ (aq) + 2KI (aq) → 2KCl(aq) + HgI₂(s)

Ionic equation:

Hg²⁺(aq) + Cl₂²⁻ (aq) + 2K⁺(aq) + 2I⁻ (aq) → HgI₂ (s) + 2K⁺(aq) + 2Cl⁻ (aq)

Net ionic equation:

Hg²⁺(aq) + 2I⁻ (aq) → HgI₂ (s)

The Cl⁻ ((aq) and K⁺ (aq) are spectator ions that's why these are not written in net ionic equation. The HgI₂ (s) can not be splitted into ions because it is present in solid form.

Spectator ions:

These ions are same in both side of chemical reaction. These ions are cancel out. Their presence can not effect the equilibrium of reaction that's why these ions are omitted in net ionic equation.

3) Ca(OH)₂ (aq) + Na₂SO₄ (aq)

Chemical equation:

Ca(OH)₂ (aq) + Na₂SO₄ (aq) → CaSO₄(s) + NaOH(aq)

Balanced chemical equation:

Ca(OH)₂ (aq) + Na₂SO₄ (aq) → CaSO₄(s) + 2NaOH(aq)

Ionic equation:

Ca²⁺(aq) + OH₂²⁻ (aq) + 2Na⁺(aq) + SO₄²⁻ (aq) → CaSO₄(s) + 2Na⁺(aq) + 2OH⁻ (aq)

Net ionic equation:

Ca²⁺(aq) + SO₄²⁻ (aq) → CaSO₄(s)

The OH⁻ ((aq) and Na⁺ (aq) are spectator ions that's why these are not written in net ionic equation. The CaSO₄ can not be splitted into ions because it is present in solid form.

Spectator ions:

These ions are same in both side of chemical reaction. These ions are cancel out. Their presence can not effect the equilibrium of reaction that's why these ions are omitted in net ionic equation.

4 0

3 years ago

How many particles of CH4 are present in a 8.25 L sample of this gas?

Snowcat [4.5K]

Is this a mole question

4 0

3 years ago

Unit 1 chemistry test

Artist 52 [7]

Ok what’s the question sir

5 0

3 years ago

2. Low-density polyethylene, used to make plastic films, is

emmainna [20.7K]

Answer:

from the spaces left between chains when the polymer is

formed.

Explanation:

Polyethylene polymer may be of high or low density. Whether the polymer is of high or low density depends on the arrangement of the polymer chains.

If the polymer chains are close together such that the resultant polymer is crystalline with chains packed closely, we have high density polyethylene.

On the other hand if the chains are not close together and there are spaces left between chains when the polymer is formed, then we have low density polyethylene.

Hence, low density polyethylene gets its name from the spaces left between chains when the polymer is formed.

6 0

3 years ago

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