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2 years ago

What happens during a lunar eclipse?

Chemistry

1 answer:

Alexeev081 [22]2 years ago

6 0

Answer:

<h2>A.EARTH PASSES BETWEEN THE SUN AND THE MOON</h2>

Explanation:

When Earth passes directly between Sun and Moon, its shadow creates a lunar eclipse.

pabrainliest po plss:)

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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

2 years ago

All crystals have the same internal arrangement of atoms. <br> a. True<br> b. False

zhenek [66]

B - False. The crystals have the same internal arrangement of atoms in the same species. For example, all crystals of quartz (Onion, Amethyst, Citrine, Smoky) have the same trigonometrical arrangement.

the atoms in salt (NaCl) are arranged in a cubic way.

3 0

2 years ago

Magnesium acetate can be prepared by a reaction involving 15.0 grams of iron(III) acetate with either 10.0 grams of Magnesium Ch

noname [10]

1) Chemical reaction:
2(CH₃COO)₃Fe + 3MgCrO₄ → Fe₂(CrO₄)₃ + 3(CH₃COO)₂Mg.
m((CH₃COO)₃Fe) = 15,0 g.
m(MgCrO₄) = 10,0 g.
n((CH₃COO)₃Fe) = m((CH₃COO)₃Fe) ÷ M((CH₃COO)₃Fe).
n((CH₃COO)₃Fe) = 15 g ÷ 233 g/mol.
n((CH₃COO)₃Fe) = 0,064 mol.
n(MgCrO₄) = m(MgCrO₄) ÷ M(MgCrO₄).
n(MgCrO₄) = 10 g ÷ 140,3 g/mol.
n(MgCrO₄) = 0,071 mol; limiting reagens.
From chemical reaction: n(MgCrO₄) : n((CH₃COO)₂Mg) = 3 : 3.
n((CH₃COO)₂Mg) = 0,071 mol.
m((CH₃COO)₂Mg) = 0,071 mol · 142,4 g/mol.
n((CH₃COO)₂Mg) = 10,11 g.

2) Chemical reaction:
2(CH₃COO)₃Fe + 3MgSO₄ → Fe₂(SO₄)₃ + 3(CH₃COO)₂Mg.
m((CH₃COO)₃Fe) = 15,0 g.
m(MgSO₄) = 15,0 g.
n((CH₃COO)₃Fe) = m((CH₃COO)₃Fe) ÷ M((CH₃COO)₃Fe).
n((CH₃COO)₃Fe) = 15 g ÷ 233 g/mol.
n((CH₃COO)₃Fe) = 0,064 mol; limiting ragens.
n(MgSO₄) = m(MgSO₄) ÷ M(MgSO₄).
n(MgSO₄) = 15 g ÷ 120,36 g/mol.
n(MgSO₄) = 0,125 mol; limiting reagens.
From chemical reaction: n(CH₃COO)₃Fe) : n((CH₃COO)₂Mg) = 2 : 3.
n((CH₃COO)₂Mg) = 0,064 mol · 3/2.
n((CH₃COO)₂Mg) = 0,096 mol.
m((CH₃COO)₂Mg) = 0,096 mol · 142,4 g/mol.
m((CH₃COO)₂Mg) = 13,7 g.

7 0

3 years ago

How many grams of Helium can be found in 3.4 moles of it.

MaRussiya [10]

one mole

If we look this number up for helium, we find that helium has a molar mass of 4.0 grams per mole. So one mole of helium has a mass of 4.0 grams. We can use the molar mass as a conversion factor. Two moles of helium times four grams of helium per mole of helium is equal to eight grams of helium.

5 0

2 years ago

In figure 21-4b in what direction is the force defeating the wire

makkiz [27]

Answer:

the one that has more power to it

6 0

2 years ago