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Sonja [21]
3 years ago
11

What mass of mgf2 is contained in 80.85 g of a 22.4% by mass solution of mgf2 in water?

Chemistry
2 answers:
Aleks [24]3 years ago
4 0
Formula: % by mass = (mass of solute / mass of solution] *100

Data:

mass of solution = 80.85 g

% by mass = 22.4%

Unknown = mass of solute

Solution

% by mass = (mass of solute / mass of solution] *100 = >

mass of solute = % by mass * mass of solution / 100

mass of solute = 22.4 * 80.85 / 100 = 18.11 g

Answer: 18.11 g
atroni [7]3 years ago
4 0

Answer : The mass of MgF_2 contained is, 18.1104 grams

Solution : Given,

Given mass of solution = 80.85 g

22.4% by mass solution of MgF_2 means that 22.4 grams of MgF_2 present in 100 gram of solution.

As, 100 grams of solution contains 22.4 grams of MgF_2

So, 80.85 grams of solution contains \frac{22.4}{100}\times 80.85=18.1104 grams of MgF_2

Therefore, the mass of MgF_2 contained is, 18.1104 grams

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gulaghasi [49]
PV=nRT will give you the answer I think. I haven’t worked with a certain unit in that problem
3 0
3 years ago
Fe(NO3)2 not sure how to get the oxidation numbers of all elements
Shtirlitz [24]
You must remember that oxidation number of hydrogen in acids is always +1, oxidation number of oxygen in oxides & acids is always -2... metals has always oxidation number on plus!

group NO3 comes from HNO3...and oxidation number of whole acid group is always on minus and equal to the amount of hydrogen atoms in this acid... so oxidation number of NO3 = -1

we have 2 NO3 groups so 2*(-1) = -2 and that is the reason why oxidation number of Fe in this formula must be +2... because sum of all elements always gives 0!

Now we could count of oxidation number for nitrogen... we write HNO3 and start counting from right to left:
3*(-2) from oxygens + 1 from hydrogen = -5
so nitrogen must have +5 oxidation number... because sum all in formula must be 0.


4 0
3 years ago
An atom (not a hydrogen atom) absorbs a photon whose associated wavelength is 300 nm and then immediately emits photon whose ass
KIM [24]

Answer:

The net energy is 2.196 eV

Explanation:

Basically, the energy of an atom increases when it absorbs a photon. In addition, the wavelength of the emitted photon is longer such that the atom absorbed a net energy in the process.

Using:

ΔE = h*c*(1/λ_{1} - 1/λ_{2})

where:

ΔE is the net energy in eV (electron-volt). 1 eV is equivalent to 1.602*10^{-19} J.

h = 4.135*10^{-15} eVs

c = 3*10^{8} m/s

λ_{1} = 300 nm = 300*10^{-9} m

λ_{2} = 640 nm = 640*10^{-9} m

Thus:

ΔE = 4.135*10^{-15} eVs*3*10^{8} m/s*(\frac{1}{300*10^{-9}m } }-\frac{1}{640*10^{-9}m })

ΔE = 4.135*10^{-15}*3*10^{8}*1.77*10^{6} eV = 2.196 eV

6 0
3 years ago
2. A student has a centrifuge tube containing 14.0 g of t-butanol and is asked to make a 1.2 m solution of ethanol/t-butanol. Ho
Vaselesa [24]

Answer:

0.774g of ethanol

0.970mL of ethanol

Explanation:

Molality is an unit of concentration defined as the ratio between moles of solute and kg of solvent.

In the problem, you need to prepare a 1.2m solution of ethanol (Solute) in t-butanol (solvent).

14.0g of butanol are <em>0.014kg </em>and as you want to prepare the 1.2m solution, you need to add:

0.014kg × (1.2moles / kg) = 0.0168 moles of solute = Moles of ethanol

To convert moles of ethanol to mass you require molar mass (Molar mass ethanol, C₂H₅OH = 46.07g/mol). Thus, mass of 0.0168 moles are:

0.0168moles Ethanol ₓ (46.07g / mol) =

<h3>0.774g of ethanol</h3>

And to convert mass in g to mL you require density of the substance (Density of ethanol = 0.798g/mL):

0.774g ₓ (1mL / 0.798g) =

<h3>0.970mL of ehtanol</h3>
8 0
3 years ago
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
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