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sasho [114]
3 years ago
13

A student performs an experiment similar to what you will be doing in lab, except using titanium metal instead of magnesium meta

l. The student weights out 0.108 g of titanium. How many moles of titanium is this?
Chemistry
1 answer:
Genrish500 [490]3 years ago
7 0

Answer:

n = 0.0022 mol

Explanation:

Moles is denoted by given mass divided by the molar mass ,  

Hence ,  

n = w / m  

n = moles ,  

w = given mass ,  

m = molar mass .  

From the information of the question ,

w = 0.108 g

As we known ,

The molar mass of titanium = 47.867 g / mol

The mole of titanium can be caused by using the above relation , i.e. ,

n = w / m  

n = 0.108 g / 47.867 g / mol

n = 0.0022 mol

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This is a combination reaction.  Look at the 2 elements on left and a compound on the right.
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3 years ago
1.- Ilumina de amarillo los verbos en pretérito y de verde los verbos en copretérito.
ivanzaharov [21]

Answer:

Pretérito (Pretérito perfecto):

<em>estiró, sudó, peleó, compartí</em>

Copretérito (Pretérito imperfecto):

<em>saludaba, pedias, leia, compraba, barrias, cargaba, estrenabas, escribian</em>

Explanation:

Los verbos en pretérito, también llamado pretérito perfecto, son los siguientes:

<em>estiró, sudó, peleó, compartí</em>

Como se aprecia, los verbos conjugados en pretérito perfecto terminan la última sílaba en vocal con acento, es decir, <em>-á</em>, <em>-é</em>, <em>-í</em> u <em>-ó</em>.

Los verbos en copretérito, también llamado pretérito imperfecto, son los siguientes:

<em>saludaba, pedias, leia, compraba, barrias, cargaba, estrenabas, escribian</em>

Como se aprecia, los verbos conjugados en pretérito imperfecto terminan la última sílaba en <em>-ía</em>, <em>ías</em>, <em>-ían</em>, <em>-aba</em>, <em>-abas</em> o <em>-aban</em>.  

3 0
2 years ago
A sample of gas X occupies 10 mº at a pressure of 120 kPa.
Mashutka [201]

Answer:

The new pressure of the gas comes out to be 400 KPa.

Explanation:

Initial volume of gas = V = 10\textrm{ m}^{3}

Initial pressure of gas = P = 120 KPa

Final volume of gas = V' = 3\textrm{ m}^{3}

Assuming temperature to be kept constant.

Assuming final pressure of the gas to be P' KPa

PV = P'V' \\120\textrm{ KPa}\times 10\textrm{ m}^{3} = \textrm{P'}\times 3\textrm{ m}^{3} \\\textrm{P'} = 400\textrm{ KPa}

New pressure of gas = 400 KPa

5 0
3 years ago
Why the is ratio of magnesium ions to chloride ions in MgCl2 is 1:2?
Monica [59]
You should read up on Proust's law, better known as the Law of Definite Proportions. This is a chemical law that defines your question more generally, on why the ratio of elements and ions are always fixed.

Basically, this compound Magnesium(II) Chloride is MgCl2 because it has the same number of protons, neutrons, and electrons all the way. This defines the properties of the compound or atom. 
5 0
3 years ago
WILL MARK BRANILEST
ahrayia [7]

Answer:

FeCl3 is the limiting reactant

O2 is in excess

Theoretical yield Cl2 = 9.84 grams

The % yield is 96.5 %

Explanation:

Step 1: Data given

Mass of FeCl3 = 15.0 grams

Moles O2 = 4.0 moles

Mass of Cl2 produced = 9.5 grams

Step 2: The balanced equation

4FeCl3 + 3O2 → 2Fe2O3 + 6Cl2

Step 3: Calculate moles FeCl3

Moles FeCl3 = mass FeCl3 / molar mass FeCl3

Moles FeCl3 = 15.0 grams / 162.2 g/mol

Moles FeCl3 = 0.0925 moles

Step 4: Calculate limiting reactant

FeCl3 is the limiting reactant. Because we have way more (more than ratio 3:4) moles O2 than FeCl3. It will completely be consumed (0.0925 moles). O2 is in excess. There will react = 0.069375 moles O2

There will remain 4.0 - 0.069375 = 3.930625 moles

Step 5: Calculate moles Cl2

For 4 moles FeCl3 we need 3 moles O2 to produce 2 moles Fe2O3 and 6 moles Cl2

For 0.0925 moles FeCl3 moles we'll have 6/4 * 0.0925 = 0.13875 moles Cl2.

Step 6: Calculate mass Cl2

Mass Cl2 = moles * molar mass

Mass Cl2 = 0.13875 moles * 70.9 g/mol

Mass Cl2 = 9.84 grams

Step 7: Calculate % yield

% yield = (actual yield / theoretical yield) * 100%

% yield = (9.5 grams / 9.84 grams ) * 100%

% yield = 96.5 %

The % yield is 96.5 %

4 0
3 years ago
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