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Masja [62]
2 years ago
3

How many grams of titanium are in an instrument gear with 5.74 x 10^22 atoms of Ti?

Chemistry
2 answers:
Tems11 [23]2 years ago
7 0

Answer:

4.562

Explanation:

Hatshy [7]2 years ago
6 0
The number of grams of titanium that are in an instrument gear with 5.74 x10^22  atoms of Ti  are calculated as below

find the number of moles of Ti that are in 5.74 x10^22 atoms

by use of Avogadro   law constant, that is 1 mole = 6.02 x10^23 atoms
                                                        what about  5.74 x10^22 atoms

by cross  multiplication
(1mole x5.74 x10^22 atoms)/6.02 x10^23  atoms = 0.0953  moles

mass of Ti =moles of Ti   molar mass of Ti

that is    0.0953 moles x 47.87 g/mol = 4.562  grams  of Ti

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Naddik [55]
So the put a lot of words to make this seem more complicated than it is. Your first equation involves the money. I’m going to use x to represent tacos and y to represent burritos.
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Y=x+2
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2 years ago
4. a student is shown a Diagram which shows gas bubbles being made by water plant that has been exposed to light. which gas is m
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A 2.50-l flask contains a mixture of methane (ch4) and propane (c3h8) at a pressure of 1.45 atm and 20°c. when this gas mixture
Cerrena [4.2K]

Answer:- Mole fraction of methane in the original gas mixture is 0.854.

Solution:- From given volume, pressure and temperature, we could calculate the total moles of the gaseous mixture of methane and propane using ideal gas law as:

PV = nRT

n=\frac{PV}{RT}

V = 2.50 L

P = 1.45 atm

T = 20 + 273 = 293 K

Let's plug in the values in the equation:

n=(\frac{1.45*2.50}{0.0821*293})

n = 0.151

Let's say the solution has X moles of methane. Then moles of propane would be = (0.151 - X)

The combustion equations of methane and propane are:

CH_4+2O_2\rightarrow CO_2+2H_2O

C_3H_8+5O_2\rightarrow 3CO_2+4H_2O

From methane balanced equation, there is 1:1 mol ratio between methane and carbon dioxide. So, X moles of methane would produce X moles of carbon dioxide.

From balanced equation of propane, there is 1:3 mol ratio between propane and carbon dioxide. So, (0.151 - X) moles of propane would give 3(0.151 - X) moles of carbon dioxide.

So, total moles of carbon dioxide that we would get from methane and propane combustion are:

X + 3(0.151 - X)

From given data, 8.60 g of carbon dioxide are formed by the combustion of gas mixture.

moles of Carbon dioxide = 8.60g(\frac{1mol}{44g})

moles of carbon dioxide = 0.195 mol

Hence, 0.195 = X + 3(0.151 - X)

Let's solve this for X as:

0.195 = X + 0.453 - 3X

0.195 = 0.453 - 2X

2X = 0.453 - 0.195

2X = 0.258

X=\frac{0.258}{2}

X = 0.129

So, there are 0.129 moles of methane in the mixture.

moles of propane = 0.151 - 0.129 = 0.022

mole fraction of methane = \frac{moles of Methane}{total moles}

mole fraction of methane = \frac{0.129}{0.151}

mole fraction of methane = 0.854

Hence, the mole fraction of methane gas in the original gas mixture is 0.854.

6 0
2 years ago
Which is the correct statement regarding the relative Rf values of the starting methyl benzoate vs the product, methyl m-nitrobe
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Answer:

1. The product has a higher Rf value on a silica gel TLC plate because it is more polar than the starting methyl benzoate.

2. False

3. True

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In comparing the electrophillic aromatic substitution of m-nitrobenzoate  and methyl benzoate, we must remember that the presence of electron withdrawing groups (such as -NO2 and -CHO) on the aromatic compound deactivates the compound towards electrophillic aromatic substitution hence, methyl m-nitrobenzoate is less reactive than methyl benzoate in Electrophilic Aromatic Substition and Methyl benzoate is less reactive than benzene in Electrophilic Aromatic Substition

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Explanation:

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