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likoan [24]
3 years ago
9

How TO DO THIS QUESTION PLEASE PLEASEEEEEEEEEeeeeeeeeeeeeeeeeeeee​

Chemistry
1 answer:
krok68 [10]3 years ago
3 0
O2 is the limited reactant

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A gas sample occupies a volume of 1.264 L when the temperature is 168.0 °C and the pressure is 946.6 torr. How many molecules ar
alukav5142 [94]

Answer:

0.26×10²³ molecules

Explanation:

Given data:

Volume of gas = 1.264 L

Temperature = 168°C

Pressure = 946.6 torr

Number of molecules of gas = ?

Solution:

Temperature = 168°C (168+273= 441 K)

Pressure = 946.6 torr (946.6/760 = 1.25 atm)

Now we will determine the number of moles.

PV = nRT

P= Pressure

V = volume

n = number of moles

R = general gas constant = 0.0821 atm.L/ mol.K  

T = temperature in kelvin

n = PV/RT

n = 1.25  atm ×1.264 L / 0.0821 atm.L/ mol.K   ×441 K

n = 1.58 /36.21 /mol

n = 0.044 mol

Now we will calculate the number of molecules by using Avogadro number.

1 mol = 6.022×10²³ molecules

0.044 mol × 6.022×10²³ molecules/ 1mol

0.26×10²³ molecules

5 0
3 years ago
A compound contains 6.0 g of carbon and 1.0 g of hydrogen and has a molar mass of 42.0 g/mol.
makvit [3.9K]

Answer:

%C = 85.71 wt%; %H = 14.29 wt%; Empirical Formula => CH₂; Molecular Formula => C₃H₆

Explanation:

%Composition

Wt C = 6 g

Wt H = 1 g

TTL Wt = 6g + 1g = 7g

%C per 100wt = (6/7)100% = 85.71 wt%

%H per 100wt = (1/7)100% = 14.29 wt % or, %H = 100% - %C = 100% - 85.71% = 14.29 wt% H

What you should know when working empirical formula and molecular formula problems.

Empirical Formula=> <u>smallest</u> whole number ratio of elements in a compound

Molecular Formula => <u>actual</u> whole number ratio of elements in a compound

Empirical Formula Weight x Whole Number Multiple = Molecular Weight

From elemental %composition values given (or, determined as above), the empirical formula type problem follows a very repeatable pattern. This is ...

% => grams => moles => ratio => reduce ratio => empirical ratio

for determination of molecular formula one uses the empirical weight - molecular weight relationship above to determine the whole number multiple for the molecular ratios.

Caution => In some 'textbook' empirical formula problems, the empirical ratio may contain a fraction in the amount of 0.25, 0.50 or 0.75. If such an issue arises, multiply all empirical ratio numbers containing 0.25 and/or 0.75 by '4'  to get the empirical ratio and multiply all empirical ration numbers containing 0.50 by '2' to get the final empirical ratio.

This problem:

Empirical Formula:

Using the % per 100wt values in part 'a' ...

              %     =>         grams                 =>                 moles

%C => 85.71% => 85.71 g* / 100 g Cpd => (85.71 / 12) = 7.14 mol C

%H => 14.29% => 14.29 g / 100 g Cpd => (14.29 / 1) = 14.29 mol H

=> Set up mole Ratio and Reduce to Empirical Ratio:

mole ratio C:H =>  7.14 : 14.29

<u>To reduce mole values to the smallest whole number ratio,  divide all mole values by the smaller mole value of the set.</u>

=> 7.14/7.14 : 14.29/7.14 => Empirical Ration=> 1 : 2

∴ Empirical Formula => CH₂

Molecular Formula:

(Empirical Formula Wt)·N = Molecular Wt => N = Molecular Wt / Empirical Wt

N = 42 / 14 = 3 => multiply subscripts of empirical formula by '3'.

Therefore, the molecular formula is C₃H₆

3 0
3 years ago
Pl help me, i need a good grade on this.
Blizzard [7]
1. Option A. Temperature and Salinity
2. Option D. (I think)
3. Option D.
Hope your test goes well!
8 0
3 years ago
The amount of matter in an object is its
andrezito [222]

The amount of matter in an object is its Mass...

7 0
3 years ago
Hydrogen gas and fluorine gas will react to form hydrogen fluoride gas. What is the standard free energy change for this reactio
Marta_Voda [28]

Answer:

\Delta G=-541.4kJ/mol

Explanation:

Hello there!

In this case, according to the given information, it turns out firstly necessary to write out the described chemical reaction as shown below:

H_2+F_2\rightarrow 2HF

Now, we set up the expression for the calculation of the standard free energy change, considering the free energy of formation of each species, specially those of H2 and F2 which are both 0 because they are pure elements:

\Delta G=2\Delta G_f^{HF}-(\Delta G_f^{H_2}+\Delta G_f^{F_2})\\\\\Delta G=2*-270.70kJ/mol-(0kJ/mol+0kJ/mol)\\\\\Delta G=-541.4kJ/mol

Regards!

4 0
3 years ago
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