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sp2606 [1]
3 years ago
5

How many moles are in 22 grams of magnesium ?

Chemistry
1 answer:
iris [78.8K]3 years ago
6 0
24.3
maybe this will help too :D
https://www.convertunits.com/from/grams+Magnesium/to/moles
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What is the relationship between.chemicals and cancer ?
timama [110]

Answer:

The relationship is like asbestos and lung cancer or benzene and leukemia because cause and effect.

Explanation:

Chemicals cause Cancer

Nicotine is chemicals so when people smoke JUULS or anything else that's associated with chemicals it causes cancer.

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3 years ago
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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
Electron configurations are a shorthand form of an orbital diagram, describing which orbitals are occupied for a given element.
kipiarov [429]

Answer:

[Ar] 3d10 4s2 4p3 is shorthand, 1s22s22p63s23p63d104s24p3 long hang

Explanation:

The shorthand is made using the lowest & closest noble gas, and picking up where it leaves off as follows, and longhand is made from a followed pattern you can easily find

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How much heat is absorbed when 447 g of water , Cp = 4.184 J/goC, goes from 25.0 oC to 35.0 oC?
Digiron [165]

Answer:

The answer to your question is  Q = 18702.5 J

Explanation:

Data

mass of water = m = 447 g

Cp = 4.184 J/g°C

Temperature 1 = T1 = 25°C

Temperature 2 = T2 = 35°C

Heat = Q = ? Joules

Process

1.- Write the formula to calculate heat

    Q = mCp(T2 - T1)

2.- Substitution

    Q = (447)(4.184)(35 - 25)

3.- Simplification

    Q = (447)(4.184)(10)

4.- Result

    Q = 18702.5 J

8 0
3 years ago
A chemist lectures about the following definition for acids and bases. The acid-base behavior is analyzed in terms of how electr
Y_Kistochka [10]

Answer:

the Lewis concept of acids and bases

Explanation:

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