Ions are atoms that have gained or lost which subatomic particles?

What changes shape but doesn't change volume

OLEGan [10]

Liquids have a definite volume, but not a definite shape. Hope this helps :)

8 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=> smallest whole number ratio of elements in a compound

Molecular Formula => actual 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

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

=> 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

2 years ago

The 500kg concrete culvert has a mean radius of 0.5m. part a if the truck has an acceleration of 3m/s2, determine the culvert's

jarptica [38.1K]

Given:

Radius of culvert, r = 0.5 m

Tangential acceleration of the truck, a = 3 m/s2

To determine:

The angular acceleration, α

Explanation:

The tangential acceleration is related to the angular acceleration through the radius as:

a = rα

α = a/r = 3 ms⁻²/0.5 m = 6 s⁻²

Ans: The angular acceleration is 6 s⁻²

6 0

2 years ago

Where is the Oort cloud located

Karolina [17]

To be honest I don’t even know

5 0

2 years ago

Read 2 more answers

Fluoride is moderately basic, by far the most basic of the halides. It is a terrible leaving group. Fluoride is many orders of m

TEA [102]

Answer: Bromide is many orders of magnitude better than fluoride in leaving group ability

Explanation:

As Size of an atom Increases, the Basicity Decreases this is because if we move downwards from the top of the periodic table to the bottom of the periodic table, the size of an atom increases. As size increases, basicity will decrease, meaning the element will be less likely to act as a base implying that the element will be less likely to share its electrons.

in the same vein. With an increase in size, basicity decreases, making the ability of the leaving group to leave increase to increase . This can be seen in the halogens going down the group from

F--- worst

Cl----fair

Br ----good

I-----excellent

with fluorine having the worst ability to leave than Bromine which is better in terms of the leaving group ability.

3 0

2 years ago