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vichka [17]
3 years ago
5

What is the formula for barium nitrate ?

Chemistry
1 answer:
brilliants [131]3 years ago
6 0

Answer:

Ba(NO3)2

Explanation:

You might be interested in
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
Which pair of elements can be used to determine the age of a fossil that is over one billion years old?
lakkis [162]

Answer:

Explanation:

There are some radioactive nuclides can be used to measure time on an archeological scale. One is the best example of this is radiocarbon dating. This process is based on the ratio of caebon-14 to carbon-12 in the atmosphere which is relatively constant.

The half time of C-14 5730 years  

Carbon-14 is a radioactive nucleus. It has a half-life of 5730 years.

All living tissues like plants and animal absorbed carbon-12 along with carbon-14 with same ratio of caebon-14 to carbon-12 in the atmosphere.

Carbon-14 dating is based on the ratio of carbon-14 to carbon-12 in the atmosphere which is relatively constant

3 0
3 years ago
Read 3 more answers
What is this plzzzzzzzzzzzzzzzzzzz help!!!!
motikmotik

Answer:

convex

converges

...............

3 0
3 years ago
Rock and ice particles make up the ____ of the Jovian planets? Ring system, craters, volcanoes, or asteroids
lara31 [8.8K]
Jovian planets are what we call the "gas giants," so immediately we can eliminate craters or volcanos because they don't have a solid surface. asteroids in space doesn't belong to any specific planet, so the answer is ring systems.
5 0
4 years ago
How does the arrangement of elements in periods relate to electron configuration
VashaNatasha [74]
Electronic Configuration of elements in a period is same because If you see the electronic Configuration of elements in a period you will notice that the valence shell electrons for all elements are present in the same Shell. For example, in first period consisting of Hydrogen and Helium, both the elements' valence electrons are present in the same Shell.
Electronic Configuration of Hydrogen,
1s^1
Electronic Configuration of Helium,
1s^2

Both elements' valance electrons are present in the 1st shell

(This is just a small example to understand the concept because other periods are long but the first period is short that's why I gave the example of the first period)
4 0
3 years ago
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