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Amanda [17]
3 years ago
10

Based on the illustration below, explain how the different organ systems work together.

Chemistry
1 answer:
Stella [2.4K]3 years ago
8 0

Answer:

this is not a proper informative question

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Predict whether each of the following bonds is ionic, polar, covalent, or nonpolar covalent:
Gekata [30.6K]

Explanation:

It is known that in order to determine a compound to be ionic, polar, covalent or non-polar covalent it is necessary to determine the electronegativity difference of the combining atoms.

So, when electronegativity difference is from 0.0 to 0.4 then bond formed between the two atoms is non-polar covalent in nature.

When electronegativity difference is greater than 0.4 and less than 1.8 then bond between the two atoms is a polar covalent bond.

When electronegativity difference is 1.8 or greater than the bond formed is ionic in nature.

Electronegativity difference of the given molecules is as follows.

Si-O = (1.90 - 3.44) = 1.54

K-Cl = (0.82 - 3.16) = 2.34

S-F = (2.58 - 3.98) = 1.4

P-Br = (2.19 - 2.96) = 0.77

Li-O = (0.98 - 3.44) = 2.46

N-P = (3.04 - 2.19) = 0.85

Therefore, given compounds are classified as follows.

Si-O, S-F, P-Br, and N-P all have a polar covalent bond. Whereas K-Cl and Li-O are ionic in nature.

7 0
3 years ago
Read 2 more answers
What speeds up a chemical reaction
storchak [24]
Catalyst can speed up chemical reaction without being used up or involved in thr reaction..
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
Which of the following expressions best represents a second order rate law?
Ksivusya [100]
I think the correct answer would be C. The expression that would best represent a second order rate law would be r =k[X][Y]. Reaction with this rate law are those that depend on the concentration of two first order reactants or a second order reactant.
8 0
3 years ago
If the students were told that one of the materials was titanium and that titanium is still a solid when heated to 1,000°C, whic
Paladinen [302]

The sample having solid substance is considered as titanium.

<h3>Which sample was titanium?</h3>

That sample is considered as Titanium which is still present as a solid substance because all the other samples turns into liquid state due to high temperature.

While on the other hand, the sample which contains titanium has a solid particles and still solid on high temperature so we can conclude that the sample having solid substance is considered as titanium.

Learn more about titanium here: brainly.com/question/11496905

5 0
2 years ago
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