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Stells [14]
3 years ago
6

Which is an angiosperm

Chemistry
2 answers:
SSSSS [86.1K]3 years ago
7 0

Answer:

idk

Explanation:

Rasek [7]3 years ago
7 0

Answer:................

Explanation:Angiosperms are plants producing flowers. Thus, they are also commonly called as the flowering plants.

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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
This chart shows descriptions of the atoms for four different substances.
Whitepunk [10]
I think it’s Object 3
Hopefully this helps
8 0
3 years ago
Write and balance molecular equations for the following reactions between aqueous solutions. You will need to decide on the form
musickatia [10]

Answer:

This is the balanced equation:

Pb(NO₃)₂ (aq) + 2NaI (aq) → 2NaNO₃ (aq)  +  PbI₂ (s) ↓    

Explanation:

This are the reactants:

PbNO₃

NaI

Iodide can react to Pb²⁺ to make a solid compound.

4 0
3 years ago
If 3.6 g of aluminum completely reacts, how much Al2O3 (in grams) can be produced
marissa [1.9K]

The amount in grams of  Al₂O₃ produced is approximately 6.80 g.

Aluminium reacts completely with oxygen(air) to produce Al₂O₃. The reaction can be represented with a chemical equation as follows:

AL + O₂ → Al₂O₃

Let's balance it

4AL + 3O₂ → 2Al₂O₃

4 moles of Aluminium reacts with 3 moles of Oxygen molecules to produce 2 moles of Aluminium oxide. Therefore,

Since, aluminium reacts completely, it is the limiting reagent in the reaction. Therefore,

Atomic mass of AL = 27 g

Molar mass of  Al₂O₃ = 101.96 g/mol

4(27 g) of AL gives 2(101.96 g) of  Al₂O₃

3.6 g of AL will give ?

cross multiply

mass of  Al₂O₃ produced = 3.6 × 203.92 / 108   = 734.112 / 108 = 6.797

mass of  Al₂O₃ produced = 6.80 g.

read more: brainly.com/question/23982245?referrer=searchResults

8 0
3 years ago
An element has two naturally-occurring isotopes. The mass numbers of these isotopes are 113 amu and 115 amu, with natural abunda
DENIUS [597]

Answer: Its average atomic mass is 114.9 amu

Explanation:

Mass of isotope 1 = 113 amu

% abundance of isotope 1 = 5% = \frac{5}{100}=0.05

Mass of isotope 2 = 115 amu

% abundance of isotope 2 = 95% = \frac{95}{100}=0.95

Formula used for average atomic mass of an element :

\text{ Average atomic mass of an element}=\sum(\text{atomic mass of an isotopes}\times {{\text { fractional abundance}})

A=\sum[(113\times 0.05)+(115\times 0.95)]

A=114.9amu

Thus its average atomic mass is 114.9 amu

5 0
3 years ago
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