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sukhopar [10]
3 years ago
8

How many molecules of O2 are there in 2.0 moles of oxygen?

Chemistry
1 answer:
nignag [31]3 years ago
5 0
28 molecul are in mol of ox
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What is the normality of a solution containing 14.8 g of Ca(OH)₂ in 250.0 mL?
aliina [53]

Answer:

Explanation:Are You From Milo?

6 0
3 years ago
How are models useful in describing how molecules are formed
Nady [450]

Answer:

The purpose of molecular modeling is to provide a three-dimensional image (either physical or software-based) that allows a chemist to better see the manner in which atoms and molecules can interact. These models can be used to interpret existing observations or to predict new chemical behavior.

Explanation:

It can describe shape and how they connect while forming the electrons.

6 0
3 years ago
A compound has a molar mass of 90. grams per mole and the empirical formula CH2O. What is the molecular formula of this compound
k0ka [10]

Answer:

C₃H₆O₃

Explanation:

Data:

EF = CH₂O

MM = 90. g/mol

Calculations:

EF Mass = (12.01 + 2.016  + 16.00) u  = 30.03 u

The molecular formula is an integral multiple of the empirical formula.

MF = (EF)ₙ

n = \dfrac{\text{MF Mass}}{\text{EF Mass }} = \dfrac{\text{90. u}}{\text{30.03 u}} = 3.00  \approx 3

MF = (CH₂O)₃ = C₃H₆O₃

7 0
3 years ago
Is material that has a density of 0.2g/ml good for building a bridge or for making couch cushions
lisov135 [29]
D = 0.2 g / ml = 0.2 g / cm³
For example, density of steel is 7.85 g / cm³.
Density of pure water is 1.0 g/cm³. An object which has a density < 1.0 g/cm³ will float in water.
Answer: Material that has a density of 0.2 g/ml ( 0.2 g/cm³ ) is good for making couch cushions. 
8 0
3 years ago
Read 2 more answers
Vanillin (used to flavor vanilla ice cream and other foods) is the substance whose aroma the human nose detects in the smallest
balu736 [363]

Answer:

Cost to supply enough vanillin is 3.2\$

Explanation:

Threshold limit of vanillin in air is 2.0\times 10^{-11}g per litre means there should be 2.0\times 10^{-11}g of vanillin in 1L of air to detect aroma of vanillin.

1ft^{3}=28.32L

So, 5.0\times 10^{7}ft^{3}=(5.0\times 10^{7}\times 28.32)L

So amount of vanillin should be present to detect = (2.0\times 10^{-11}\times 5.0\times 10^{7}\times 28.32)g

As cost of 50 g vanillin is 112\$ therefore cost of  (2.0\times 10^{-11}\times 5.0\times 10^{7}\times 28.32)gvanillin = (2.0\times 10^{-11}\times 5.0\times 10^{7}\times 28.32\times 112)\$ = 3.2\$

5 0
3 years ago
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