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3 years ago

1)

Chemistry

1 answer:

zaharov [31]3 years ago

3 0

Answer:

B. what happens when iron rusts in the presence of water

Explanation:

Chemistry is a branch of science involving the composition and changes of matter. In other words, a chemist, who is a trained specialist in the science of chemistry, seeks to answer questions related to properties of matter.

In this question, a chemist is most likely to study "what happens when iron rusts in the presence of water" because it involves a change in the chemical properties of a substance (iron) i.e. a chemical change. However, what happens when ice melts to liquid water is a physical change.

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draw the structure of two acyclic compounds with 3 or more carbons which exhibits one singlet in the 1H-NMR spectrum

loris [4]

Answer:

attached below

Explanation:

Structure of two acyclic compounds with 3 or more carbons that exhibits one singlet in 1H-NMR spectrum

a) Acetone CH₃COCH₃

Attached below is the structure

b) But-2-yne (CH₃C)₂

Attached below is the structure

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2 years ago

In terms of bonds, what would the molecule C₆H₁₂ be classified as?

Colt1911 [192]

Answer:

Alkene

Explanation:

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3 years ago

How do I solve the following word problem: The half-life of Phosphorus - 32 is 14.3 days. It is used to study a plant's use of f

BaLLatris [955]

Half life is the time that it takes for half of the original value of some amount of a radioactive element to decay.

We have the following equation representing the half-life decay:

$A=A_o\times2^{(-\frac{t}{t_{half}})_{}_{}}$

A is the resulting amount after t time

Ao is the initial amount = 50 mg

t= Elapsed time

t half is the half-life of the substance = 14.3 days

We replace the know values into the equation to have an exponential decay function for a 50mg sample

$A=\text{ 50 }\times2^{\frac{-t}{14.3}}$

That would be the answer for a)

To know the P-32 remaining after 84 days we have to replace this value in the equation:

$\begin{gathered} A=\text{ 50 }\times2^{\frac{-84}{14.3}} \\ A=0.85\text{ mg} \end{gathered}$

So, after 84 days the P-32 remaining will be 0.85 mg

4 0

1 year ago

30 its ASAP!!!! What is the volume, in liters, occupied by 0.485 moles of Oxygen gas at 23.0 oC and 0.980 atm?

USPshnik [31]

Use the universal gas formula

PV=nRT
where
P=pressure ( 0.980 atm)
V=volume (L)
T=temperature ( 23 ° C = 23+273.15 = 296.15 ° K)
n=number of moles of ideal gas (0.485 mol)
R=universal gas constant = 0.08205 L atm / (mol·K)

Substitute values,
Volume, V (in litres)
=nRT/P
=0.485*0.08205*296.15/0.980
= 12.0256 L
= 12.0 L (to three significant figures)

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3 years ago

How many formula units are in 35.0 g KNO3? (KNO3, 101.11 g/mol) [?]×10¹²] fun KNO3

alina1380 [7]

mol=35:101.11 g/mol=0.346

formula units = 0.346 x 6.02 x 10²³ = 2.082 x 10²³

3 0

1 year ago