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Whitepunk [10]
3 years ago
13

Which objects will most likely float

Chemistry
1 answer:
storchak [24]3 years ago
6 0
Sponges, apples, light weight balls, etc.
they’re less dense than water
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5. All of the following statement describes compounds, EXCEPT?
Sveta_85 [38]

Answer:

C

Explanation:

Compounds cannot be separated by any physical means.

3 0
3 years ago
Read 2 more answers
Calculate the initial rate for the formation of c at 25 ∘c, if [a]=0.50m and [b]=0.075m
Nataly_w [17]
The rate of formation of a product depends on the the concentrations of the reactants in a variable way.

If two products, call them A and B react together to form product C, a general equation for the formation of C has the form:

rate = k*[A]^m * [B]^n

Where the symbol [ ] is the concentration of each compound.

Then, plus the concentrations of compounds A and B you need k, m and n.

Normally you run controled trials in lab which permit to calculate k, m and n .

Here the data obtained in the lab are:

<span>Trial      [A]      [B]         Rate </span><span>
<span>            (M)     (M)          (M/s) </span>
<span>1         0.50    0.010      3.0×10−3 </span>
<span>2         0.50    0.020       6.0×10−3 </span>
<span>3         1.00 0  .010       1.2×10−2</span></span>


Given that for trials 1 and 2 [A] is the same you can use those values to find n, in this way

rate 1 = 3.0 * 10^ -3 = k [A1]^m * [B1]^n

rate 2 = 6.0*10^-3 = k [A2]^m * [B2]^n

divide rate / rate 1 => 2 = [B1]^n / [B2]^n

[B1] = 0.010 and [B2] = 0.020 =>

6.0 / 3.0  =( 0.020 / 0.010)^n =>

2 = 2^n => n = 1

 
Given that for data 1 and 3 [B] is the same, you use those data to find m

rate 3 / rate 1 = 12 / 3.0   = (1.0)^m / (0.5)^m =>

4 = 2^m => m = 2

Now use any of the data to find k

With the first trial: rate = 3*10^-3 m/s = k (0.5)^2 * (0.1) =>

k = 3.0*10^-3 m/s / 0.025 m^3 = 0.12 m^-2 s^-1

Now that you have k, m and n you can use the formula of the rate with the concentrations given

rate = k[A]^2*[B] = 0.12 m^-2 s^-1 * (0.50m)^2 * (0.075m) = 0.0045 m/s = 4.5*10^=3 m/s

Answer: 4.5 * 10^-3 m/s
8 0
3 years ago
Read 2 more answers
Use Lewis diagrams to show how electrons are shared to form covalent bonds in compounds wit the following atoms. Write the formu
bixtya [17]
The Lewis structure/diagram for CH2O (aka Formaldehyde) can be written in either of the following ways shown in the picture.
The dots represent electrons in the valence shell of the atom (the outermost shell). The green dots are electrons that belong to the Oxygen atom, the blue belong to the Carbon atom, and the pink belong to the Hydrogen atoms.
Covalent bonds are bonds between atoms where atoms share electrons with each other. Atoms bond because they obey the octet rule ( the rule states that most atoms of main-group elements tend to want 8 electrons in their valence shells). 
Oxygen has 6 valence electrons, Carbon has 4, and Hydrogen has 1. H does not follow the octet rule, but C and O do, so the atoms are arranged in this way so that the O and C atoms have a full octet of electrons in their valence.

8 0
2 years ago
An artifact originally had 16 grams of​ carbon-14 present. the decay model upper a equals 16 e superscript negative 0.000121 ta=
xeze [42]

<u>Given:</u>

Initial amount of carbon, A₀ = 16 g

Decay model = 16exp(-0.000121t)

t = 90769076 years

<u>To determine:</u>

the amount of C-14 after 90769076 years

<u>Explanation:</u>

The radioactive decay model can be expressed as:

A = A₀exp(-kt)

where A = concentration of the radioactive species after time t

A₀ = initial concentration

k = decay constant

Based on the given data :

A = 16 * exp(-0.000121*90769076) = 16(0) = 0

Ans: Based on the decay model there will be no C-14 left after 90769076 years

3 0
3 years ago
Read 2 more answers
Percent concentration is one of the most common and basic concentration measurement used by general public. true or false?
Aleksandr [31]

Answer:

  • <u>TRUE:</u> <em>Percent concentration is one of the most common and basic concentration measurement used by general public</em>

Explanation:

In chemistry there are many <em>concentration measurements</em> used to describe the mixtures. Some of them are, percent, molarity, molality, and molar fraction, among others.

Percent concentration is a popular one because it is commonly understood and used by the non specialist people, i.e. general public.

The percent concentration of a component is defined as: (amount of component in the mixture / amount of mixture) × 100.

The amounts may be measured in mass units (e.g grams) or volume units (e.g. mililiters).

For solutions, mass percent concentration is:

  • % = (mass of solute / mass of solution) × 100.

And voluem percen contration is:

  • % = (volume of solute / volume of solution) × 100

Since percentage is used in many profesional and personal activities, most persons use it.

For example, rubbing alcohol, that everyone buys in pharmacies, is 70%; vinager, used in the food, is acetic acid at 5% - 8%.

5 0
3 years ago
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