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Deffense [45]
3 years ago
6

Which option describes the most cost-effective way for a skateboard company to purchase plywood?

Chemistry
2 answers:
marta [7]3 years ago
6 0
A pallet of 500 sheets for $3,500 is the most cost-effective option.
This is because:
- if you buy 100 for $1,000, the price per sheet is $10.
- If you buy 200 for $1,800, the price per sheet is $9
- If you buy 1,000 for $8,000, the price per sheet is $8
- If you buy 500 for $3,500, the price per sheet is only $7.
Sati [7]3 years ago
6 0

Answer: C

Explanation:

The unit rate of the first problem is $10

The unit rate of the second problem is $9

The unit rate of the third problem is $7

The unit rate of the fourth problem is $8

The answer would be C because it has the lowest unit rate.

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A student has an unknown sample. How can spectroscopy be used to identify the sample?
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Answer: A. It can identify the elements in the sample.

Explanation: on edge

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Select the correct answer. Which of these elements is a transition metal?
PtichkaEL [24]

I don't see the options for an answer, so here is a list of all of the transition metals lol

  • <em>Scandium</em>
  • <em>Titanium</em>
  • <em>Vanadium</em>
  • <em>Chromium</em>
  • <em>Manganese</em>
  • <em>Iron</em>
  • <em>Cobalt</em>
  • <em>Nickel</em>
  • <em>Copper</em>
  • <em>Zinc</em>
  • <em>Yttrium</em>
  • <em>Zirconium</em>
  • <em>Niobium</em>
  • <em>Molybdenum</em>
  • <em>Technetium</em>
  • <em>Ruthenium</em>
  • <em>Rhodium</em>
  • <em>Palladium</em>
  • <em>Silver</em>
  • <em>Cadmium</em>
  • <em>Lanthanum</em>
  • <em>Hafnium</em>
  • <em>Tantalum</em>
  • <em>Tungsten</em>
  • <em>Rhenium</em>
  • <em>Osmium</em>
  • <em>Iridium</em>
  • <em>Platinum</em>
  • <em>Gold</em>
  • <em>Mercury</em>
  • <em>Actinium</em>
  • <em>Rutherfordium</em>
  • <em>Dubnium</em>
  • <em>Seaborgium</em>
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8 0
2 years ago
If you start with 6 moles of N2 and 6 moles of H2 (meaning you won't have enough of 1 of the ingredients), how many moles of NH3
Ivahew [28]

Answer:

4molNH_3

Explanation:

Hello there!

In this case, according to the given information it will be firstly necessary to set up the chemical equation taking place:

N_2+3H_3\rightarrow 2NH_3

We infer we need to calculate the moles of NH3 by using both of the moles of N2 and H2 at the beginning, in order to identify the limiting reactant:

n_{NH_3}=6molN_2*\frac{2molNH_3}{1molN_2}=12molNH_3\\\\ n_{NH_3}=6molH_2*\frac{2molNH_3}{3molH_2}=4molNH_3\\

Thus, since hydrogen yields the fewest moles of ammonia, we conclude that we are just able to yield 4 moles of NH3.

Regards!

8 0
3 years ago
Nitrogen effuses through a pinhole 1.7 times as fast as another gaseous element under the same conditions. Estimate the other el
marysya [2.9K]

Answer:

80.92, Krypton

Explanation:

<u>What is effusion?</u>

• It is a process where gas escapes through a pinhole (a very small hole) into a region of low pressure or vacuum

<u>Graham's law of effusion of </u><u>gas</u>

• states that at a given constant temperature and pressure, the rate of effusion of gases is inversely proportional to the square root of their molar masses

\boxed{ \frac{Rate_1}{Rate_2} =  \sqrt{ \frac{M_2}{M_1} } }

<u>Calculations</u>

Nitrogen exist as N₂ at room temperature, thus its molar mass is 2(14)= 28.

Let the rate and molar mass of unknown gas be Rate₂ and M₂ respectively.

Since N₂ effuses 1.7 times as fast as the unknown gas,

Rate₁= 1.7(Rate₂)

\frac{Rate_1}{Rate_2} = 1.7

1. 7 =  \sqrt{ \frac{M_2}{28} }

Square both sides:

2.89  = \frac{M_2}{28}

Multiply both sides by 28:

2.89(28)= M₂

M₂= 80.92

<u>Identity of </u><u>gas</u>

The molar mass of 80.92 lies between Bromine and Krypton. However since Bromine exist as Br₂, the value of it's molar mass would be 159.8 instead. Hence, Bromine is eliminated.

If the gas is a diatomic element, the atomic weight is 80.92 ÷2= 40.46. Thus, we are now considering if Argon could be its identity. However, Argon is a noble gas and will not exist as a diatomic element. Argon is therefore eliminated too.

Thus based on the above reasoning, its probable identity is Krypton.

7 0
3 years ago
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