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Ede4ka [16]
3 years ago
13

(1.7 x 10-5 m) x (3.72 x 10-4 m) *

Chemistry
1 answer:
rosijanka [135]3 years ago
7 0

Answer:

The answer will be 1992/5 or 398.4

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First person with the right answer gets brainliest thanks (btw the numbers on the right are the answers choose the right one)
a_sh-v [17]

Answer:

there isnt anything

Explanation:

7 0
2 years ago
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Fix the formula for KF2<br> Write out the full correct formula?
Korvikt [17]

Given :

A chemical compound KF_2 .

To Find :

Fix the formula for KF_2 and write the full correct formula.

Solution :

We know, Potassium( K ) and Fluorine( F ) both have a valency of 1 i.e potassium can donate one electron and Fluorine can accept one electron only.

So, the chemical formula KF_2 is wrong because no element has filled electron.

Therefore, to stabilise the molecule, 1 Potassium atom should make a bond from 1 Fluorine i.e KF ( correct formula ) .

Hence, this is the required solution.

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2 years ago
The first ionization energy of mg is 735 kj/mol. calculate zeff.
Novay_Z [31]
2.25 I believe

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STSN
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3 years ago
At a cost of $1600/oz, how much would you have to pay for a solid cubic foot of gold?
Alex777 [14]
<h3>You will pay $ 30876800</h3>

We'll begin by calculating the mass in ounce (oz) of a cube foot (ft³) of gold. This can be obtained as follow:

<h3 />

Density of gold = 19298 oz/ft³

Volume of gold = 1 ft³

<h3>Mass of gold =?</h3>

Density = mass /volume

19298 = mass / 1

<h3>Mass of gold = 19298 oz</h3>

Finally, we shall determine the cost of 19298 oz of gold. This can be obtained as follow:

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Therefore,

19298 oz = 19298 × 1600

19298 oz = $ 30876800

Therefore, a solid cube foot of gold (i.e 19298 oz) will cost $ 30876800

Learn more: brainly.com/question/15407624

4 0
2 years ago
if you are told to get 100 mL of stock solution to use to prepare smaller size sample for an experiment, which piece of glasswar
fgiga [73]

Answer:

A beaker  

Step-by-step explanation:

Specifically, I would use a 250 mL graduated beaker.

A beaker is appropriate to measure 100 mL of stock solution, because it's easy to pour into itscwide mouth from a large stock bottle.

You don't need precisely 100 mL solution.

If the beaker is graduated, you can easily measure 100 mL of the stock solution.

Even if it isn't graduated, 100 mL is just under half the volume of the beaker, and that should be good enough for your purposes (you will be using more precise measuring tools during the experiment).

6 0
2 years ago
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