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4 years ago

PLEASE HELP ASAP! I WILL MARK AS BRAINLIEST

Chemistry

2 answers:

Triss [41]4 years ago

8 0

From my knowledge i believe the young human is correct the answer is 37 root

olchik [2.2K]4 years ago

5 0

Answer:

Explanation:

the awnser is 37square root to the bse of 3

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Platinum crystallizes with the face-centered cubic unit cell. The radius of a platinum atom is 139 pm. Calculate the edge length

horrorfan [7]

Answer:

Length = 393pm, Density = 21.3 g/cm^3.

Explanation:

From the question above, we have the following parameters or data which is going to aid in solving the above Question.

=> The radius of a platinum atom = 139 pm.

Therefore, the length can be calculated by making use of the formula given below:

Length = 2 √( 2r) = 2 × √ (2 × 139 × 10^-12m ) = 393 × 10^-10 m = 393pm.

The density can be calculated by making use of the chemical formula given below:

Density = mass ÷ volume = (195.064/ 6.02 × 10^23) ÷ (3.93 × 10^-10/ 10^-2) = 21.3 g/cm^3.

3 0

3 years ago

How many moles of Al are necessary to form 23.6 g of AlBr₃ from this reaction: 2 Al(s) + 3 Br₂(l) → 2 AlBr₃(s) ?

Gnoma [55]

Answer:

0.088 mole of Al.

Explanation:

First, we shall determine the number of mole in 23.6 g of AlBr₃.

This is illustrated below:

Mass of AlBr₃ = 23.6 g

Molar Mass of AlBr₃ = 27 + 3(80) = 267 g/mol

Mole of AlBr₃ =.?

Mole = mass/Molar mass

Mole of AlBr₃ = 23.6 / 267

Mole of AlBr₃ = 0.088 mol

Next, we shall writing the balanced equation for the reaction.

This is given below:

2Al(s) + 3Br₂(l) → 2AlBr₃(s)

From the balanced equation above,

2 moles of Al reacted with 3 mole of Br₂ to 2 moles AlBr₃.

Finally, we shall determine the number of mole of Al needed for the reaction as follow:

From the balanced equation above,

2 moles of Al reacted to 2 moles AlBr₃.

Therefore, 0.088 mole of Al will also react to produce 0.088 mole of AlBr₃.

4 0

3 years ago

Which procedure must be completed before switching to high power objective on a microscope

Elis [28]

1.turn the revolving turret so that low power lens come into position
2 place the microscope slide on the stage
3 look at the objective lens and the stage from the side as you turn adjust knob so that stage can move upward
4 look through the eyepiece and move the focus knob until image come into focus
5.adjust the condenser to regulate thee amount of light
6 move the microscope slide around until the sample is in the center of field of view

7 0

4 years ago

How many atoms are present in 179.0 g of iridium? include units please:)!

denpristay [2]

Answer:

179.0 g of iridium (1 mol / 192.217 g) ( 6.022 x 10^23 atoms / 1 mol ) = 5.61 x 10^23 atoms of iridium

Explanation:

5 0

4 years ago

Read 2 more answers

Scintillatium has a halflife of 16 minutes. if a sample has 800 grams, find a formula for its mass after t minutes.

hichkok12 [17]

Halflife is the time needed for a radioactive molecule to decay half of its current mass. If t is the time elapsed, the formula for the halflife would be:
final mass= original mass * $\frac{1}{2} ^{\frac{t}{halflife}$

If you put the information of the problem into the formula, the equation will be:
final mass= original mass * $\frac{1}{2} ^{\frac{t}{halflife}$
final mass= 800g * $\frac{1}{2} ^{\frac{t}{16 min}$

8 0

4 years ago