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kondor19780726 [428]
2 years ago
15

PLEASE HELP

Chemistry
2 answers:
Marysya12 [62]2 years ago
6 0

Answer:

Military

Explanation:

I may be wrong but military seems most likely

Maru [420]2 years ago
5 0

Answer: c or 3, "To react to Noriega's attacks".

Explanation: I got right in assignment 2021

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Question 7
chubhunter [2.5K]
This answer is D or c
5 0
2 years ago
An oxide of aluminum contains 0.545 g of Al and 0.485 g of O. Find the empirical formula
uysha [10]

Answer:

The answer is: <u>Al2O3</u>

Explanation:

The data they give us is:

  • 0.545 gr Al
  • 0.485 gr O.

To find the empirical formula without knowing the grams of the compound, we find it per mole:

  • 0.545 g Al * 1 mol Al / 27 g Al = 0.02 mol Al
  • 0.485 g O * 1 mol O / 16 g O = 0.03 mol O

Then we must divide the results obtained by the lowest result, which in this case is 0.02:

  • 0.02 mol Al / 0.02 = 1  Al
  • 0.03 mol O / 0.02 = 1.5  O

Since both numbers have to give an integer, multiply by 2 until both remain integers:

  • 1Al * 2 = 2Al
  • 1.5O * 2 = 3O

Now the answer is given correctly:

  • Al2O3

8 0
2 years ago
Describe the relationship between the radius of a cation and that of the atom from which forms.
vichka [17]

The radius of the cation is much smaller than the corresponding neutral atom.(b) The radius of an anion is much larger than the corresponding neutral atom.Explanation:The size of the atom or ion is inversely proportional to the nuclear charge experienced by the electrons.(a)The size of the cation is smaller than the size of the corresponding neutral atom. This is because after removal of an electron from the highest principle energy level the nuclear charge experienced by the valence electrons increases resulting in the decrease in size.(b)The size of an anion is larger than the size of the corresponding neutral atom. In an anion, an extra electron is added to the highest principle energy level but the effective nuclear charge pulling the electrons towards the nucleus is still same. The net effective nuclear charge experienced by the electrons present in the outermost shell decrease. Moreover, due to the added electron, the repulsion between the electrons also increases resulting in the increase in size

Make since? i hope this helps

4 0
3 years ago
What would be the volume in millilitres of a blood sample of 2.15 microliters (ul)?​
Zanzabum

2.15 x 10⁻³mL

Explanation:

Given parameter:

    Volume of blood sample in uL = 2.15uL

Conversion           uL → mL

   micro- and milli-  are both prefixes of sub-units.

liter is a unit of volume of a substance.

       micro - is 10⁻⁶

       milli- is of the order 10⁻³

The problem is converting from micro to milli:

     if we multiply  10⁻⁶ by 10³ we would have our milli;

  1000uL = 1mL

  2.15uL :   2.15uL x \frac{1mL}{1000uL} = 2.15 x 10⁻³mL

learn more:

Volume brainly.com/question/5055270

#learnwithBrainly

4 0
3 years ago
(4) Calculate the % of a compound that can be removed from liquid phase 1 by using ONE to FOUR extractions with a liquid phase 2
maksim [4K]

Answer:

One extraction: 50%

Two extractions: 75%

Three extractions: 87.5%

Four extractions: 93.75%

Explanation:

The following equation relates the fraction q of the compound left in volume V₁ of phase 1 that is extracted n times with volume V₂.

qⁿ = (V₁/(V₁ + KV₂))ⁿ

We also know that V₂ = 1/2(V₁) and K = 2, so these expressions can be substituted into the above equation:

qⁿ = (V₁/(V₁ + 2(1/2V₁))ⁿ = (V₁/(V₁ + V₁))ⁿ =  (V₁/(2V₁))ⁿ = (1/2)ⁿ

When n = 1, q = 1/2, so the fraction removed from phase 1 is also 1/2, or 50%.

When n = 2, q = (1/2)² = 1/4, so the fraction removed from phase 1 is (1 - 1/4) = 3/4 or 75%.

When n = 3, q = (1/2)³ = 1/8, so the fraction removed from phase 1 is (1 - 1/8) = 7/8 or 87.5%.

When n = 4, q = (1/2)⁴ = 1/16, so the fraction removed from phase 1 is (1 - 1/16) = 15/16 or 93.75%.

5 0
3 years ago
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