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Dahasolnce [82]
3 years ago
13

I NEED HELP ASAP!! I DOMT KNOW HOW TO DO THIS

Chemistry
1 answer:
KatRina [158]3 years ago
3 0
What part of it are you confused about
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A chemist prepares a solution of sodium chloride(NaCl) by measuring out 25.4g of sodium chloride into a 100ml volumetric flask a
labwork [276]

Answer: 4 molL-1

Explanation:

Detailed solution is shown in the image attached. The number of moles of NaCl is first obtained. Since the molarity must be in units of molL-1, the volume is divided by 1000 and the formula stated in the solution is applied and the answer is given to one significant figure.

5 0
3 years ago
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In the scientific method, analysis should follow right after
alina1380 [7]
Observation/ question
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3 years ago
Prokaryotic is an example of a/an-<br><br> A:tissue <br> B:cell type<br> C:organ<br> D:organism
Leto [7]

Answer:

B. Cell Type

Explanation:

that is the answer i believe

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3 years ago
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Solid aluminum (AI) and oxygen (O_2) gas react to form solid aluminum oxide (Al_2O_3). Suppose you have 7.0 mol of Al and 9.0 mo
Nimfa-mama [501]

Answer:

There will be formed 3.5 moles of Al2O3 ( 356.9 grams)

Explanation:

Step 1: Data given

Numbers of Al = 7.0 mol

Numbers of mol O2 = O2

Molar mass of Al = 26.98 g/mol

Molar mass of O2 = 32 g/mol

Step 2: The balanced equation

4Al(s) + 3O2(g) → 2Al2O3(s)  

Step 3: Calculate limiting reactant

For 4 moles Al we need 3 moles O2 to produce 2 moles Al2O3

Al is the limiting reactant, it will be consumed completely (7 moles).

O2 is in excess.  There will react 3/4 * 7 = 5.25 moles

There will remain 9-5.25 = 3.75 moles

Step 4: Calculate moles Al2O3

For 4 moles Al we'll have 2moles Al2O3

For 7.0 moles of Al we'll have 3.5 moles of Al2O3 produced

Step 5: Calculate mass of Al2O3

Mass Al2O3 = moles Al2O3 * molar mass Al2O3

Mass Al2O3 = 3.5 moles* 101.96 g/mol

Mass Al2O3 = 356.9 grams

There will be formed 3.5 moles of Al2O3 ( 356.9 grams)

3 0
3 years ago
Which of the following statements about connecting paragraphs is correct?
Reika [66]
B.

And maybe put your question in the English/Literature tag next time lol
4 0
2 years ago
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