Chemistry : asked on tink921
 30.08.2022

(Popcorn Stand) A salt shaker has 2.91 moles of NaCl, how many grams of NaCl are in the salt shaker?

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24.06.2023, solved by verified expert
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170 grams NaCl

Explanation:

Find the mass of NaCl by adding the two elements atomic mass

Na: 22.99

Cl: 35.45

22.99+35.45=58.44 g

Convert moles to grams.

2.91 mol NaCl * 58.44 g NaCl= 170.0604 g

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Chemistry
Step-by-step answer
P Answered by PhD

170 grams NaCl

Explanation:

Find the mass of NaCl by adding the two elements atomic mass

Na: 22.99

Cl: 35.45

22.99+35.45=58.44 g

Convert moles to grams.

2.91 mol NaCl * 58.44 g NaCl= 170.0604 g

Chemistry
Step-by-step answer
P Answered by Master
Answer: 0.0045 mol
Explanation: Convert 30 ml to l: 30 mL = 0.03 L
Molarity = mol/l
mol = molarity * L
mol = 0.15 * 0.03 = 0.0045 mol
Answer: 0.0045 mol
Explanation: Convert 30 ml to l: 30 mL = 0.03 L
Molarity = mol/l
mol = molarit
Chemistry
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P Answered by PhD

Answer:

52.6 gram

Step-by-step explanation:

It is clear by the equation 2(27+3×35.5)= 267 gm of AlCl3 reacts with 6× 80 = 480 gm of Br2 . So 29.2 gm reacts = 480× 29.2/267= 52.6 gm

Chemistry
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P Answered by PhD

glycoproteins

Explanation:

A positive reaction for Molisch's test is given by almost all carbohydrates (exceptions include tetroses & trioses). It can be noted that even some glycoproteins and nucleic acids give positive results for this test (since they tend to undergo hydrolysis when exposed to strong mineral acids and form monosaccharides).

Chemistry
Step-by-step answer
P Answered by PhD
15 moles.Explanation:Hello,In this case, the undergoing chemical reaction is:Clearly, since carbon and oxygen are in a 1:1 molar ratio, 15 moles of carbon will completely react with 15 moles of oxygen, therefore 15 moles of oxygen remain as leftovers. In such a way, since carbon and carbon dioxide are also in a 1:1 molar ratio, the theoretical yield of carbon dioxide is 15 moles based on the stoichiometry:Best regards.
Chemistry
Step-by-step answer
P Answered by PhD
Answer: 25 g
Explanation: Given:
Original amount (N₀) = 100 g
Number of half-lives (n) = 11460/5730 = 2
Amount remaining (N) = ?
N = 1/2ⁿ × N₀
N = 1/2^2 × 100
N = 0.25 × 100
N = 25 g
Chemistry
Step-by-step answer
P Answered by PhD
Answer: 7.8125 g
Explanation: Given:
Original amount (N₀) = 500 g
Number of half-lives (n) = 9612/1602 = 6
Amount remaining (N) = ?
N = 1/2ⁿ × N₀
N = 1/2^6 × 500
N = 0.015625 × 500
N = 7.8125 g
Chemistry
Step-by-step answer
P Answered by PhD
Answer: The product formed is potassium chloride.
Explanation:
Precipitation reaction is defined as the chemical reaction in which an insoluble salt is formed when two solutions are mixed containing soluble substances. The insoluble salt settles down at the bottom of the reaction mixture.

The chemical equation for the reaction of potassium phosphate and magnesium chloride follows (look at the picture)

2 moles of aqueous solution of potassium phosphate reacts with 3 moles of aqueous solution of magnesium chloride to produce 1 mole of solid magnesium phosphate and 6 moles of aqueous solution of potassium chloride.
Answer: The product formed is potassium chloride.
Explanation:
Precipitation reaction is defined a
Chemistry
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P Answered by PhD
Answer: B. carbon tetrachloride, CCI4
Explanation: The other options are incorrect. Let's write the correct formulas:
A. Diarsenic pentoxide - As2O5
C. Sodium dichromate - Na2Cr2O7
D. magnesium phosphide - Mg3P2
Chemistry
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P Answered by PhD
Answer: a. basic
b. basic
c. acidic
d. neutral

Explanation: Acids and bases can be classified in terms of hydrogen ions or hydroxide ions, or in terms of electron pairs. (look at the picture)
Let us note that from the pH scale, a pH of;
0 - 6.9 is acidic
7 is neutral
8 - 14 is basic

But pH= - log [H^+]
pOH = -log [OH^-]
Then;
pH + pOH = 14
Hence;
pH = 14 - pOH

a. [H+] = 6.0 x 10-10M
pH= 9.22 is basic
b. [OH-] = 30 × 10-2M
pH = 13.5 is basic
C. IH+1 = 20× 10-7M
pH = 0.56 is acidic
d. [OH-] = 1.0 x 10-7M
pH = 7 is neutral
Answer: a. basic
b. basic
c. acidic
d. neutral

Explanation: Acids and bases can be classified

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