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Lina20 [59]
3 years ago
8

Vinegar is a solution of acetic acid in water. If certain vinegar has a concentration of 1.5% v / v.

Chemistry
1 answer:
lina2011 [118]3 years ago
3 0
2 There is about 1000ml in a liter. 1 multiply 1050 by 0.15.
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New questionWhat mass of calcium chloride (CaCl₂) would beproduced from the reaction of 125.9 g of hydrochloriacid (HCI) with ex
marshall27 [118]

Answer:

191.6 g of CaCl₂.

Explanation:

What is given?

Mass of HCl = 125.9 g.

Molar mass of CaCl₂ = 110.8 g/mol.

Molar mass of HCl = 36.4 g/mol.

Step-by-step solution:

First, we have to state the chemical equation. Ca(OH)₂ react with HCl to produce CaCl₂:

Ca(OH)_2+2HCl\rightarrow CaCl_2+2H_2O.

Now, let's convert 125.9 g of HCl to moles using the given molar mass (remember that the molar mass of a compound can be found using the periodic table). The conversion will look like this:

125.9\text{ g HCl}\cdot\frac{1\text{ mol HCl}}{36.4\text{ g HCl}}=3.459\text{ moles HCl.}

Let's find how many moles of CaCl₂ are being produced by 3.459 moles of HCl. You can see in the chemical equation that 2 moles of HCl reacted with excess Ca(OH)₂ produces 1 mol of CaCl₂, so we state a rule of three and the calculation is:

3.459\text{ moles HCl}\cdot\frac{1\text{ mol CaCl}_2}{2\text{ moles HCl}}=1.729\text{ moles CaCl}_2.

The final step is to find the mass of CaCl₂ using the molar mass of CaCl₂. This conversion will look like this:

1.729\text{ moles CaCl}_2\cdot\frac{110.8\text{ g CaCl}_2}{1\text{ mol CaCl}_2}=191.6\text{ g CaCl}_2.

The answer would be that we're producing a mass of 191.6 g of CaCl₂.

4 0
1 year ago
What is the electron structure of sodium
yarga [219]
Electron structure of sodium:
₁₁Na: 1s²2s²2p⁶3s¹
8 0
3 years ago
Read 2 more answers
How many molecules are in 41.8 g of sulfuric acid
Anton [14]

Answer

× 10²³ molecules are in 41.8 g of sulfuric acid

Explanation

The first step is to convert 41.8 g of sulfuric acid to moles by dividing the mass of sulfuric acid by its molar mass.

Molar mass of sulfuric acid, H₂SO₄ = 98.079 g/mol

Mole=\frac{Mass}{Molar\text{ }mass}=\frac{41.8\text{ }g}{98.079\text{ }g\text{/}mol}=0.426187053\text{ }mol

Finally, convert the moles of sulfuric acid to molecules using Avogadro's number.

Conversion factor: 1 mole of any substance = 6.022 × 10²³ molecules.

Therefore, 0.426187053 moles of sulfuric acid is equal

\frac{0.426187053\text{ }mol}{1\text{ }mol}\times6.022×10²³\text{ }molecules=2.57\times10^{23}\text{ }molecules

Thus, 2.57 × 10²³ molecules are in 41.8 g of sulfuric acid.

3 0
1 year ago
A Calorie unit used in food is equal to the amount of energy necessary to raise the temperature of 1 kilogram of water ________
Kitty [74]

A Calorie unit used in food is equal to the amount of energy necessary to raise the temperature of 1 kilogram of water by <u>1</u> degrees Celsius.

<h3>What is One Calorie ?</h3>

The amount of heat energy required to raise the temperature by 1 gram of water through 1°C is known as One Calorie.

1 Calorie = 4.18 J

Thus from the above conclusion we can say that A Calorie unit used in food is equal to the amount of energy necessary to raise the temperature of 1 kilogram of water by <u>1</u> degrees Celsius.

Learn more about the One calorie here: brainly.com/question/1061571

#SPJ4

7 0
2 years ago
Draw the major organic substitution product(s) for (2R,3S)-2-bromo-3-methylpentane reacting with the given nucleophile. Indicate
Ede4ka [16]

Answer:

The final product of the reaction is (<em>2S,3S</em>)-2-ethoxy-3-methylpentane.

Explanation:

The given reaction undergoes S_{N}2  mechanism in which the nucleophile attacks the backside and it is substituted by the elimination of bromine.

Due to the backside attack of nucleophile , the inverse in stereo-chemistry is observed.

After the substitution of ethoxy group, the configuration is assigned according to the priority it shows clock wise direction(R) - configuration.

When hydrogen faces the front side , it results shows inverse configuration i.e, S- configuration.

The chemical reaction is as follows.

 

6 0
3 years ago
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