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Ivenika [448]
3 years ago
15

Hey guys! How is the answer C? I need steps plz

Chemistry
1 answer:
oksian1 [2.3K]3 years ago
5 0
So basically the answer is c bc if your dividing then multiplying you’d boils get your answer
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In standardizing a naoh solution a student found that 25.55cm of base neutralize exactly 21.35cm of 0.12M HCl find the molarity
nalin [4]

Answer:

O.1M

Explanation:

First let's generate a balanced equation for the reaction

NaOH + HCl —>NaCl + H2O

From the equation,

The ratio of the acid to base is 1:1.

From the question, we obtained the following:

Ma = Molarity of acid = 0.12M

Va = volume of acid = 21.35cm3

Vb = volume of base = 25.55cm3

Mb = Molarity of base =?

We obtained nA(mole of acid) and nB(mole of base) to be 1

The molarity of the base can be calculated for using:

MaVa/ MbVb = nA / nB

0.12x21.35 / Mb x 25.55 = 1

Cross multiply to express in linear form

Mb x 25.55 = 0.12x21.35

Divide both side by 25.55

Mb = (0.12x21.35) / 25.55

Mb = 0.1M

The molarity of the base is 0.1M

6 0
3 years ago
Silicon is prepared by the reduction of K₂SiF6 with Al. Write the equation for this reaction. (Hint: Can F⁻ be oxidized in this
BlackZzzverrR [31]

4Al + 3K2SiF6 = 6KF + 3Si + 4AIF3 is the reaction for preparation of silicon by the reduction of K₂SiF6 with Al.

AlF3xH2O-based inorganic compounds are referred to as aluminium fluoride. They are all solids without colour. Aluminium fluoride is a crystalline (sand-like), odourless, white, or colourless powder. In addition to being used to make aluminium, it also functions as a flux in welding processes and in ceramic glazes and enamels.

Silicon (Si) is created by reducing potassium silicofluoride with aluminium as the reducing agent (K2SIF6). While K2SiF6 is reduced to Si in this equation, aluminium is oxidised to aluminium fluoride. As a result, the balanced equation describing aluminum's reduction of K2SiF6 to silicon non-metal is as follows: 4Al + 3K2SiF6 = 6KF + 3Si + 4AIF3

Learn more about aluminium fluoride here:

brainly.com/question/17131529

#SPJ4

3 0
1 year ago
Somebody answer please a picture is attached!??
DanielleElmas [232]
9.2 that is the answer
7 0
3 years ago
How many moles of oxygen are present in 33.6l of the gas at 1atm and 0c
andreev551 [17]

The answer is: 1.5 moles of oxygen are present.

V(O₂) = 33.6 L; volume of oxygen.

p(O₂) = 1.0 atm; pressure of oxygen.

T = 0°C; temperature.

Vm = 22.4 L/mol; molar volume at STP (Standard Temperature and Pressure).

At STP one mole of gas occupies 22.4 liters of volume.

n(O₂) = V(O₂) ÷ Vm.

n(O₂) = 33.6 L ÷ 22.4 L/mol.

n(O₂) = 1.50 mol; amount of oxygen.

5 0
3 years ago
Xa4D6A/viewform?hr_submission=ChglisyXkhUSEAj9|d2MkQsSBwjg24qS6woo
Juli2301 [7.4K]
It’s should be option 4
4 0
3 years ago
Read 2 more answers
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