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Ipatiy [6.2K]
4 years ago
14

What type of mixtures are these?

Chemistry
2 answers:
Annette [7]4 years ago
7 0

Answer:

not sure

Explanation:

Anestetic [448]4 years ago
3 0
Food coloring dissolved in water
> Homogeneous Mixture

Peanuts and almonds mixed together in a bowl
> Heterogeneous Mixture

A bucket full of sand and gravel
> Heterogeneous Mixture

A cup of tea and sugar
> Homogeneous Mixture
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What volume of an hcl solution with a ph of 1.3 can be neutralized by one dose of milk of magnesia?
kvv77 [185]

274 mL H3 O+ and fully neutralized

It will take one teaspoon of Mg(OH)2 to completely neutralize 2.00×10^2mL  of H3O+.

<h3>What is the purpose of milk of magnesia?</h3>
  • For a brief period of time, this medicine is used to relieve sporadic constipation.
  • It is an osmotic laxative, which means that it works by drawing water into the intestines, which aids in causing bowel movement.
<h3>What dosage of milk of magnesia is recommended for constipation?</h3>
  • Take Milk of Magnesia once day, preferably before bed, in divided doses, or as prescribed by a physician.
  • suggested dosage: 30 mL to 60 mL for adults and kids 12 years of age and older. 15 mL to 30 mL for children aged 6 to 11 years.

learn more about milk of magnesia here

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the question you are looking for is

People often take milk of magnesia to reduce the discomfort associated with acid stomach or heartburn. The recommended dose is 1 teaspoon, which contains 4.00x 10^{2} mg of Mg(OH)_2. What volume of an HCl solution with a pH of 1.3 can be neutralized by one dose of milk of magnesia? If the stomach contains 2.00x10^{2}mL of pH 1.3 solution, is all the acid neutralized? If not, what fraction is neutralized?

7 0
1 year ago
1 How is the density of a substance calculated? 
user100 [1]
Sorry for the scribbles lol

4 0
3 years ago
In the laboratory, a volume of 100 mL of sulfuric acid (H2SO4) is recorded. How many g are there of the liquid if its density is
ser-zykov [4K]

Answer:

\large \boxed{\text{183 g}}  

Explanation:

\begin{array}{rcl}\text{Density} & = & \dfrac{\text{Mass}}{\text{Volume}}\\\\\rho & = &\dfrac{m}{V}\\\\1.83 \text{ g$\cdot$ cm}^{-3} & = & \dfrac{m}{\text{100 cm}^{3}}\\\\m & = & \text{183 g}\\\end{array}\\\text{There are $\large \boxed{\textbf{183 g}}$ of sulfuric acid.}

8 0
4 years ago
Calculate the molarity of a solution prepared by dissolving 0.2 mol sucrose in enough water to make a 100 ml solution.
Lelechka [254]

Taking into account the definition of molarity, the molarity of a solution prepared by dissolving 0.2 mol sucrose in enough water to make a 100 mL solution is 2 \frac{moles}{liter}.

<h3>Definition of molarity</h3>

Molar concentration or molarity is a measure of the concentration of a solute in a solution and indicates the number of moles of solute that are dissolved in a given volume.

The molarity of a solution is calculated by dividing the moles of solute by the volume of the solution:

molarity=\frac{number of moles}{volume}

Molarity is expressed in units \frac{moles}{liter}.

<h3>Molarity in this case</h3>

In this case, you have:

  • number of moles= 0.2 moles
  • volume= 100 mL= 0.1 L

Replacing in the definition of molarity:

molarity=\frac{0.2 mole}{0.1 L}

Solving:

<u><em>molarity= 2 </em></u>\frac{moles}{liter}

Finally, the molarity of a solution prepared by dissolving 0.2 mol sucrose in enough water to make a 100 mL solution is 2 \frac{moles}{liter}.

Learn more about molarity:

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8 0
2 years ago
Calculate the equilibrium constant at 25 ∘C for the reaction Fe(s)+2Ag+(aq)→Fe2+(aq)+2Ag(s)
Murrr4er [49]

Answer:

1.7 × 10 ^42

Explanation:

Using Nernst equation

E°cell = RT/nF Inq

at equilibrium

Q=K

E°cell  = 0.0257 /n Ink= 0.0592/n log K

Fe2+(aq)+2e−→Fe(s)     E∘= −0.45 V

Ag+aq)+e−→Ag(s)         E∘= 0.80 V

Fe(s)+2Ag+(aq)→Fe2+(aq)+2Ag(s)

balance the reaction

Fe → Fe²⁺ + 2e⁻  reversing for oxidation E° = 0.45 v

2 Ag⁺ +2e⁻ → 2Ag

n = 2 moles  and K = equilibrium constant

E° cell = 0.80 + 0.45 = 1.25 V

E° cell = (0.0592 / n) log K  

substitute the value into the equations and solve for K

(1.25 × 2) / 0.0592  = log K

42.23 = log K

k = 10^ 42.23

K = 1.7 × 10 ^42

8 0
3 years ago
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