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3 years ago

It is a type of mixture that separates on standing

Chemistry

2 answers:

vaieri [72.5K]3 years ago

8 0

B suspensions :) sorry if I’m wrong

erma4kov [3.2K]3 years ago

6 0

Answer:

B) Suspensions

In suspension the size of particles are big as compared to solution

Colloids shows Tyndall effect (,for ex milk)

You might be interested in

What is woven into farbic?

LenaWriter [7]

Woven fabric is any textile formed by weaving. Woven fabrics are often created on a loom, and made of many threads woven on a warp and a weft. Technically, a woven fabric is any fabric made by interlacing two or more threads at right angles to one another.

4 0

3 years ago

A 50.0-mL volume of 0.15 M HBr is titrated with 0.25 M KOH. Calculate the pH after the addition of 13.0 mL of KOH. Express your

Mazyrski [523]

Answer:

pH= 1.17

Explanation:

The neutralization reaction between HBr (acid) and KOH (base) is given by the following equation:

HBr(aq) + KOH(aq) → KBr(aq) + H₂O(l)

According to this equation, 1 mol of HBr reacts with 1 mol of KOH. Then, the moles can be expressed as the product between the molarity of the acid/base solution (M) and the volume in liters (V). So, we calculate the moles of acid and base:

Acid:

M(HBr) = 0.15 M = 0.15 mol/L

V(HBr) = 50.0 mL x 1 L/1000 mL = 0.05 L

moles of HBr = M(HBr) x V(HBr) = 0.15 mol/L x 0.05 L = 7.5 x 10⁻³ moles HBr

Base:

M(KOH) = 0.25 M = 0.25 mol/L

V(HBr) = 13.0 mL x 1 L/1000 mL = 0.013 L

moles of HBr = M(HBr) x V(HBr) = 0.25 mol/L x 0.013 L = 3.25 x 10⁻³ moles KOH

Now, we have: 7.5 x 10⁻³ moles HBr > 3.25 x 10⁻³ moles KOH

HBr is a strong acid and KOH is a strong base, so they are completely dissociated in water: the acid produces H⁺ ions and the base produces OH⁻ ions. So, the difference between the moles of HBr and the moles of KOH is equal to the moles of remaining H⁺ ions after neutralization:

moles of H⁺ = 7.5 x 10⁻³ moles HBr - 3.25 x 10⁻³ moles KOH = 4.25 x 10⁻³ moles H⁺

From the definition of pH:

pH = -log [H⁺]

The concentration of H⁺ ions is calculated from the moles of H⁺ divided into the total volume:

total volume = V(HBr) + V(KOH) = 0.05 L + 0.013 L = 0.063 L

[H⁺] = (moles of H⁺)/(total volume) = 4.25 x 10⁻³ moles/0.063 L = 0.067 M

Finally, we calculate the pH after neutralization:

pH = -log [H⁺] = -log (0.067) = 1.17

3 0

2 years ago

What is the molarity of a solution prepared by dissolving 8 grams of BaCl2 in enough

olga55 [171]

Answer:

Explanation:

molar mass of BaCl2 = 208.23

mol of BaCl2 = 8/208.22

mol of BaCl2 = 0.03841905

Molarity = 0.03841905/0.450

Molarity = 0.085 M

6 0

2 years ago

How many moles do you have in 37.3 g of Co(CrO4)3

shutvik [7]

Answer:

0.0917 mol Co(CrO₄)₃

General Formulas and Concepts:

Chemistry - Atomic Structure

Reading a Periodic Table
Using Dimensional Analysis

Explanation:

Step 1: Define

37.3 g Co(CrO₄)₃

Step 2: Identify Conversions

Molar Mass of Co - 58.93 g/mol

Molar Mass of Cr - 52.00 g/mol

Molar Mass of O - 16.00 g/mol

Molar Mass of Co(CrO₄)₃ - 58.93 + 3(52.00) + 12(16.00) = 406.93 g/mol

Step 3: Convert

$37.3 \ g \ Co(CrO_4)_3(\frac{1 \ mol \ Co(CrO_4)_3}{406.93 \ g \ Co(CrO_4)_3} )$ = 0.091662 mol Co(CrO₄)₃

Step 4: Check

We are given 3 sig figs. Follow sig fig rules and round.

0.091662 mol Co(CrO₄)₃ ≈ 0.0917 mol Co(CrO₄)₃

8 0

2 years ago

How many seconds does it take a race horse to run six furlongs at 40.7 miles per hour if: I furlong = 40 rods; 5.5 yards = 1 rod

lana [24]

Distance traveled by the race horse=6 furlongs

Converting the distance from furlongs to rods: 40 rods =1 furlong

$6 furlongs*\frac{40rods}{1furlong} =240rods$

Converting the distance from rods to yards, feet and miles: 5.5 yards = 1 rod, 3foot =1 yard, 1 mile = 5280 feet.

$240rods*\frac{5.5yards}{1rod}*\frac{3foot}{1yard}*\frac{1mi}{5280feet}=0.75mi$

The given speed of race horse = 40.7mi/hr

Calculating the time required:

$0.75mi*\frac{1hr}{40.7mi}*\frac{60min}{1hr}*\frac{60s}{1min}= 66.34s$

Therefore, 66.34 s is required for a race horse to run six furlongs at 40.7 miles per hour.

8 0

3 years ago