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3 years ago

Give example of a homogeneous mixture

2 answers:

8 0

Hello There

Answer: An example of a homogeneous mixture is water. It seems to only have one substance, but it contains minerals and <span>gases, but these are dissolved in the water.

I hope this helps
-Chris</span>

White raven [17]3 years ago

4 0

Fluid illustrations incorporate unadulterated water, white vinegar, sugar water, corn oil, and blood plasma. Shockingly, homogeneous blends are not restricted to fluids, they can likewise be gases and solids. Another kind of homogeneous blend is known as a colloid.

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The reaction between ethylene and hydrogen bromide to form ethyl bromide is carried out in a continuous reactor. The product str

iren2701 [21]

Answer:

Explanation:

The first thing to do is to write out the chemical equation showing the reaction between the two chemical compounds.

C2H4 + HBr ------------------------> C2H5Br.

The chemical species/compounds reacted in the ratio of 1 : 1 : 1. That is one mole of C2H4 reacted with one mole of HBr to give one mole of C2H5Br.

So, we are given; C2H5Br = 58.7 mol%, HBr = 18.3 mol%. Thus, for C2H4, we have [ 100 - (58.7 + 18.3)] = 23 mol%. Therefore, the limiting reagent is HBr.

So, if we have 100 mol/s ;

C2H5Br = 100 × 0.587 = 58.7 mol/s, HBr = 100 × 0.183 = 18.3 mol/s and C2H4 = 100 × 0.23 = 23 mol/s.

Therefore, the total value of the flow rate in the feed = (HBr in feed) mol/s + (C2H4 in feed) mol/s = (58.7 + 18.3)mol/s + (58.7 + 23) mol/s =( 77 + 81.7) mol/s = 158.7 mol/s. .

So, 100/158.7 × 30.5 = 192.2.

Also the feed composition in HBr and C2H4 is;

HBr = 77/158.7 = 0.4852, C2H4 = 81.7/158.7 = 0.515.

Therefore, the number of HBr in feed and C2H4 in feed = 160m/s and 200m/s respectively (approx.)

Therefore, the fraction conversion is;

8 0

3 years ago

HELP! ASAP!!! WILL MARK BRAINLIEST!!

san4es73 [151]

Answer:

Alkali Metals are Group 1 so

Rb it isnt Hydrogen because it is a gas

Explanation:

8 0

3 years ago

Read 2 more answers

Which is a difference between global and local winds?

valentina_108 [34]

Explanation:

local winds are considered breezes while global winds produce mostly storms

3 0

3 years ago

Calculate the freezing point (0°C) of a 0.05500 m aqueous solution of glucose.

Rainbow [258]

Answer:

-0.1767°C (Option A)

Explanation:

Let's apply the colligative property of freezing point depression.

ΔT = Kf . m. i

i = Van't Hoff factot (number of ions dissolved). Glucose is non electrolytic so i = 1

m = molality (mol of solute / 1kg of solvent)

We have this data → 0.095 m

Kf is the freezing-point-depression constantm 1.86 °C/m, for water

ΔT = T° frezzing pure solvent - T° freezing solution

(0° - T° freezing solution) = 1.86 °C/m . 0.095 m . 1

T° freezing solution = - 1.86 °C/m . 0.095 m . 1 → -0.1767°C

4 0

3 years ago

A compound consisting of B, N, and H undergoes elemental analysis. The % composition by mass is found to be 40.28% B, 52.20% N,

swat32

Empirical formula is the simplest ratio of components making up a compound.

The percentage composition of each element has been given

therefore the mass present of each element in 100 g of compound is

B N H

mass 40.28 g 52.20 g 7.53 g

number of moles

40.28 g / 11 g/mol 52.20 g / 14 g/mol 7.53 g / 1 g/mol

= 3.662 mol = 3.729 mol = 7.53 mol

divide the number of moles by the least number of moles, that is 3.662

3.662 / 3.662 3.729 / 3.662 7.53 / 3.662

= 1.000 = 1.018 = 2.056

the ratio of the elements after rounding off to the nearest whole number is

B : N : H = 1 : 1 : 2

therefore empirical formula for the compound is B₁N₁H₂

that can be written as BNH₂

3 0

3 years ago