Ask question

Ask question

All categories

2 years ago

5

complete the puzzle below by identifying the four type of speech context and two types of interpersonal communication evident in

the presented situation

2 answers:

rusak2 [61]2 years ago

6 0

You forgot to attach the puzzle

Alinara [238K]2 years ago

5 0

The puzzle isn’t there

You might be interested in

1) Aluminum sulphate can be made by the following reaction: 2AlCl3(aq) + 3H2SO4(aq) Al2(SO4)3(aq) + 6 HCl(aq) It is quite solubl

kolezko [41]

Answer:

88.9%

Explanation:

Step 1:

The balanced equation for the reaction. This is given below:

2AlCl3(aq) + 3H2SO4(aq) —> Al2(SO4)3(aq) + 6HCl(aq)

Step 2:

Determination of the masses of AlCl3 and H2SO4 that reacted and the mass of Al2(SO4)3 produced from the balanced equation.

Molar mass of AlCl3 = 27 + (35.5x3) = 133.5g/mol

Mass of AlCl3 from the balanced equation = 2 x 133.5 = 267g

Molar mass of H2SO4 = (2x1) + 32 + (16x4) = 98g/mol

Mass of H2SO4 from the balanced equation = 3 x 98 = 294g

Molar mass of Al2(SO4)3 = (27x2) + 3[32 + (16x4)]

= 54 + 3[32 + 64]

= 54 + 3[96] = 342g/mol

Mass of Al2(SO4)3 from the balanced equation = 1 x 342 = 342g

Summary:

From the balanced equation above,

267g of AlCl3 reacted with 294g of H2SO4 to produce 342g of Al2(SO4)3.

Step 3:

Determination of the limiting reactant. This is illustrated below:

From the balanced equation above,

267g of AlCl3 reacted with 294g of H2SO4.

Therefore, 25g of AlCl3 will react with = (25 x 294)/267 = 27.53g of H2SO4.

From the calculations made above, we see that only 27.53g out 30g of H2SO4 given were needed to react completely with 25g of AlCl3.

Therefore, AlCl3 is the limiting reactant and H2SO4 is the excess.

Step 4:

Determination of the theoretical yield of Al2(SO4)3.

In this case we shall be using the limiting reactant because it will produce the maximum yield of Al2(SO4)3 since all of it is used up in the reaction.

The limiting reactant is AlCl3 and the theoretical yield of Al2(SO4)3 can be obtained as follow:

From the balanced equation above,

267g of AlCl3 reacted to produce 342g of Al2(SO4)3.

Therefore, 25g of AlCl3 will react to produce = (25 x 342) /267 = 32.02g of Al2(SO4)3.

Therefore, the theoretical yield of Al2(SO4)3 is 32.02g

Step 5:

Determination of the percentage yield of Al2(SO4)3.

This can be obtained as follow:

Actual yield of Al2(SO4)3 = 28.46g

Theoretical yield of Al2(SO4)3 = 32.02g

Percentage yield of Al2(SO4)3 =..?

Percentage yield = Actual yield /Theoretical yield x 100

Percentage yield = 28.46/32.02 x 100

Percentage yield = 88.9%

Therefore, the percentage yield of Al2(SO4)3 is 88.9%

3 0

2 years ago

Consider a culture medium on which only gram-positive organisms such as Staphylococcus aureus colonies can grow due to an elevat

Olin [163]

<h2>Selective & Differential Medium</h2>

Explanation:

Selective media allow specific types of organisms to develop, and inhibit the development of different living beings. The selectivity is cultivated in a few ways.For model, living beings that can use a given sugar are handily screened by making that sugar the main carbon source in the medium. On the other hand,selective hindrance of certain sorts of microorganisms can be accomplished by adding dyes, anti-infection agents, salts or explicit inhibitors which influence the digestion or enzyme systems of the living beings
Differential media are utilized to separate firmly related life forms or groups of living beings. owing to the pre of specific colors or synthetic compounds in the media, the creatures will deliver trademark changes or development designs that are utilized for ID or separation. An assortment of particular and differential media are utilized in clinical, demonstrative and water contamination research facilities, and in food and dairy laboratories
Selective media because elevated NaCI level is designed to help grow selective bacteria.differential media because the fermented sugar gives off a yellow halo which allows for differentiate between bacteria

4 0

2 years ago

Substitution of an amino group on the para position of acetophenone shifts the cjo frequency from about 1685 to 1652 cm−1 , wher

cluponka [151]

Answer:

Here's what I get.

Explanation:

The frequency of a vibration depends on the strength of the bond (the force constant).

The stronger the bond, the more energy is needed for the vibration, so the frequency (f) and the wavenumber increase.

Acetophenone

Resonance interactions with the aromatic ring give the C=O bond in acetophenone a mix of single- and double-bond character, and the bond frequency = 1685 cm⁻¹.

p-Aminoacetophenone

The +R effect of the amino group increases the single-bond character of the C=O bond. The bond lengthens, so it becomes weaker.

The vibrational energy decreases, so wavenumber decreases to 1652 cm⁻¹.

p-Nitroacetophenone

The nitro group puts a partial positive charge on C-1. The -I effect withdraws electrons from the acetyl group.

As electron density moves toward C-1, the double bond character of the C=O group increases.

The bond length decreases, so the bond becomes stronger, and wavenumber increases to 1693 cm¹.

6 0

2 years ago

The chemical reaction Na+ + Cl- NaCl is best described as a(n) _____.

svlad2 [7]

<span>Na + Cl = NaCl

answer : </span><span>synthesis reaction .

hope this helps!

</span>

5 0

2 years ago

Read 2 more answers

If a swimming pool contains 2,850 kiloliters (kL) of water, how many gallons (gal) of water does it contain? (1 gal = 3.785 L)

siniylev [52]

we are given

a swimming pool contains 2,850 kiloliters (kL) of water

2,850 kiloliters (kL)=2850000L

we know that

$1gal=3.785L$

$3.785L=1gal$

Firstly , we will find for 1L

$1L=\frac{1}{3.785}gal$

now, we can multiply both sides by 2850000

$2850000*1L=2850000*\frac{1}{3.785}gal$

$2850000L=2850000*\frac{1}{3.785}gal$

$2850000L=752972.25892gal$

so,

a swimming pool contains 752972.25892gal of water...........Answer

6 0

2 years ago

Read 2 more answers