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3 years ago

What does L represent? When did it form?

Chemistry

1 answer:

frosja888 [35]3 years ago

4 0

Answer:

All I can say is that it seems to go through the other layers. I would say this is an intrusive rock and probably the youngest

Explanation:

I suggest d i k e s. Put it together. somehow I cant post it when put together

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Raw materials used in the manufacturing of indigenous soap

Harrizon [31]

Answer:

Fat

Alkali

Explanation:

Fat and alkali are the two primary raw materials needed to manufacture soap.

Sodium hydroxide or potassium hydroxide is generally used as an alkali. The use of alkali depends on the intended application of the soap.

Raw animal fat was used in the past but these days, processed fat is used in the soap manufacturing process. Vegetable fats ( e.g, palm oil, olive oil, coconut oil) are also being used in soap manufacturing.

Additives are also used to enrich the color and texture of the soap.

5 0

3 years ago

A 1.10 g sample contains only glucose and sucrose. When the sample is dissolved in water to a total solution volume of 25.0L, th

olga_2 [115]

Answer:

$\large \boxed{79 \, \%}$

Explanation:

I assume the volume is 2.50 L. A volume of 25.0 L gives an impossible answer.

We have two conditions:

(1) Mass of glucose + mass of sucrose = 1.10 g

(2) Osmotic pressure of glucose + osmotic pressure of sucrose = 3.78 atm

Let g = mass of glucose

and s = mass of sucrose. Then

g/180.16 = moles of glucose, and

s/342.30 = moles of sucrose. Also,

g/(180.16×2.50) = g/450.4 = molar concentration of glucose. and

s/(342.30×2.50) = s/855.8 = molar concentration of sucrose.

1. Set up the osmotic pressure condition

Π = cRT, so

$\begin{array}{rcl}\Pi_{\text{g}} +\Pi_{\text{s}}&=&\Pi_{\text{tot}}\\\dfrac{g}{450.4}\times8.314\times298 + \dfrac{s}{855.8}\times8.314\times298 & = & 3.78\\\\5.501g + 2.895s & = & 3.78\\\end{array}$

Now we can write the two simultaneous equations and solve for the masses.

2. Calculate the masses

$\begin{array}{lrcl}(1)& g + m & = & 1.10\\(2) &5.501g +2.895s & = & 3.78\\(3) & m & = &1.10 - g\\&5.501g + 2.895(1.10 - g) & = & 3.78\\&2.606g + 3.185 & = & 3.78\\ &2.606g & = & 0.595\\(4) & g & = & \mathbf{0.229}\\&0.229 + s & = & 1.10\\& s & = & \mathbf{0.871}\\\end{array}$

We have 0.229 g of glucose and 0.871 g of sucrose.

3. Calculate the mass percent of sucrose

$\text{Mass percent} = \dfrac{\text{Mass of component}}{\text{Total mass}} \times \, 100\%\\\\\text{Percent sucrose} = \dfrac{\text{0.871 g}}{\text{1.10 g}} \times \, 100\% = 79 \, \%\\\\\text{The mixture is $\large \boxed{\mathbf{79 \, \%}}$ sucrose}$

6 0

3 years ago

Which is an example of chemical weathering?

kobusy [5.1K]

Answer:

b. the layer of oxide formed on statues made of metal

Explanation:

Chemical weathering is a process whereby rocks are disintegrated or weathered causing a chemical reaction thereby leading to change in the parent rock constituents .

In chemical weathering a new substance is usually formed after the weathering takes place.

From the question the layer of statue made by metal will likely form oxides of those metals. Example Silver will react with oxygen to form an entirely new substance like silver(i)oxide.

7 0

3 years ago

How many moles of water are produced from 19.2 g of B2H6

lbvjy [14]

The Balanced chemical equation of reaction of Borane with oxygen is as follow,
B₂H₆ + 3O₂ -----> 2HBO₂ + 2H₂O
According to this equation 27.66 g (1 mole) of B₂H₆ reacts with oxygen to produce 36 g (2 moles) of water.
The amount of water produced when 19.2 g of B₂H₆ reacted is calculated as follow,
$\frac{27.66 g B2H6 produced}{19.2 g of B2H6 will produce}$ = $\frac{36 g of H2O}{x g of water}$
Solving for x,
x = (36 g of H₂O ₓ 19.2 g of H₂B₆) / 27.66 g of B₂H₆

x = 24.98 g of H₂O

Result:
24.98 g of water is produced when 19.2 g of B₂H₆ is reacted with excess of oxygen.

5 0

3 years ago

The element chlorine has two stable isotopes, chlorine-35 with a mass of 34.97 amu and chlorine-37 with a mass of 36.95 amu. Fro

sergejj [24]

<u>Answer: </u>One isotope has a percentage abundance of 75.75 % and the percentage abundance of another isotope is 24.24%.

<u>Explanation:</u>

We are given the two stable isotopes of chlorine with their respective masses. The average atomic mass of chlorine is also given.

Average atomic mass of chlorine = 35.45 amu.

Let us assume the fractional abundance of one isotope be 'x' and the fractional abundance for another isotope will be (1 - x) because the total fractional abundance is always equal to 1.

For Chlorine-35 isotope:

Fractional abundance = x

Mass = 34.97 amu

For Chlorine-37 isotope:

Fractional abundance = 1 - x

Mass = 36.95 amu

The formula for the calculation of average atomic mass is given by:

$\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i$

Putting values in above equation, we get:

$35.45=34.97(x)+36.95(1-x)\\\\35.45=34.97x+36.95-36.95\\\\-1.5=-1.98x\\\\x=\frac{-1.5}{-1.98}=0.7575$

$1-x=1-0.7575=0.2424$

Converting these two fractional abundances into percentage abundances by multiplying it with 100.

$x=0.7575\times 100=75.75\%\\\\1-x=0.2424\times 100=24.24\%$

Hence, one isotope has a percentage abundance of 75.75 % and the percentage abundance of another isotope is 24.24%.

7 0

3 years ago