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3 years ago

which of the following oils (coconut or safflower)would you use to prepare a soap that is a soft and why?

1 answer:

6 0

Coconut oil because coconut oil is known for its softening , repairing abilities and Coconut oil increases the healthy cholesterol (known as HDL) in your body,
is high in natural saturated fats ,Saturated fats not only increase the healthy cholesterol (known as HDL cholesterol) in your body, but also help convert the LDL “bad” cholesterol into good cholesterols.

hoped it helped!!!

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Given the following sets of values, calculate the unknown

weqwewe [10]

Answer:

3.91 L

Explanation:

Using the ideal gas law equation as follows:

PV = nRT

Where:

P = pressure (atm)

V = volume (L)

n = number of moles (mol)

R = gas law constant (0.0821 Latm/molK)

T = temperature (K)

Based on the information given in this question,

P = 5.23 atm

V= ?

n = 0.831 mol

T = 27°C = 27 + 273 = 300K

Using PV = nRT

V = nRT/P

V = (0.831 × 0.0821 × 300) ÷ 5.23

V = 20.47 ÷ 5.23

V = 3.91 L

8 0

3 years ago

2 points

erica [24]

9 grams of hydrogen gas (H2) will SC Johnson need to react in order to make 1 bottle of Windex.

Explanation:

Balance equation for the formation of ammonia from H2 gas.

N2 + 3H2 ⇒ 2 $NH_{3}$

Given

mass of ammonia in 1 bottle of windex = 51 gram

atomic mass of ammonia 17.01 gram/mole

number of moles = $\frac{mass}{atomic mass of 1 mole}$

number of moles = $\frac{51}{17.01}$

= 3 moles of ammonia is formed.

in 1 bottle of windex there are 3 moles of ammonia 0r 51 grams of ammonia.

From the equation it can be found that:

3 moles of hydrogen reacted to form 2 moles of ammonia

so, x moles of hydrogen will react to form 3 moles of ammonia.

$\frac{2}{3}$ = $\frac{3}{x}$

x = 4.5 moles of hydrogen will be required.

to convert moles into gram formula used:

mass = atomic mass x number of moles (atomic mass of H2 is 2grams/mole)

= 2 x 4.5

= 9 grams of hydrogen.

7 0

3 years ago

A package contains 1.33 lb of ground round. If it contains 29% fat, how many grams of fat are in the ground round? The book is s

Effectus [21]

Answer:

To obtain the grams of fat that the ground round has, knowing that it weighs 1.33 pounds we must pass this value to grams. Since 1 pound equals 453.59 grams, 1.33 pounds equals 603.27 (453.59 x 1.33).

Now, to obtain 29 percent of 603.27, we must make the following calculation: 603.27 / 100 x 29, which gives a total of 174.94 grams.

In this way, your reasoning is correct and it is probably a mistake in the book.

6 0

3 years ago

Given the reaction _K(s) +_ Cl2(g) → _KCl(s) what is the amount of K, in grams, needed to completely react with 2 moles of Cl2(g

damaskus [11]

Answer:

156.4g K

Explanation:

I'm not sure if it is correct but I think it should be this

What do we know so far?: 2K + 1Cl2 -> 2KCl, 2 mol of Cl2

What are we looking for?: #g of K

What is the ratio of K to Cl2?: 2:1

Set up equation: 2molCl2 x $\frac{2mol K}{1 mol Cl2}$

Cancel unwanted units: 2 x $\frac{2mol K}{1}$

Answer we got: 2 x 2mol K = 4mol K

Converting moles to grams: 4 x 39.1 (molar mass of K) = 156.4g K

3 0

3 years ago

Unit: Chemical Quantities

Vaselesa [24]

Answer:

(See explanation for further details)

Explanation:

1) The quantity of moles of sulfur is:

$n = \frac{1.20\times 10^{24}\,atoms}{6.022\times 10^{23}\,\frac{atoms}{mol} }$

$n = 1.993\,moles$

2) The number of atoms of helium is:

$x = (1.5\,moles)\cdot \left(6.022\times 10^{23}\,\frac{atoms}{mole} \right)$

$x = 9.033\times 10^{23}\,atoms$

3) The quantity of moles of carbon monoxide is:

$n = \frac{4.15\times 10^{23}\,molecules}{6.022\times 10^{23}\,\frac{molecules}{mol} }$

$n = 0.689\,moles$

4) The number of molecules of sulfur dioxide is:

$x = (2.25\,moles)\cdot \left(6.022\times 10^{23}\,\frac{molecules}{mole} \right)$

$x = 1.355\times 10^{24}\,molecules$

5) The quantity of moles of sodium chloride is:

$n = \frac{2.4\times 10^{23}\,molecules}{6.022\times 10^{23}\,\frac{molecules}{mol} }$

$n = 0.399\,moles$

6) The number of formula units of magnesium iodide is:

$x = (1.8\,moles)\cdot \left(6.022\times 10^{23}\,\frac{f.u.}{mole} \right)$

$x = 1.084\times 10^{24}\,f.u.$

7) The quantity of moles of potassium permanganate is:

$n = \frac{3.67\times 10^{23}\,f.u.}{6.022\times 10^{23}\,\frac{f.u.}{mol} }$

$n = 1.214\,moles$

8) The number of molecules of carbon tetrachloride is:

$x = (0.25\,moles)\cdot \left(6.022\times 10^{23}\,\frac{molecules}{mole} \right)$

$x = 1.506\times 10^{23}\,molecules$

9) The quantity of moles of aluminium is:

$n = \frac{3.67\times 10^{23}\,atoms}{6.022\times 10^{23}\,\frac{atoms}{mol} }$

$n = 0.609\,moles$

10) The number of molecules of oxygen difluoride is:

$x = (3.52\,moles)\cdot \left(6.022\times 10^{23}\,\frac{molecules}{mole} \right)$

$x = 2.120\times 10^{24}\,molecules$

3 0

3 years ago