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Artemon [7]
3 years ago
14

Imagine you had prepared a 0.1 M sodium phosphate solution. Would you expect it to have a

Chemistry
1 answer:
Setler79 [48]3 years ago
4 0

Answer:

idk

Explanation:

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If there are 10 navy beans,3 pinto beans,and 17 lentils in a container,what is the percent composition of the container by bean?
Stells [14]

The percent abundance in the container by bean will be 43.33 %  if there are 10 navy beans, 3 pinto beans, and 17 lentils in a container

<h3>What is Percentage composition ?</h3>

The percentage composition of a given compound is defined as the ratio of the amount of each element to the total amount of individual elements present in the compound multiplied by 100.

% composition by bean = number of beans / total number of objects in container x 100

  • Total object = 10 ( navy beans) + 3 (pinto beans) + 17 ( lentils) = 30 objects
  • Total beans = 10 ( navy beans) + 3 (pinto beans) = 13 beans

hence ;

  • % composition by beans = 13/30 x 100 = 43.33 %

Therefore, the percent composition of the container by bean is 43.33 %

Learn more about Percentage composition here ;

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2 years ago
Do you think it is good to bring torches into the cave ​
kompoz [17]

Answer:

no

Explanation:

better use flashlichts

6 0
3 years ago
Gold alloys are _____.
chubhunter [2.5K]
Gold alloys consist of most commonly Nickel! 
Most commonly used in things like jewelry so it doesn't have such a heavy weight.
7 0
3 years ago
1. If you run into a curb on a skateboard, you will fall forward off your skateboard because
uysha [10]

Answer:

D)applied

Explanation:

4 0
3 years ago
Read 2 more answers
Sunflower oil contains 0.080 mol palmitic acid (C16H32O2)/mol, 0.060 mol stearic acid (C18H36O2)/mol, 0.27 mol oleic acid (C18H3
Cerrena [4.2K]

Answer:

x_H=0.882

x_N=0.118

Explanation:

In this reactor, oleic and linoleic acid react with hydrogen to form stearic acid. This reactions can be represented by:

Oleic: C_{18}H_{34}O_2 (l) + H_2 (g) \longrightarrow C_{18}H_{36}O_2 (l)

Linoleic: C_{18}H_{32}O_2 (l) + 2 H_2 (g) \longrightarrow C_{18}H_{36}O_2 (l)

Having this reactions in mind, the first thing is to determine the moles of hydrogen required:

<u>Base of caculation: 1 mol of sunflower oil</u>

For oleic acid: n_{Holeic}=\frac{1 mol H_2}{1 mol oleic}*\frac{0.27 mol oleic}{1 mol oil}*\frac{335 mol oil}{hr}

n_{Holeic}=frac{90.45 mol H_2}{hr}

For linoleic acid: n_{Hlinoleic}=\frac{2 mol H_2}{1 mol linoleic}*\frac{0.59 mol linoleic}{1 mol oil}*\frac{335 mol oil}{hr}

n_{Holeic}=frac{395.3 mol H_2}{hr}

n_{Htotal}=\frac{90.45 mol H_2}{hr}+\frac{395.3 mol H_2}{hr}

n_{Htotal}=frac{485.75 mol H_2}{hr}

Applying the excess:

n_{Htotal}=frac{485.75 mol H_2}{hr}*1.65=801.48 mol

Nitrogen: n_N= 801.48 mol*\frac{0.05 mol N}{0.95 mol}

n_N= 42.2 mol N

<u>After the reactions</u>:

n_H=801.48 mol-485.75mol=315.73 mol

and the nitrogen is inert.

Purge stream:

n_total=42.2+315.73 mol=357.93 mol

x_H=\frac{315.73mol}{357.93mol}=0.882

x_N=\frac{42.2mol}{357.93mol}=0.118

4 0
3 years ago
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