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dedylja [7]
3 years ago
7

In the image to the right, a student is completing an experiment. What should the student do in order to follow the safety guide

lines for this lab? Check all that apply.
Adjust the Bunsen burner flame so it is orange.

Wear gloves when handling chemicals.

Make sure that none of the liquids near the Bunsen burner are flammable.

Keep papers and books in the work area for reference.
Chemistry
2 answers:
attashe74 [19]3 years ago
7 0

Wear gloves when handling chemicals.

Make sure that none of the liquids near the Bunsen burner are flammable.

Explanation:

The safety precautions the student must apply while completing his experiment is that he/she must wear gloves while handling the chemicals and they must make sure that none of the liquids near the Bunsen burner are flammable.

  • Safety guidelines provides a detailed and careful way to carry out experiment and observations in the laboratory.
  • This is all in a bid to accidents in the workspace.
  • It is a good practices to wear gloves and laboratory coat while handling chemicals.
  • Also make sure to put out the Bunsen burner when not in use.
  • Keep papers and books away from the work area. Chemicals and papers are not friends.
  • Ensure that flammable liquids are not close to the Burner.

learn more:

Safety cautions brainly.com/question/4543399

#learnwithBrainly

Brilliant_brown [7]3 years ago
3 0

Answer: B,C

Explanation:

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Identify which sets of quantum numbers are valid for an electron. Note that each set is ordered (n,l,ml,ms).
Fynjy0 [20]

Answer:

The answer to your question is:

Explanation:

Quantum numbers have the numbers

n = from 1 to 7

l = s = 0 p = 1 d = 2  f= 3

m = ± l

s = ±1/2

Group of answer choice.

A) 2, -1. 1, -1/2     The option is incorrect "l" could not be negative

B) 2, 1, 0 , 1/2       The option is posible

C) 3, 2, 1, -1          This option is incorrect "s" can not value -1

D) 4, 3, 2, 1/2       This option is correct

E) 4, 3, -4, 1/2      This option is incorrect "m2 can not value -4

F) 1, 1, 0, 1/2         This option is incorrect, the second number is not posible

G) 1, 1, 0, -1/2        This option is incorrect the second number could not be 1.

H) 3, 0, 0, 1/2       This option is incorrect "l" can not value 0

I) 0, 2, 0, 1/2         This option is incorrect, n can not value 0

J) 3, 2, 2, 1/2        This option is correct

K) 1, 2, 0, -1/2       This option is incorrect "l" can not value 2.

3 0
3 years ago
Question 34 (1 point)
Grace [21]

Answer:

8.33 atm

Explanation:

Xe is   5  out of (4+5)   or   5 / 9 ths of the gas present

  5/9  * 15 atm = 8.33 atm

4 0
1 year ago
What is the hydrogen ion concentration of a solution with a pH 5? *
nekit [7.7K]

Answer:

In the same way, a solution with a pH of 5 contains 10-5mol/l of hydrogen ions, a solution with a pH of 6 contains 10-6mol/l of hydrogen ions, while the solution with a pH of 7 contains 10-7mol/l of hydrogen ions.

Explanation:

5 0
3 years ago
For the following reaction, 4.31 grams of iron are mixed with excess oxygen gas . The reaction yields 5.17 grams of iron(II) oxi
natka813 [3]

<u>Answer:</u> The theoretical yield of iron (II) oxide is 5.53g and percent yield of the reaction is 93.49 %

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}       ....(1)

  • <u>For Iron:</u>

Given mass of iron = 4.31 g

Molar mass of iron = 53.85 g/mol

Putting values in above equation, we get:  

\text{Moles of iron}=\frac{4.31g}{53.85g/mol}=0.0771mol

For the given chemical reaction:

2Fe(s)+O_2(g)\rightarrow 2FeO(s)

By Stoichiometry of the reaction:

2 moles of iron produces 2 moles of iron (ii) oxide.

So, 0.0771 moles of iron will produce = \frac{2}{2}\times 0.0771=0.0771mol of iron (ii) oxide

Now, calculating the theoretical yield of iron (ii) oxide using equation 1, we get:

Moles of of iron (II) oxide = 0.0771 moles

Molar mass of iron (II) oxide = 71.844 g/mol

Putting values in equation 1, we get:  

0.0771mol=\frac{\text{Theoretical yield of iron(ii) oxide}}{71.844g/mol}=5.53g

To calculate the percentage yield of iron (ii) oxide, we use the equation:

\%\text{ yield}=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100

Experimental yield of iron (ii) oxide = 5.17 g

Theoretical yield of iron (ii) oxide = 5.53 g

Putting values in above equation, we get:

\%\text{ yield of iron (ii) oxide}=\frac{5.17g}{5.53g}\times 100\\\\\% \text{yield of iron (ii) oxide}=93.49\%

Hence, the theoretical yield of iron (II) oxide is 5.53g and percent yield of the reaction is 93.49 %

7 0
3 years ago
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