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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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Degger [83]
To determine a planet's mass, astronomers typically measure the minuscule movement of the star caused by the gravitational tug of an orbiting planet. For planets the massof Earth detecting such a tiny tug is extraordinarily challenging with current technology
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3 years ago
How many grams of agcl would be needed to make a 4. 0 m solution with a volume of 0. 75 l?
devlian [24]

430 g of AgCl would be needed to make a 4.0m solution with a volume of 0.75 L.

<h3>What is Molarity?</h3>
  • The amount of a substance in a specific volume of solution is known as its molarity (M).
  • The number of moles of a solute per liter of a solution is known as molarity.
<h3>Calculation of Required amount of AgCl</h3>

Remember that mol/L is the unit of molarity (M).

We can compute the necessary number of moles of solute by multiplying the concentration by the liters of solution, according to dimensional analysis.

0.75L×4.0M=3.0mol

Then, using the periodic table's molar mass for AgCl, convert from moles to grams:

3.0mol×143.321gmol=429.963g

The final step is to round to the correct significant figure, which in this case is two: 430g.

Hence, 430 g of AgCl would be needed to make a 4.0m solution with a volume of 0.75 L.

Learn more about Molarity here:
brainly.com/question/8732513

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8 0
1 year ago
The activation energy (E*) for 2N2O ---&gt; 2N2 + O2 is 250 KJ. If the k for this reaction is 0.380/M at 1001oK, what will k be
Sedbober [7]

Answer:

Explanation:

GIven that:

The activation energy = 250 kJ

k₁ = 0.380 /M

k₂ = ???

Initial temperature T_1 = 1001 K

Final temperature T_2 = 298 K

Applying the equation of Arrhenius theory.

In \dfrac{k_2}{k_1 }= \dfrac{Ea}{R}( \dfrac{1}{T_1 }- \dfrac{1}{T_2})

where ;

R gas constant = 8.314  J/K/mol

In \dfrac{k_2}{0.380 }= \dfrac{250 * 10^3}{8,314}( \dfrac{1}{1001 }- \dfrac{1}{298})

In \dfrac{k_2}{0.380 }= -70.8655

\dfrac{k_2}{0.380 }= e^{-70.8655}

\dfrac{k_2}{0.380 }= 1.67303256 \times 10^{-31}

{k_2}= 1.67303256 \times 10^{-31} \times {0.380 }

{k_2}= 6.3575 \times 10^{-32}  /M .sec

Half life:

At 1001 K.

t_{1/2} = \dfrac{In_2}{k_1}

t_{1/2} = \dfrac{0.693}{0.38}

t_{1/2} = 1.82368 secc

At 298 K:

t_{1/2} = \dfrac{0.693}{6.3575 \times 10^{-32}}

t_{1/2} =1.09 \times 10^{31} \ sec

3 0
3 years ago
Which type of bond is salt
tigry1 [53]

Answer:

An Ionic Bond

Explanation:

Salt compounds are formed by the "donating" of electrons.

8 0
3 years ago
Read 2 more answers
How many moles is 50g of CaCO3?
Neko [114]

\huge\underline\mathbb\pink{♡Your Answer♡}

Formula mass of CaCO3 is

40 + 12 + 3-100amu....

100g CaCO3 = 1 mole..

50g of CaCO3 = 1÷ 100x 5 = 0.5mole...

<h3>Hence ,answer is 0.5mole...</h3>

Hope it helps you..

Thanks...

3 0
3 years ago
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