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3 years ago

Which lab safety rule do you think is the most important to follow in chemistry lab?

Chemistry

1 answer:

Flauer [41]3 years ago

7 0

Answer:

Practice good personal hygiene. Wash your hands after removing gloves, before leaving the laboratory, and after handling a potentially hazardous material. While working in the laboratory, wear personal protective equipment - eye protection, gloves, laboratory coat - as directed by your supervisor.

Explanation:

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Help guys plz help me especially with 2 and 3

german

Answer:

1.D 2.D

Explanation:

3 0

3 years ago

Read 2 more answers

What is the pH of a KOH solution that has [H ] = 1. 87 × 10–13 M? What is the pOH of a KOH solution that has [OH− ] = 5. 81 × 10

vlada-n [284]

pH is the hydrogen ion concentration and pOH is the hydroxide ion concentration in the solution. pH KOH is 12.73, pOH KOH is 2.24 and pH NaCl is 7.

pH is the negative log of the hydrogen ion concentration and pOH is the negative log of the hydroxide ion concentration.

The relation between the pH and pOH can be given as, $\rm pOH = 14 - pH$

The pH of KOH can be calculated by the formula,

$\rm pH = \rm -log [H^{+}]$

In the first case, the concentration of the KOH is $1. 87 \times 10^{-13}\;\rm M$

Substituting values in the equation:

$\begin{aligned} \rm pH &= \rm -log [H^{+}]\\\\&= \rm -log [1. 87 \times 10^{-13}\;\rm M ]\\\\&= 12.73\end{aligned}$

Hence, the pH of KOH is 12.73.

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pOH of KOH can be calculated by the formula,

$\rm pOH = \rm -log [OH^{-}]$

The hydroxide concentration of the KOH solution is $5. 81 \times 10^{-3}\;\rm M$

Substituting value in the equation:

$\begin{aligned} \rm pOH &= \rm -log [OH^{-}]\\\\&= \rm -log [5. 81 \times 10^{-3}\;\rm M ]\\\\&= 2.24 \end{aligned}$

Hence, the pOH of KOH is 2.24

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The pH of NaCl can be calculated by the formula,

$\rm pH = \rm -log [H^{+}]$

In the third case, the concentration of the NaCl is $1. 00\times 10^{-7}\;\rm M$

Substituting values in the equation:

$\begin{aligned} \rm pH &= \rm -log [H^{+}]\\\\&= \rm -log [1. 00 \times 10^{-7}\;\rm M ]\\\\&= 7 \end{aligned}$

Hence, the pH of KOH is 7.0.

Therefore, KOH is basic and NaCl is approximately neutral.

Learn more about pH and pOH here:

brainly.com/question/13885794

3 0

3 years ago

Energy removal is illustrated in. A. Changing water ice to waterB. Changing water to steamC. Boiling of gasolineD. Evaporation o

Kryger [21]

Answer:

D. Evaporation of sea water

Explanation:

6 0

2 years ago

Only a small fraction of a weak acid ionizes in aqueous solution. What is the percent ionization of a 0.100-M solution of acetic

Mademuasel [1]

Answer:

1.33%

Explanation:

In an aqueous solution, a weak acid such as acetic acid, will be in equilibrium with its conjugate base, acetate ion, thus:

CH₃CO₂H(aq) + H₂O(l) ⇌ H₃O⁺(aq) + CH₃CO₂⁻(aq )

Where dissociation constant, ka, is defined as the ratio of concentrations of products and reactants:

Ka = 1.8x10⁻⁵ = [H₃O⁺] [CH₃CO₂⁻] / [CH₃CO₂H]

<em>H₂O is not taken into account in the equilibrium because is a pure liquid</em>

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When a solution of acetic acid becomes to equilibrium, the original concentration of the acid decreases producing more H₃O⁺ and CH₃CO₂⁻.

The concentrations at equilibrium when a 0.100M solution of acetic acid reaches this state, is:

[CH₃CO₂H] = 0.100M - X

[H₃O⁺] = X

[CH₃CO₂⁻] = X

<em>Where X is reaction coordinate.</em>

Replacing in Ka expression:

1.8x10⁻⁵ = [H₃O⁺] [CH₃CO₂⁻] / [CH₃CO₂H]

1.8x10⁻⁵ = [X] [X] / [0.100M - X]

1.8x10⁻⁶ - 1.8x10⁻⁵X = X²

1.8x10⁻⁶ - 1.8x10⁻⁵X - X² = 0

Solving for X:

X = -0.00135 → False solution. There is no negative concentrations.

X = 0.00133 → Right solution.

That means concentration of acetate ion is:

[CH₃CO₂⁻] = 0.00133M.

Now, percent ionization is defined as 100 times the ratio between weak acid that is ionizated, [CH₃CO₂⁻] = 0.00133M, per initial concentration of the acid, [CH₃CO₂H] = 0.100M. Replacing:

% Ionization = 0.00133M / 0.100M × 100 =

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4 0

3 years ago

Look into the picture for the question.

ira [324]

Answer:

Option D. 1092K

Explanation:

T1 = 273K

V1 = V

P1 = P

V2 = 2V( i.e double of original volume)

P2 = 2P (i.e double of original pressure)

T2 =?

P1V1/T1 = P2V2/T2

PV/273 = 2P x 2V / T2

Cross multiply to express in linear form

PV x T2 = 273 x 2P x 2V

Divide both side by PV

T2 = (273 x 2P x 2V) / PV

T2 = 273 x 2 x 2

T2 = 1092K

3 0

4 years ago