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3 years ago

Write down five laboratory safety rules.

Chemistry

1 answer:

OLEGan [10]3 years ago

5 0

Answer:

Five Laboratory Safety Rules:

1). Do not eat in the laboratory.

2). Do not touch any chemical or reagent unless you are told to do so.

3). Neither play in lab, nor sit on the table.

4). Don't remove labels on any reagent.

5). Don't taste anything in the laboratory, no matter how familiar it appears.

Hope it helps.

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Which of earths spheres contains most of its mass?

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The geosphere is about 99.94% of Earth's mass, so C is the answer.

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3 years ago

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What happens to these physical properties as the strength of intermolecular forces increases?Increase or decrease?a) melting poi

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Boiling Point, Melting Point, Viscosity, Surface Tension. Decrease: Vapor Pressure.

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3 years ago

Si la eficiencia de una máquina de calor es del 50% significa que ?

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No toques el link

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2 years ago

2. Suppose that 21.37 mL of NaOH is needed to titrate 10.00 mL of 0.1450 M H2SO4 solution.

AysviL [449]

Answer:

0.1357 M

Explanation:

(a) The balanced reaction is shown below as:

$2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O$

(b) Moles of $H_2SO_4$ can be calculated as:

$Molarity=\frac{Moles\ of\ solute}{Volume\ of\ the\ solution}$

Or,

$Moles =Molarity \times {Volume\ of\ the\ solution}$

Given :

For $H_2SO_4$ :

Molarity = 0.1450 M

Volume = 10.00 mL

The conversion of mL to L is shown below:

1 mL = 10⁻³ L

Thus, volume = 10×10⁻³ L

Thus, moles of $H_2SO_4$ :

$Moles=0.1450 \times {10\times 10^{-3}}\ moles$

Moles of $H_2SO_4$ = 0.00145 moles

From the reaction,

1 mole of $H_2SO_4$ react with 2 moles of NaOH

0.00145 mole of $H_2SO_4$ react with 2*0.00145 mole of NaOH

Moles of NaOH = 0.0029 moles

Volume = 21.37 mL = 21.37×10⁻³ L

Molarity = Moles / Volume = 0.0029 / 21.37×10⁻³ M = 0.1357 M

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3 years ago

Last week you reacted magnesium with a hydrochloric acid aqueous solution and hydrogen gas was produced. Let's say that you coll

miv72 [106K]

Answer:

69.8 kilo Pasacl is the pressure of the hydrogen gas.

Explanation:

$Mg+2HCl\rightarrow MgCl_2+H_2$

Pressure at which hydrogen gas collected = p = 101.2 kilo Pascals

Vapor pressure water = $p^o$ = 31.4 kilo Pascals

The pressure of hydrogen gas = P

The pressure at which gas was collected was sum of vapor pressure of water and hydrogen gas.

$p=P+p^o$

$P =p-p^o=101.2 kPa-31.4 kPa=69.8 kPa$

69.8 kilo Pasacl is the pressure of the hydrogen gas.

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3 years ago