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3 years ago

In the chemistry laboratory why do scientists use a mortal and pestle

1 answer:

4 0

To increase surface area of a solute.
The mortar and pestle can make the solute from bulk to powder. 
So the surface area will increase and the reaction rate or dissolving rate will also increase.

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A 2.00 liter evacuated container has a mass of 1050.0 g. When the container is filled with an unknown gas at 800. mm Hg pressure

Andrei [34K]

Answer:

(A) 28

Explanation:

To solve this problem we use the PV=nRT equation, where:

P = 800 mmHg ⇒ 800/760 = 1.05 atm

V = 2.00 L

n = ?

R = 0.082 atm·L·mol⁻¹·K⁻¹

T = 25.0 °C ⇒ 25.0 + 273.16 = 298.16 K

We input the data:

1.05 atm * 2.00 L = n * 0.082 atm·L·mol⁻¹·K⁻¹ * 298.16 K

And solve for n:

n = 0.086 mol

Now we calculate the gas' mass:

Gas Mass = (Mass of Container w/ Gas) - (Mass of Empty Container)

Gas Mass = 1052.4 g - 1050.0 g = 2.4 g

Finally we calculate the unknown gas' molar mass, using its mass and its number of moles:

Molar Mass = mass / moles

Molar Mass = 2.4 g / 0.086 mol = 27.9 g/mol

So the answer is option (A).

5 0

3 years ago

10. Write the formula for the compounds: copper (1) chloride and copper (II) chloride

Natali [406]

Answer:

Copper(II) chloride (CuCl2) reacts with several metals to produce copper metal or copper(I) chloride (CuCl) with oxidation of the other metal.

Explanation:

3 0

3 years ago

For a particular reaction, Δ=−111.4 kJ and Δ=−25.0 J/K.

vichka [17]

Answer:

$\Delta G =-103.95kJ$

Explanation:

Hello there!

In this case, since the thermodynamic definition of the Gibbs free energy for a change process is:

$\Delta G =\Delta H-T\Delta S$

It is possible to plug in the given H, T and S with consistent units, to obtain the correct G as shown below:

$\Delta G =-111.4kJ-(298K)(-25.0\frac{J}{K}*\frac{1kJ}{1000J} )\\\\\Delta G =-103.95kJ$

Best regards!

6 0

3 years ago

At 500K, the equilibrium constant for the reaction N2O4D2NO2 is 1.5 x 10^3. What is the equilibrium constant for the reaction: 6

Veronika [31]

Answer:

K = 2.96x10⁻¹⁰

Explanation:

Based on the initial reaction:

N2O4 ⇄ 2NO2; K = 1.5x10³

Using Hess's law, we can multiply this reaction changing K:

3 times this reaction:

3N2O4 ⇄ 6NO2; K = (1.5x10³)³ =3.375x10⁹

The inverse reaction has a K of:

6NO2 ⇄ 3N2O4 K = 1/3.375x10⁹;

3 0

3 years ago

Which of these are examples of passive transport? A. osmosis B.diffusion c. endocytosis d.8 mitosis

AfilCa [17]

Answer:

Passive Transport

Explanation:

The three examples of passive transport are

Diffuison

Osmosis

facilated diffuison

So the answer can be A or B

4 0

2 years ago