Convert 25 kilograms to pounds.

Describe how to use a transfer pipet to deliver 10.00 mL of liquid by placing the steps in the correct order.

nexus9112 [7]

This question is incomplete, the complete question is;

Describe how to use a transfer pipet to deliver 10.00 mL of liquid by placing the following steps in the correct order;

- Replace the bulb with your index finger and wipe excess liquid off the outside of the pipette.

- Place the tip of the pipette against the side of the beaker and drain the liquid from the pipette until the bottom of the meniscus reaches the calibration mark.

- Transfer the pipette to the receiving vessel.

- Drain the pipette by gravity while holding the tip against the side of the receiving vessel.

- Use a rubber bulb to suck liquid up past the 10.00 ml calibration mark.

- Suck up a third volume of liquid past the 10.00 ml calibration mark.

- Discard the first two pipette volumes of the liquid to rinse the pipette.

Answer:

- Use a rubber bulb to suck liquid up past the 10.00 ml calibration mark.

- Discard the first two pipette volumes of the liquid to rinse the pipette.

- Suck up a third volume of liquid past the 10.00 ml calibration mark.

- Replace the bulb with your index finger and wipe excess liquid off the outside of the pipette.

- Place the tip of the pipette against the side of the beaker and drain the liquid from the pipette until the bottom of the meniscus reaches the calibration mark.

- Transfer the pipette to the receiving vessel.

- Drain the pipette by gravity while holding the tip against the side of the receiving vessel.

Explanation:

First of all, We use a suction device to suck up liquid into pipet. Then we will fill the pipet up to the 10 mL marks and will discard the initial two volumes of liquid and will take a final third volume. We will replace the bulb with index finger and will drain it by placing the tip of pipet at the wall of beaker and drain the liquid.

Arranged in the following steps correctly;

- Use a rubber bulb to suck liquid up past the 10.00 ml calibration mark.

- Discard the first two pipette volumes of the liquid to rinse the pipette.

- Suck up a third volume of liquid past the 10.00 ml calibration mark.

- Replace the bulb with your index finger and wipe excess liquid off the outside of the pipette.

- Place the tip of the pipette against the side of the beaker and drain the liquid from the pipette until the bottom of the meniscus reaches the calibration mark.

- Transfer the pipette to the receiving vessel.

- Drain the pipette by gravity while holding the tip against the side of the receiving vessel.

5 0

3 years ago

Define the term compound

dalvyx [7]

Answer:

A compound is a substance which contains two or more elements chemically combined together.

A compound is formed as a result of a chemical change.

Explanation:

7 0

3 years ago

What is the molar mass of an unknown gas with a density of 2.00 g/L at 1.00 atm and 25.0 °C?

soldier1979 [14.2K]

Answer:

Explanation:Explanation:

Your starting point here will be the ideal gas law equation

∣

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

a

P

V

=

n

R

T

a

∣

−−−−−−−−−−−−−−−

, where

P

- the pressure of the gas

V

- the volume it occupies

n

- the number of moles of gas

R

- the universal gas constant, usually given as

0.0821

atm

⋅

L

mol

⋅

K

T

- the absolute temperature of the gas

Now, you will have to manipulate this equation in order to find a relationship between the density of the gas,

ρ

, under those conditions for pressure and temperature, and its molar mass,

M

.

You know that the molar mass of a substance tells you the mass of exactly one mole of that substance. This means that for a given mass

m

of this gas, you can express its molar mass as the ratio between

m

and

n

, the number of moles it contains

∣

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

a

M

=

m

n

a

∣

−−−−−−−−−−−−−

(

1

)

Similarly, the density of the substance tells you the mass of exactly one unit of volume of that substance.

This means that for the mass

m

of this gas, you can express its density as the ratio between

m

and the volume it occupies

∣

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

a

ρ

=

m

V

a

∣

−−−−−−−−−−−

(

2

)

Plug equation

(

1

)

into the ideal gas law equation to get

P

V

=

m

M

⋅

R

T

Rearrange to get

P

V

⋅

M

=

m

⋅

R

T

P

⋅

M

=

m

V

⋅

R

T

M

=

m

V

⋅

R

T

P

Finally, use equation

(

2

)

to write

M

=

ρ

⋅

R

T

P

Convert the temperature of the gas from degrees Celsius to Kelvin then plug in your values to find

M

=

1.02

g

L

⋅

0.0821

atm

⋅

L

mol

⋅

K

⋅

(

273.15

+

37

)

K

0.990

atm

M

=

∣

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

a

26.3 g mol

−

1

a

∣

−−−−−−−−−−−−−−−−

I'll leave the answer rounded to three

7 0

3 years ago

5. A penny weighs about 2.5 g. How many moles of pennies would be required to equal the mass of the moon (7.3x10^24 kg)

schepotkina [342]

5. 1.16 x $10^{26}$ moles moles of pennies would be required to equal the mass of the moon.

6. 12.86 moles of ethanol are in a 750 ml bottle of vodka.

Explanation:

5 .Data given:

mass of penny = 2.5 grams

atomic mass of penny = 62.93 grams/mole

moles present in mass of the moon given as = 7.3 x $10^{24}$ kg

number of moles = $\frac{mass}{atomic mass of 1 mole}$

number of moles = $\frac{2.5}{62.93}$

0.039 moles of penny is present in 2.5 grams

0.039 moles of penny in 2.5 grams of it

so, x moles in 7.3 X $10^{27}$ grams

$\frac{0.039}{2.5} =\frac{x}{7.3 X 10^{27} }$

x = 1.16 x $10^{26}$ moles

so when the mass of the penny given is equal to the mass of moon, number of moles of penny present is 1.1 x $10^{26}$ .

6.

Given:

vodka = 40% ethanol

volume of vodka bottle = 750 ml

moles of ethanol =?

density of ethanol =0.79 g/ml

atomic mass of ethanol = 46.07 grams/mole

so, from the density of ethanol given we can calculate how much ethanol is present in the solution.

density = $\frac{mass}{volume}$

density x volume = mass

0.79 x 750 = 592.5 grams

number of moles = $\frac{mass}{atomic mass of 1 mole}$

number of moles of ethanol = $\frac{592.5}{46.07}$

= 12.86 moles of ethanol

6 0

3 years ago

Define the following: Bronsted-Lowry acid - Lewis acid- Strong acid - (5 points) Problem 6: Consider the following acid base rea

Allushta [10]

Answer: Yes, $HCl$ is a strong acid.

acid = $HCl$ , conjugate base = $Cl^-$ , base = $H_2O$ , conjugate acid = $H_3O^+$

Explanation:

According to the Bronsted-Lowry conjugate acid-base theory, an acid is defined as a substance which looses donates protons and thus forming conjugate base and a base is defined as a substance which accepts protons and thus forming conjugate acid.

Yes $HCl$ is a strong acid as it completely dissociates in water to give $H^+$ ions.