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3 years ago

What piece of lab equipment would you use to conduct a small chemical experiment?

Chemistry

2 answers:

Scrat [10]3 years ago

6 0

test tube

Explanation:

you state it is small and a test tube is the smallest container out of the answers you listed

Svetradugi [14.3K]3 years ago

3 0

Answer:

You would use a test tube to conduct the experiment.

Explanation:

All of the other options are to big for a smaller experiment.

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How many ml of naoh are needed to neutralize 15.0 ml of 0.350m ch3cooh?

horsena [70]

I dont know im sorry

5 0

2 years ago

A freezer compartment is covered with a 2-mm-thick layer of frost at the time it malfunctions. If the compartment is in ambient

ruslelena [56]

Answer:

The time required to melt the frost is 3.25 hours.

Explanation:

The time required to melt the frost dependes on the latent heat of the frost and the amount of heat it is transfered by convection to the air .

The heat transferred per unit area can be expressed as:

$q=h_c*A*\Delta T\\\\q/A=h_c*\Delta T$

being hc the convective heat transfer coefficient (2 Wm^-2K^-1) and ΔT the difference of temperature (20-0=20 °C or K).

$q/A=h_c*\Delta T=2\frac{W}{m^2K}*20K=40\frac{W}{m^2}$

If we take 1 m^2 of ice, with 2 mm of thickness, we have this volume

$V=T*A = 0.002 m * 1 m^2=0.002m^3$

The mass of the frost can be estimated as

$M=\rho * V=700\frac{kg}{m^3}*0.002m^3= 1.4 kg$

Then, the amount of heat needed to melt this surface (1 m²) of frost is

$Q=L*M=334\frac{kJ}{kg}*1.4kg= 467.6kJ$

The time needed to melt the frost can be calculated as

$t=\frac{Q}{(q/A)}=\frac{467.6kJ/m2}{40W/m2} = 11.69\frac{kJ}{W}*\frac{1W*s}{1J}*\frac{1000J}{1kJ}= 11690s=3.25h$

7 0

2 years ago

What is the radius of a hydrogen atom whose electron is bound by 0.544 ev? express your answer with the appropriate units?

insens350 [35]

First, we need to calculate the principal quantum number n for this electron, using the equation:
E = (-13.60 eV) / (n x n)
where E is the energy that is used to bound the electron (here, E = - 0.544 eV).
- 0.544 eV = (-13.60 eV) / (n x n)
n x n = (- 13.60 eV) / (- 0.544 eV)
n x n = 25
n = 5

The orbital radius that is equal to the radius of a hydrogen atom is calculated using the equation:
r = 0.053 nm x n x n
r = 0.053 nm x 5 x 5
r = 0.053 nm x 25
r = 1.325 nm

6 0

3 years ago

4Al(s) + 30₂(g) → 2Al2O3 (s)

Margaret [11]

1) 1 molecules
2) 2 oxygen atoms
3)2 moles of Al2O3 are formed
4)4:3

6 0

2 years ago

A buffer is prepared by adding 12.0g of ammonium chloride (NH4Cl) to 250mL of 1.00 M NH3 solution?The pH is 9.3 Write the net io

solniwko [45]

Answer:- $NH_3(aq)+H^+(aq)\rightleftharpoons NH_4^+(aq)$

Explanations:- The solution we have is a buffer solution and we know that a buffer solution resists a change in its pH if a strong acid or base is added to it.

Here, the buffer solution we have is of a weak base and it's conjugate acid. So, a strong acid(nitric acid) is added to this buffer then it reacts with the base present in the buffer so that the acid could be neutralized. This is called buffer action.

The net ionic equation is written as:

$NH_3(aq)+H^+(aq)\rightleftharpoons NH_4^+(aq)$

Note that $HNO_3$ is a strong acid and nitrate ion is the spectator ion so it is not included in the net ionic equation.

8 0

2 years ago