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3 years ago

Which unit is not a measurement of volume? cm L ml cm3

Chemistry

2 answers:

Xelga [282]3 years ago

8 0

Which unit is not a measurement of volume?

<u>cm </u>

cm³

Cm is a measurement for length

MariettaO [177]3 years ago

7 0

The correct answer is cm

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A plastic 2L soda bottle is flexible enough that the volume of the bottle can change even without opening it. If you have an emp

Aneli [31]

Answer: The new volume be if you put it in your freezer is 1.8 L

Explanation:

To calculate the final temperature of the system, we use the equation given by Charles' Law. This law states that volume of the gas is directly proportional to the temperature of the gas at constant pressure.

Mathematically,

$\frac{V_1}{T_1}=\frac{V_2}{T_2}$

where,

$V_1\text{ and }T_1$ are the initial volume and temperature of the gas.

$V_2\text{ and }T_2$ are the final volume and temperature of the gas.

We are given:

$V_1=2L\\T_1=25^oC=(25+273)K=298K\\V_2=?\\T_2=-4.0^oC=((-4)+273)K=269K$

Putting values in above equation, we get:

$\frac{2}{298}=\frac{V_2}{269}\\\\V_2=1.8L$

The new volume be if you put it in your freezer is 1.8 L

8 0

3 years ago

What is the main energy conversion that occurs when you plug in a lamp

MrRa [10]

Elecrtical or Radiantal

4 0

3 years ago

Read 2 more answers

What cells do you think is in metaphase? why?

Tcecarenko [31]

Etaphase is a stage during the process of cell division (mitosis or meiosis). Usually, individual chromosomes cannot be observed in the cell nucleus. However, during metaphase of mitosis or meiosis the chromosomes condense and become distinguishable as they align in the center of the dividing cell.

3 0

2 years ago

Aluminum reacts with sulfur gas to produce aluminum sulfide. a) What is the limiting reactant? What is the excess reagent? b) Ho

Sophie [7]

Answer:

a) Limiting: sulfur. Excess: aluminium.

b) 1.56g Al₂S₃.

c) 0.72g Al

Explanation:

Hello,

In this case, the initial mass of both aluminium and sulfur are missing, therefore, one could assume they are 1.00 g for each one. Thus, by considering the undergoing chemical reaction turns out:

$2Al(s)+3S_2(g)\rightarrow 2Al_2S_3(s)\\$

a) Thus, considering the assumed mass (which could be changed based on the one you are given), the limiting reagent is identified as shown below:

$n_S^{available}=1.00gS_2*\frac{1molS_2}{64gS_2} =0.0156molS_2\\n_S^{consumed\ by \ Al}=1.00gAl*\frac{1molAl}{27gAl}*\frac{3molS_2}{2molAl}=0.0556molS_2$

Thereby, since there 1.00g of aluminium will consume 0.0554 mol of sulfur but there are just 0.0156 mol available, the limiting reagent is sulfur and the excess reagent is aluminium.

b) By stoichiometry, the produced grams of aluminium sulfide are:

$m_{Al_2S_3}=0.0156molS_2*\frac{2molAl_2S_3}{3molS_2} *\frac{150gAl_2S_3}{1molAl_2S_3} =1.56gAl_2S_3$

c) The leftover is computed as follows:

$m_{Al}^{excess}=(0.0556-0.0156)molS_2*\frac{2molAl}{3molS_2}*\frac{27gAl}{1molAl} =0.72 gAl\\$

NOTE: Remember I assumed the quantities, they could change based on those you are given, so the results might be different, but the procedure is quite the same.

Best regards.

7 0

3 years ago

9) After lab, all of Darrel’s friends looked at his data and laughed and laughed. They told him that he was 30.8% too low in the

zaharov [31]

If he was 30.8% too low, it means that he was at 69.2% of the boiling point needed. So 50o C is 69.2% of total.

In order to know what 100% is, you can divide the number by it's percentage and then multiply it by a hundred.

So: 50/30.8=1.623
1.623*100=162.3
So the correct boiling point of the liquid he was working with in the lab is 162.3 oC

3 0

3 years ago

Read 2 more answers