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lana66690 [7]
3 years ago
15

1) If you transfer energy into two solid substances, which do you think would melt first: a substance with stronger molecular at

traction or a substance with weaker molecular attraction?
2) If you transfer energy out of two substances in the gas phase, which do you think would condense first: a substance with stronger molecular attraction or a substance with weaker molecular attraction?

PLEASE HELP
Chemistry
1 answer:
skad [1K]3 years ago
8 0

Answer:

1) A substance with a weaker molecular attraction would melt first.

2) A substance with stronger molecular attraction would condense first.

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On the axes provided, label pressure on the horizontal axis from O mb to 760 mb and volume on the vertical axis from O to 1 mL.
Delicious77 [7]

Answer:

  • Please, find the graph with the labels and points located on the axes in the picture attached.

Explanation:

This is how you meet all the instructions and some important comments to understand how this kind of graphs word:

<u>1) Label pressure on the horizontal axis from O mb to 760 mb and volume on the vertical axis from O to 1 mL. </u>

The horizontal axis is used to record the independent variable and the vertical axis is used to record the dependent variable. The axes most be properly labeled with the name of the variable and the units.

In this case the origin is the point (0,0) which means that the axes cross each other, perpendicularly, at a pressure of 0 mb and a volume of 0.0 mililiters.

<u>2) Assign values to axes divisions in such a way that you occupy almost all the space on both axes. </u>

A good graph searches to occupy the whole space on both cases; to do that, find the maximum value for each variable, pressure and volume, and choose the values of the marks.

The range of the pressure (horizontal axis) is [90, 760 mb], so you should choose big divisions (marks) of 100 mb, and assign 800 mb to the right most mark on the horizontal axis. Then, you can divide each interval of 100 mb into 10 spaces, with small divisions of 10 mb (my graph uses 4 spcaes, with small divisions of 25 mb, but I recommend you use small divisions of 10 mb).

The range of the volume (vertical axis) is [0.1, 0.8], so you should choose only divisions with value of 0.1 ml.

<u>3) Now locate and label the points: </u>

  • (90, 0.9) ⇒ 90 mb, 0.9 ml
  • (100, 0.8) ⇒ 100 mb, 0.8 ml
  • (400, 0.2) ⇒ 400 mb, 0.2 ml
  • (600, 0.15) ⇒ 600 mb, 0.15 ml
  • (760, 0.1) ⇒ 760 mb, 0.1 ml

The points of the kind (x, y) are called ordered pairs, which means that the order matters, because it has a meaning: the first number represents the independent variable and the second number represents the dependent variable.

So, in the point (90, 0.9), 90 is a pressure of 90 mb and 0.9 is a volume of 0.9 ml.

To locate (600, 0.15), since the horizontal marks have value of 0.1, you must locate the second coordinate of your point between the marks 0.1 and 0.2 ml.

With that you can now locate each point on your graph.

5 0
3 years ago
Read 2 more answers
4. A student started with a 0.032 g sample of copper which he took through the series of reactions described in this experiment.
Elodia [21]

Answer:

Y=48.6\%

Explanation:

Hello,

In this case, we can consider the following chemical reaction for the oxidation of copper which only occurs at high temperatures:

2Cu+O_2\rightarrow 2CuO

In such a way, for 0.032 grams of copper, the following grams of copper (II) oxide (black product) are yielded:

m_{CuO}=0.032gCu*\frac{1molCu}{63.546gCu} *\frac{2molCuO}{2molCu}*\frac{79.546gCuO}{1molCuO}  =0.078gCuO

Therefore, the percent yield is:

Y=\frac{0.038g}{0.078g}*100\%\\ \\Y=48.6\%

Best regards.

6 0
3 years ago
The picture below shows a flower experiencing phototropism, a type of growth movement.
lakkis [162]
The correct answer is A
4 0
3 years ago
Read 2 more answers
The daily production of carbon dioxide from an 880 MW coal-fired power plant is estimated to be 31,000 tons. A proposal has been
miskamm [114]

Answer:

1.3578\times 10^{8} m^3 volume of carbon dioxide gas would be collected during a one-year period.

Explanation:

Mass of carbon dioxide gas collected in a day = 31,000 Tonn

1 Tonn = 1000 kilogram

31,000 Tonn = 31,000\times 1000 kg=31,000,000 kg

Specific volume of the carbon dioxide gas = V_s

V_s=0.012 m^3/kg

Volume of carbon dioxide gas collected in a day = V

V=V_s\times 31,000,000 kg=0.012 m^3/kg\times 31,000,000

V=372,000 m^3

1 year = 365 days

Volume of carbon dioxide gas collected in a year = V'

V'=372,000\times 365 m^3=1.3578\times 10^{8} m^3

1.3578\times 10^{8} m^3 volume of carbon dioxide gas would be collected during a one-year period.

5 0
4 years ago
A florist prepares a solution of nitrogen-phosphorus fertilizer by dissolving 5.66 g of NH4NO3 and 4.42 g of (NH4)3PO4 in enough
motikmotik

Answer:

0.00725 M

Explanation:

Considering for NH_4NO_3

Mass = 5.66 g

Molar mass of NH_4NO_3 = 80.043 g/mol

Moles = Mass taken / Molar mass

So,  

<u>Moles = 5.66 / 80.043 moles = 0.0707 moles</u>

NH_4NO_3 will dissociate as:

NH_4NO_3\rightarrow NH_4^++NO_3^-

Thus 1 mole of NH_4NO_3 yields 1 mole of ammonium ions. So,

<u>Ammonium ions furnished by NH_4NO_3 = 1 × 0.0707 moles = 0.0707 moles</u>

Considering for (NH_4)_3PO_4

Mass = 4.42 g

Molar mass of (NH_4)_3PO_4 = 149.09 g/mol

Moles = Mass taken / Molar mass

So,  

Moles = 4.42 / 149.09 moles = 0.0296 moles

(NH_4)_3PO_4 will dissociate as:

(NH_4)_3PO_4\rightarrow 3NH_4^++PO_4^{3-}

Thus 1 mole of NH_4NO_3 yields 3 moles of ammonium ions. So,

<u>Ammonium ions furnished by (NH_4)_3PO_4 = 3 × 0.0296 moles = 0.0888 moles</u>

<u>Total moles of the ammonium ions = 0.0707 + 0.0888 moles = 0.1595 moles</u>

Given that:

Volume = 22.0 L

So, Molarity of the NH_4^+ is:

<u>Molarity = Moles / Volume = 0.1595 / 22 M = 0.00725 M</u>

8 0
3 years ago
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