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hjlf
3 years ago
15

What is a line graph

Chemistry
1 answer:
Ymorist [56]3 years ago
7 0
A line graph is a graph that shows data connected by line segments.
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Calculate the average atomic mas of an unknown element that has two naturally-occurring isotopes.
professor190 [17]
The average atomic mass if the element above is calculated by the sum of the product of the isotope abundance and its atomic mass unit. It is expressed as:

Average atomic mass = Σ xi(Mi)
<span>Average atomic mass = (.7547 x 248.7) + (.2453 x 249.4) = 248.87
</span>
Hope this helps.
8 0
3 years ago
Read 2 more answers
You have 100 mL of a 12M solution of HCl, and you need to dilute it to 1.5M for an experiment. How many liters will your new sol
sashaice [31]

Answer:

800.0 mL.

Explanation:

  • To solve this problem; we must mention the rule states the no. of millimoles of a substance before and after dilution is the same.

<em>(MV)before dilution of HCl = (MV)after dilution of HCl</em>

M before dilution = 12.0 M, V before dilution = 100.0 mL.

M after dilution = 1.5 M, V after dilution = ??? mL.

∵ (MV)before dilution of HCl = (MV)after dilution of HCl

∴ (12.0 M)(100.0 mL) = (1.5 M)(V after dilution of HCl)

<em>∴ V after dilution of HCl = (12.0 M)(100.0 mL)/(1`.5 M) = 800.0 mL.</em>

8 0
3 years ago
Explain how percent yield is calculated
Alex73 [517]
Actual yield/theoretical yield x 100
4 0
3 years ago
Warm air and water both tend to rise while cooler air and water sink. When different parts of the oceans are heated unevenly, th
ale4655 [162]

Answer:

a

Explanation:

3 0
2 years ago
Does adding 1 mol of NaCl to 1 kg of water lower the vapor pressure of water to the same extent, a lesser extent, or a greater e
igor_vitrenko [27]

Answer:

Adding 1 mol of NaCl to 1 kg of water lower the vapor pressure of water <em><u>to the same extent</u></em>  by adding 1 mol of C_06H_{12}O_6 to 1 kg of water.

Explanation:

1) Moles of NaCl ,n_1=1 mol

Mass of water = m= 1 kg = 1000 g

Moles of water = n_2=\frac{1000 g}{18 g/mol}=55.55 mol

Vapor pressure of the solution = p

Vapor pressure of the pure solvent that is water = p_o=17.5 Torr

Mole fraction of solute(NaCl)= \chi_1=\frac{n_1}{n_1+n_2}

\frac{p_o-p}{p_o}=\frac{n_1}{n_1+n_2}

\frac{17.5 Torr-p}{17.5 Torr}=\frac{1 mol}{1 mol+55.55 mol}

p=17.19 Torr

The vapor pressure for the NaCl solution at 17.19 Torr.

2) Moles of sucrose ,n_1=1mol

Mass of water = m  = 1 kg = 1000 g

Moles of water = n_2=\frac{1000 g}{18 g/mol}=55.55 mol

Vapor pressure of the solution = p'

Vapor pressure of the pure solvent that is water = p_o=17.5 Torr

Mole fraction of solute ( glucose)= \chi_1=\frac{n_1}{n_1+n_2}

\frac{p_o-p}{p_o}=\frac{n_1}{n_1+n_2}

\frac{17.5 Torr-p}{17.5 Torr}=\frac{1 mol}{1 mol+55.55 mol}

p'=17.19 Torr

The vapor pressure for the glucose solution at 17.19 Torr.

p = p' = 17.19 Torr

Adding 1 mol of NaCl to 1 kg of water lower the vapor pressure of water to the same extent  by adding 1 mol of C_06H_{12}O_6 to 1 kg of water.

3 0
3 years ago
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