All categories

2 years ago

An empty beaker weighs 32.4257 grams. A 10 ml pipet sample of an unknown liquid is

Chemistry

1 answer:

pishuonlain [190]2 years ago

5 0

I thrush udnendhnd nsndndndnd 10

You might be interested in

How many moles are in 17 grams of magnesium chloride (MgCl2) ?

seraphim [82]

Answer:

Molar mass = (24.31 + 2 × 35.45) = 95.21 g mol–1 i.e. 95.21 g of MgCl2 is exactly 1 mole.

Explanation:

7 0

2 years ago

What do scientists make to help them make a hypothesis or collect data during an experiment

Veronika [31]

They will most likely make a table, or some sort of graphing chart

5 0

2 years ago

Which of the following statements describes the correct method of preparation of 1.00 L of a 2.0 M urea solution?

Zolol [24]

Answer:

To prepare 1.00 L of 2.0 M urea solution, we need to dissolve 120 g of urea in enough water to produce a total of 1.00 L solution

Explanation:

Molarity of a solute in a solution denotes number of moles of solute dissolved in 1 L of solution.

So, moles of urea in 1.00 L of a 2.0 M urea solution = 2 moles

We know, number of moles of a compound is the ratio of mass to molar mass of that compound.

So, mass of 2 moles of urea = $(2\times 60.06)g=120 g$

Therefore to prepare 1.00 L of 2.0 M urea solution, we need to dissolve 120 g of urea in enough water to produce a total of 1.00 L solution

So, option (C) is correct.

8 0

3 years ago

Which statement best describes the work and energy in these examples?

kari74 [83]

Answer:

The gas and food are examples of energy.

Explanation:

The reason why is that food is energy as proteins and electrolytes. Gas is a energy of fossil fuels.

3 0

2 years ago

Mount Everest rises to a height of 8,850 m above sea level. At a base camp on the mountain the atmospheric pressure is measured

yan [13]

Answer:

Water will boil at $76^{0}\textrm{C}$ .

Explanation:

According to clausius-clapeyron equation for liquid-vapor equilibrium:

$ln(\frac{P_{2}}{P_{1}})=\frac{-\Delta H_{vap}^{0}}{R}[\frac{1}{T_{2}}-\frac{1}{T_{1}}]$

where, $P_{1}$ and $P_{2}$ are vapor pressures of liquid at $T_{1}$ (in kelvin) and $T_{2}$ (in kelvin) temperatures respectively.

Here, $P_{1}$ = 760.0 mm Hg, $T_{1}$ = 373 K, $P_{2}$ = 314.0 mm Hg

Plug-in all the given values in the above equation:

$ln(\frac{314.0}{760.0})=\frac{-40.7\times 10^{3}\frac{J}{mol}}{8.314\frac{J}{mol.K}}\times [\frac{1}{T_{2}}-\frac{1}{373K}]$

or, $T_{2}=349 K$

So, $T_{2}=349K=(349-273)^{0}\textrm{C}=76^{0}\textrm{C}$

Hence, at base camp, water will boil at $76^{0}\textrm{C}$

6 0

2 years ago