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gtnhenbr
2 months ago
10

A 1.20 g sample of water is injected into an evacuated 5.00 l flask at 65°c. part of the water vaporizes and creates a pressure

of 187.5 mmhg. what percentage of the water vaporized?
Chemistry
1 answer:
alisha [2.9K]2 months ago
3 0
Assuming the water vapor behaves as an ideal gas,


PV = nRT

For conversions, 760 mmHg = 101325 Pa and 1,000 L = 1 m³
(187.5 mmHg)(101325 Pa/760 mmHg)(5 L)(1 m³/1,000 L) = n(8.314 m³Pa/molK)(65+273 K)
Calculating for n,
n = 0.0445 mole of water

Considering the molar mass of water is 18 g/mol,
The mass of the vaporized water = 0.0445 * 18 = 0.8 g of water evaporated

Therefore,
The fraction of water that vaporized = 0.8/1.2 * 100 = 66.7%
You might be interested in
Suppose you have a bucket of sand containing 5.5 billion grains of sand ( 5.5×109 grains). Most of the grains of sand are white,
alisha [2963]

Explanation:

5.5 billion grains of sand (5.5×10^9 grains)

Assuming the amount of brown sand is 6.0%, how many brown grains are found in the bucket?

Grains of brown sand = Percentage of brown sand * Total sand in the bucket

Calculated brown grains = 0.06 * 5.5×10^9 = 0.33 x 10^9 = 3.3 x 10^8 grains

If the concentration of brown sand is 6.0 ppm, what is the number of brown grains in the bucket?

Grains of brown sand = Concentration of brown sand * Total sand in the bucket

6 ppm = 6 / 1,000,000 = 0.000006

Calculated brown grains = 0.000006 * 5.5×10^9 = 3.3 x 10^4 grains

If the concentration of brown sand is 6.0 ppb, how many brown grains exist in the bucket?

Grains of brown sand = Concentration of brown sand * Total sand in the bucket

6 ppb = 6 / 1,000,000,000 = 0.000000006

Calculated brown grains = 0.000000006 * 5.5×10^9 = 3.3 x 10^1 = 33 grains

5 0
2 months ago
In order to use a pipet, place a ____________ at the top of the pipet. Use this object to fill the pipet such that the _________
Tems11 [2777]

Answer:

The right responses are "bulb or pump; meniscus; outside".

Explanation:

Pipets are essential tools in laboratory settings. They are designed for transferring liquids from one vessel to another. First, a bulb or pump is attached to the top to empty the pipet completely. Next, fill the pipet until the meniscus (the curved top of the liquid) aligns with the measurement line corresponding to the volume needed. Finally, dispense the liquid into a second container and make sure to eliminate the last drop beyond the pipet tip.

3 0
2 months ago
What volume of co2 gas at 645 torr and 800. k could be produced by the decomposition of 45.0 g of caco3? caco3(s) → cao(s) + co2
KiRa [2933]
In the reaction: <span>caco3(s) → cao(s) + co2(g), it is evident that
1 mol (which is 100 g) of CaCO3 yields 1 mol (which is 44 g) of CO2
Now, the molarity of CaCO3 present in the reaction system is
</span>= \frac{weight of CaCO3 (g)}{gram molecular weight}
= \frac{45}{100} = 0.45 mol

Thus, 0.45 mol of CaCO3 leads to the formation of 0.45 mol of CO2.

According to the ideal gas equation, we have PV = nRT
V = \frac{nRT}{P}.
Considering P = 645 torr = 0.8487 atm (because 1 atm = 760 torr)
In that case, V = \frac{0.45 X 0.08206 X 800}{0.8487}

= 34.8 l
5 0
1 month ago
Calculate ΔH and ΔStot when two copper blocks, each of mass 10.0 kg, one at 100°C and the other at 0°C, are placed in contact in
eduard [2782]

Clarification:

The pertinent information is outlined as follows.

m = 10.0 kg = 10,000 g (since 1 kg = 1000 g)

Starting temperature of block 1, T_{1} = 100^{o}C = (100 + 273) K = 373 K

Starting temperature of block 2, T_{2} = 0^{o}C = (0 + 273) K = 273 K

Therefore, the heat lost by block 1 equals the heat received by block 2

mC \Delta T = mC \times \Delta T

10000 g \times 0.385 \times (T_{f} - 100)^{o}C = 10000 g \times 0.385 \times (0 - T_{f})^{o}C

T_{f} - 100^{o}C = 0^{o}C - T_{f}

2T_{f} = 100^{o}C

T_{f} = 50^{o}C

It's important to convert the temperature into Kelvin as (50 + 273) K = 323 K.

Additionally, the relationship between enthalpy and temperature change is as follows.

\Delta H = mC \Delta T

= 10000 g \times 0.385 J/K g \times 323 K

= 1243550 J

or, = 1243.5 kJ

Next, determine the entropy change for block 1 as follows.

\Delta S_{1} = mC ln \frac{T_{f}}{T_{i}}

= 10000 g \times 0.385 J/K g \times ln \frac{323}{373}

= 10000 g \times 0.385 J/K g \times -0.143

= -554.12 J/K

Now, the entropy change for block 2 is as follows.

   \Delta S_{2} = mC ln \frac{T_{f}}{T_{i}}

           = 10000 g \times 0.385 J/K g \times ln \frac{323}{273}

           = 10000 g \times 0.385 J/K g \times 0.168

           = 647.49 J/K

Thus, the total entropy is the sum of the entropy changes of both blocks.

                   = -554.12 J/K + 647.49 J/K\Delta S_{total} = \Delta S_{1} + \Delta S_{2}

           = 93.37 J/K

In conclusion, for this reaction, the outcome is 1243.5 kJ and \Delta S_{total} is 93.37 J/K.

6 0
2 months ago
How many protons neutrons and electrons are there in a neutral atom of 43k (potassium-43)?
castortr0y [3046]

Answer:

             Protons: 19

             Neutrons: 25

             Electrons: 19

Explanation:

Protons:

                   The atomic number determines the number of protons in an atom. Consequently, with Potassium's atomic number being 19, it contains 19 protons.

Neutrons:

                   The formula to find neutrons is:

                           # of Neutrons  =  Atomic Mass - # of Protons

Given:

                                Atomic Mass  =  43

                                # of Protons  =  19

Thus,

                                # of Neutrons  =  43 - 19

                                # of Neutrons  =  24

Electrons:

                                   In a neutral atom, the quantity of electrons matches that of protons. Therefore, a neutral Potassium atom with 19 protons must equally have 19 electrons.

3 0
2 months ago
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