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salantis [7]
3 years ago
15

All of the following are true about science except (2 points)

Chemistry
1 answer:
ycow [4]3 years ago
5 0
The answer is B. it can be disproven by philosophy or religion

hope this helps !
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How would you make a 2.00L of 0. 500M sodium chloride solution. (assume you have a fully equipped lab with water); sketch, calc
Gwar [14]

Answer:

Sodium chloride solution:

First you need to calculate the mass of salt needed (done in the explanation), which is 58.44g. Then it have to be weighted in an analytical balance in a weighting boat and then transferred into a 2L volumetric flask that is going to be filled until the mark with distilled water.

Sulfuric acid dilution:

First you need to calculate the volume needed (done in the explanation), it is 16.6 mL. Using a graduated pipette one measures this volume and transfer it into a 2L volumetric flask that is already half filled with distilled water, and then one fills it until its mark.

Explanation:

Sodium chloride solution:

Each liter of a 0.500M solution has half mol, so 2L of said solution has 1 mol of salt. Sodium chloride molar mass is 58.44g/mol, so in 2L of solution there is 58.44g of salt. That`s the mass that`s going to be weighted and transferred to a 2L volumetric flask.

Sulfuric acid dilution:

This is the equation for dilution of solutions:

c_{1} v_{1} =c_{2} v_{2}

Where "c1" stands for the initial concentration (stock solution concentration), "v1" for the initial volume (volume of stock solution used), "c2" for the desired concentration and "v2" for the desired volume.

When we are diluting from a stock solution we want to know how much do we have to pipette from the stock solution into our volumetric flask. We do so by isolating the "v1" term from the dilution equation:

v_{1} =\frac{c_{2} v_{2} }{c_{1} }

in this case that would be:

v_{1} =\frac{0.100 x2.0 }{12.0 }=0.0166L=16.6mL

5 0
3 years ago
Match the statements with the atmospheric layer they describe.
vekshin1

The correct matches are:

1. Exosphere - Temperatures reach as high as 2000 C yet it feels very cold

This is the top layer of the atmosphere. The atoms are so dispersed that despite it having very high temperature it doesn't feel like it at all.

2. Thermosphere - Particles that have enough energy can escape into space

The thermosphere is the fourth highest layer of the atmosphere. The atoms in this layer are relatively distant from one another, so the particles that have enough energy manage to escape easily into the exosphere and then the space.

3. Mesosphere - It is the coldest region of the atmosphere

The mesosphere is the third highest layer. In this layer the temperatures constantly drop, and they go down to -85 degrees, making it the coldest layer by far.

4. Stratosphere - Ninety percent ozone is in this layer

The startosphere has a separte zone in it which is dominated by only one gas, the ozone. It is called the ozone layer, the one that protects the Earth from too intense UV radiation, and in fact over 90% of this gas is locate here.

5. Troposphere - It is warm due to the heat from Earth's surface

The troposphere is the densest and lowest of the layers. It is the one that also has Greenhouse gases which manage to trap the heat that is radiated from the surface of the Earth, thus keeping this layer relatively warm.

6 0
3 years ago
Read 2 more answers
Where light is hitting an object
tiny-mole [99]
When lights hits an object, the light either get absorbed, reflected or refracted
3 0
3 years ago
In methane combustion, the following reaction pair is important: At 1500 K, the equilibrium constant Kp has a value of 0.003691
andre [41]

Explanation:

Let us assume that the value of K_{r} = 2.82 \times 10^{5} \times e^({\frac{-9835}{T}}) m^{6}/mol^{2}s

Also at 1500 K, K_{r} = 2.82 \times 10^{5} \times e^({\frac{-9835}{1500}}) m^{6}/mol^{2}s

                     K_{r} = 400.613 m^{6}/mol^{2}s

Relation between K_{p} and K_{c} is as follows.

                  K_{p} = K_{c}RT

Putting the given values into the above formula as follows.

                  K_{p} = K_{c}RT

         0.003691 = K_{c} \times 8.314 \times 1500

                K_{c} = 2.9 \times 10^{-7}

Also,     K_{c} = \frac{K_{f}}{K_{r}}

or,                K_{f} = K_{c} \times K_{r}

                               = 2.9 \times 10^{-7} \times 400.613

                               = 1.1617 \times 10^{-4} m^{6}/mol^{2}s

Thus, we can conclude that the value of K_{f} is 1.1617 \times 10^{-4} m^{6}/mol^{2}s.

6 0
3 years ago
Find the density of a cube on Earth that weighs 1.5 kg and has a side-length of 10 cm.
Inga [223]

Answer:

1.5g/cm³

Explanation:

density=mass÷volume

mass= 1.5kg (<em>c</em><em>h</em><em>a</em><em>n</em><em>g</em><em>e</em><em> </em><em>i</em><em>n</em><em>t</em><em>o</em><em> </em><em>g</em>) = 1500g

volume of the cube = 10×10×10 = 1000cm³

density= divide 1500g÷1000cm = 1.5g/cm³

<h2>Density= 1.5g/cm³</h2>

YOUR WELCOME!

4 0
3 years ago
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