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3 years ago

Which statements accurately describe the polarity and electronegativity of water?

1 answer:

5 0

The statement that most accurately and effectively described the polarity and electronegativity in water is that the covalent bonds within these water molecules bind with the single oxygen atom in the molecule, as well as the two hydrogen atoms that it holds as well.

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Which of the following may suggest a catalyst has been used in a reaction, given the energy diagram for the same reaction withou

anzhelika [568]

The answer I think would be C

5 0

3 years ago

in the reaction Mg (s) + 2HCl (aq) H2 (g) + MgCl2 (aq), how many moles of hydrogen gas will e produced from 75.0 milliliters of

Anna35 [415]

This question uses mole ratios. firstly, we look at how many moles of H2 are produced for every mole of HCl: from 1 Hcl, we get half a mole of H2. therefore, however many moles of HCl are present in the 0.075 L of 1.0M HCl, half that number of H2 moles will be produced.

n=cv
=0.0750 x 1.0
=0.075 mol HCl
n(H2)= 0.5 x n(HCl)
=0.0375 moles of H2

6 0

3 years ago

How do you calculate the number of photons having a wavelength of 10.0 micrometers required to produce 1.0 kilojoules of energy

shtirl [24]

To calculate this,

We know that energy is 1 photon
E = hc/wavelenth
wavelength of 10.0 m

Solution:
h = 6.626 x 10^-34 Jsec
C = 2.9979 x 10^8 m/sec
E = 6.626 10^-34 * 2.9979 10^8 / 10 = 1.9864 10^-26J

Then, the number of photons is computed by:

n = 1000 / 1.9864 10^-26 = 5.04 10^28 photons

4 0

3 years ago

A student ran the following reaction in the laboratory at 311 K:CH4(g) + CCl4(g) 2CH2Cl2(g)When she introduced 4.10×10-2 moles o

statuscvo [17]

Answer: The value of $K_{eq}$ is 0.044

Explanation:

We are given:

Initial moles of methane = $4.10\times 10^{-2}mol=0.0410moles$

Initial moles of carbon tetrachloride = $6.51\times 10^{-2}mol=0.0651moles$

Volume of the container = 1.00 L

Concentration of a substance is calculated by:

$\text{Concentration}=\frac{\text{Number of moles}}{\text{Volume}}$

So, concentration of methane = $\frac{0.0410}{1.00}=0.0410M$

Concentration of carbon tetrachloride = $\frac{0.0651}{1.00}=0.0651M$

The given chemical equation follows:

$CH_4(g)+CCl_4(g)\rightleftharpoons 2CH_2Cl_2(g)$

Initial: 0.0410 0.0651

At eqllm: 0.0410-x 0.0651-x 2x

We are given:

Equilibrium concentration of carbon tetrachloride = $6.02\times 10^{-2}M=0.0602M$

Evaluating the value of 'x', we get:

$\Rightarrow (0.0651-x)=0.0602\\\\\Rightarrow x=0.0049M$

Now, equilibrium concentration of methane = $0.0410-x=[0.0410-0.0049]=0.0361M$

Equilibrium concentration of $CH_2Cl_2=2x=[2\times 0.0049]=0.0098M$

The expression of $K_{eq}$ for the above reaction follows:

$K_{eq}=\frac{[CH_2Cl_2]^2}{[CH_4]\times [CCl_4]}$

Putting values in above expression, we get:

$K_{eq}=\frac{(0.0098)^2}{0.0361\times 0.0603}\\\\K_{eq}=0.044$

Hence, the value of $K_{eq}$ is 0.044

8 0

3 years ago

what is the kinetic energy of the emitted electrons when cesium is exposed to uv rays of frequency 1.4×1015hz?

Anika [276]

The equation you used is KE=hv-hv0, where h=6.63*10^-34 (constant). You multiply h by 1.5*10^15. Multiply h by the threshold freq of cesium (from part A). Subtract the second answer from the first answer, and you get the kinetic energy. Hope this helps.

7 0

3 years ago