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3 years ago

Why is it important to conserve or save our energy resources?

Chemistry

1 answer:

earnstyle [38]3 years ago

5 0

answer:

reducing energy use limits the number of carbon emissions in the environment
conserving energy produces a higher quality of life. Reduced emissions result in cleaner air quality
it helps create a healthier planet, or at least helps sustain the resources we already have

explanation:

credits: online research

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Thich of the following is a chemical property2

Dimas [21]

Answer:

"Freezing point and ability to react with oxygen" are chemical properties

Explanation:

The change of liquid into solid is the freezing point. The melting point is more than the freezing point in certain cases of mixtures for certain organic compounds like fats. As soon as the mixture freezes it becomes solid and which results in change in the composition from the liquid and solid in this way the it drastically reduces the freezing point. The melting point gets higher due to the pressure. This happens due to the release of heat of which results in the rise of temperature to the freezing point .Also the reaction of elements with oxygen which leads to formation of new substance is also an chemical property

4 0

4 years ago

When liquid silver nitrate and liquid sodium chloride are combined, solid silver chloride forms along with a new liquid, sodium

Shalnov [3]

B Silver Chloride
when you mix AgNO3- and NaCl- they make AgCl which is insoluble and settle to the bottom of the test tube, and NaNO3 which remains as a liquid solution

5 0

4 years ago

Read 2 more answers

Calculate the mass of glucose metabolized by a 60.0 −kg person in climbing a mountain with an elevation gain of 1550 m . Assume

lbvjy [14]

Answer:

Mass of glucose = 515.34 g

Explanation:

We are given;

Mass; m = 60 kg

Elevation; h = 1550 m

Acceleration due to gravity; 9.8 m/s²

Now, work performed to lift 60kg by 1550m is given by the formula;

W = mgh

W = 60 × 9.8 × 1550

W = 911400 J

We are told the actual work is 4 times the one above.

Thus;

Actual work = 4W = 4 × 911400 = 3,645,600 J

Now,

Molar mass of Glucose(C6H12O6) = 180 g/mol

We are given standard enthalpy of combustion = -1273.3 KJ/mol = -1273300

Moles of glucose = 3645600/1273300 = 2.863mol

Mass of glucose = 2.863 mol × 180 g/mol

Mass of glucose = 515.34 g

4 0

4 years ago

_____ occurs when the product of the ion concentrations exceeds the Ksp.

arlik [135]

Precipitation occurs when the product of the ion concentration exceeds the Ksp.

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3 years ago

Read 2 more answers

An atom has a diameter of 2.50 Å and the nucleus of that atom has a diameter of 9.00×10−5 Å . Determine the fraction of the volu

chubhunter [2.5K]

Answer:

The fraction of the volume of the atom that is taken up by the nucleus is $4.6656\times 10^{-14}$ .

The density of a proton is $6.278\times 10^{14} g/cm^3$ .

Explanation:

Diameter of the atom ,d = 2.50 Å

Radius of the atom ,r = 0.5 d=0.5 × 2.50 Å = 1.25Å

Volume of the sphere= $\frac{4}{3}\pi r^3$

Volume of atom = V

$V=\frac{4}{3}\pi r^3$ ..[1]

Diameter of the nucleus ,d' = $9.00\times 10^{-5}\AA$

Radius of the nucleus ,r' = 0.5 d'= $0.5\times 9.00\times 10^{-5}\AA=4.5\times 10^{-5}\AA$

Volume of nucleus = V'

$V=\frac{4}{3}\pi r'^3$ ..[2]

Dividing [2] by [1]

$\frac{V'}{V}=\frac{\frac{4}{3}\pi r'^3}{\frac{4}{3}\pi r^3}$

$=\frac{r'^3}{r^3}=\frac{(4.5\times 10^{-5}\AA)^3}{(1.25 \AA)^3}$

$\frac{V'}{V}=4.6656\times 10^{-14}$

The fraction of the volume of the atom that is taken up by the nucleus is $4.6656\times 10^{-14}$ .

Diameter of the proton ,d = $1.72\times 10^{-15} m = 1.72\times 10^{-13} cm$

1 m = 100 cm

Radius of the proton,r = 0.5 d= $0.5\times 1.72\times 10^{-13} cm=8.6\times 10^{-14} cm$

Volume of the sphere= $\frac{4}{3}\pi r^3$

Volume of atom = V

$V=\frac{4}{3}\times 3.14\times (8.6\times 10^{-14} cm)^3=2.664\times 10^{-39}cm^3$

Mass of proton, m = 1.0073 amu = $1.0073\times 1.66054\times 10^{-24} g$

$1 amu = 1.66054\times 10^{-24} g$

Density of the proton : d

$d=\frac{m}{V}=\frac{1.0073\times 1.66054\times 10^{-24} g}{2.664\times 10^{-39}cm^3}=6.278\times 10^{14} g/cm^3$

The density of a proton is $6.278\times 10^{14} g/cm^3$ .

5 0

4 years ago