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3 years ago

Calculate the density of gold (in units of g/mL) if a 5.00 cm cube has a mass of 2.41kg

2 answers:

6 0

Turn 2.41 kg into grams :

2.41 * 1000 => 2,410 g

1 cm³ = 1 mL , so:

5.00 cm³ = 5.00 mL

Therefore:

D = mass / volume

D = 2,410 / 5.00

D = 482 g/mL

levacccp [35]3 years ago

5 0

The volume of a 5cm cube is 0.5dm*0.5dm*0.5dm = 0.125dm^3
1dm^3 = 1l
So 0.125dm^3 = 125mL
This gives us 2410/125 = 19.28
So the density of gold is 19.28g/mL
Final check: Gold is indeed 19.3g/mL

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3. Heating by direct contact between particles is called

Yanka [14]

I believe it is “Conduction” im not 100% sure tho, I haven’t done this since the 7th grade.

8 0

3 years ago

Water vapor is condensed steam. True False

adelina 88 [10]

False, because water vapor, water vapour or aqueous vapor, is the gaseous phase of water. It is one state of water within the hydrosphere. Water vapor can be produced from the evaporation or boiling of liquid water or from the sublimation of ice. Unlike other forms of water, water vapor is invisible.

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3 years ago

5. A beam of photons with a minimum energy of 222 kJ/mol can eject electrons from a potassium surface. Estimate the range of wav

torisob [31]

Answer: The range of wavelengths of light that can be used to cause given phenomenon is $8.953 \times 10^{21} m$ .

Explanation:

Given: 222 kJ/mol (1 kJ = 1000 J) = 222000 J

Formula used is as follows.

$E = \frac{hc}{\lambda}$

where,

E = energy

h = Planck's constant = $6.625 \times 10^{-25} Js$

c = speed of light = $3 \times 10^{8} m/s$

Substitute the values into above formula as follows.

$E = \frac{hc}{\lambda}\\222000 J = \frac{6.625 \times 10^{-34}Js \times 3 \times 10^{8} m/s}{\lambda}\\\lambda = 8.953 \times 10^{21} m$

Thus, we can conclude that the range of wavelengths of light that can be used to cause given phenomenon is $8.953 \times 10^{21} m$ .

7 0

3 years ago

X-rays had a frequency of about 3x10^18 determine it's wavelength (in nm) and energy per photon

Serggg [28]

Answer:

E = 19.89×10⁻¹⁶ J

λ = 1×10⁻¹ nm

Explanation:

Given data:

Frequency of xray = 3×10¹⁸ Hz

Wavelength of xray = ?

Energy of xray = ?

Solution:

speed of wave = wavelength × frequency

speed = 3×10⁸ m/s

3×10⁸ m/s = λ ×3×10¹⁸ s⁻¹

λ = 3×10⁸ m/s / 3×10¹⁸ s⁻¹

λ = 1×10⁻¹⁰m

m to nm:

λ = 1×10⁻¹⁰m×10⁹

λ = 1×10⁻¹ nm

Energy of x-ray:

E = h.f

h = plancks constant = 6.63×10⁻³⁴ Js

by putting values,

E = 6.63×10⁻³⁴ Js ×3×10¹⁸ s⁻ ¹

E = 19.89×10⁻¹⁶ J

7 0

3 years ago

Give the IUPAC name for the following structure: 3-chloro-6-methylcyclohexanol 2-methyl-5-chlorocyclohexanol 1-chloro-4-methylcy

Shtirlitz [24]

Answer:

5-chloro-2-methylcyclohexanol

Explanation:

There is no structure for the compound, but we can analyze the proposed options using the IUPAC rules to name organic compounds.

IUPAC rules state that to name an organic compound, first we have to identify the priorities for the functional groups present in the compound. In this case, the priority functional group is the alcohol group, so we will start the counting of the carbons in this group. Then, the counting of carbon atoms is followed by the next substituents so they have the lowest possible numbers, in this case, we can assign the number 2 to the methyl group and 5 to the chloride group, and name the compound in alphabetical order, using commas to separate the words from the numbers and with no space between the words.

Since the other options involve: high countings for the susbtituents groups (3-chloro-6-methylcyclohexanol), wrong assignation of priority functional group (1-chloro-4-methylcyclohexanol), wrong sequence of counting in the compound (2-methyl-3-chlorocyclohexanol) and no alphabetical order to name the compound (2-methyl-5-chlorocyclohexanol), the correct option is:

5-chloro-2-methylcyclohexanol

Have a nice day!

4 0

4 years ago