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goldenfox [79]
3 years ago
5

Suggest why some manufacturers of car and jewellery would use metal alloys instead of pure metals?

Chemistry
1 answer:
nika2105 [10]3 years ago
6 0
I think using metal alloys would be better than using the pure metal would cut down on the amount of gases that are released in the air.
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Hydroxylamine is a weak molecular base with kb = 6.6 x 10-9. what is the ph of a 0.0500 m solution of hydroxylamine?
7nadin3 [17]
Answer is: pH of hydroxylamine solution is 9,23.

Kb(NH₂OH) = 1,8·10⁻⁵<span>.
c</span>₀(NH₂OH)<span> = 0,0500 M = 0,05 mol/L.
c(NH</span>₂⁺) = c(OH⁻) = x.

c(NH₂OH<span>) = 0,05 mol/L - x.
Kb = c(NH</span>₂⁺) · c(OH⁻) / c(NH₂OH).

0,0000000066 = x² /  (0,05 mol/L - x). 

solve quadratic equation: x = c(OH⁻) = 0,000018 mol/L.<span>
pOH = -log(</span>0,000018 mol/L) = 4,74.<span>
pH = 14 - 4,74 = 9,23.</span>


4 0
3 years ago
2. A rock occupies a volume of 20 cm and has a mass of 54 grams. Find the density of<br> this rock.
vredina [299]
The answer is 2.7gm/cm^3 . Density is mass divided by volume .
6 0
2 years ago
All of the halogens, group 17, have seven valence electrons. which of these would represent the oxidation number of the halogens
fomenos
It depends if it occurs naturally it has oxidation number of 0
but when it react with other element it has an oxidation number of -1
6 0
2 years ago
CAN SOMEONE PLEASE HELP!
katrin [286]

Answer:C

Explanation: they break apart into ions as they dissolve into solution this process is known as dissociation.

3 0
2 years ago
Read 2 more answers
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
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