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vlada-n [284]
3 years ago
5

Students in science are studying solutions and the dissolving rate of solids. One group was given some instant lemonade mix to

Chemistry
1 answer:
UkoKoshka [18]3 years ago
7 0

Answer: A E D C

Explanation:

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The diameter of a hydrogen atom is approximately 1.06 ✕ 10−10 m, as defined by the diameter of the spherical electron cloud arou
mylen [45]

Answer:

a) 10.0 mm

b) 8.7 x 10¹³ times

Explanation:

Atom diameter = 1.06 x 10⁻¹⁰ m  ________________ 100%

Nucleus diameter = 2.40 x 10⁻¹⁵ m ______________   x

x = 2.26 x 10⁻³ %

The nucleus diameter is equivalent to 2.26 x 10⁻³ % of the total atom size.

a) The Empire State Building model:

1 ft = 304.8 mm

1454 ft = 443179.2 mm

443179.2 mm______ 100%

        y           ______   2.26 x 10⁻³ %

              y = 10.0 mm

In this model,  the diameter of the nucleus would be 10.0 mm.

b) Sphere volume: V =(4 · π · r³ )/3

V atom = (4 . π .( 0.53x10⁻¹⁰)³ )/3

V atom = 6.2 x 10⁻³¹ m³

V nucleus = (4 . π .( 1.2x10⁻¹⁵)³ )/3

V nucleus = 7.2 x 10⁻⁴⁵ m³

V atom / V nucleus = 6.2 x 10⁻³¹ m³ / 7.2 x 10⁻⁴⁵ m³

V atom / V nucleus = 8.7 x 10¹³

The atom is  times 8.7 x 10¹³ larger in volume than its nucleus.        

6 0
4 years ago
What type of particles move to create electicity?<br> electrons<br> protons<br> neutrons
sladkih [1.3K]

the answer is electrons.....

5 0
3 years ago
Explain the voyage of water molecules as it boils in water
Ierofanga [76]
Hot water rises cold water sinks. So the warm water will stay at the bottom. Transfer of heat through the molecules will make all the water boil
3 0
3 years ago
How is mercury obtained and separated from other nearby substances?​
VLD [36.1K]

Answer:

Most mercury forms in a sulfide ore called cinnabar, but mercury is also frequently found in small amounts in other ores. A common method for separating mercury from cinnabar is to crush the ore and then heat it in a furnace in order to vaporize the mercury. This vapor is then condensed into liquid mercury form.

Explanation:

7 0
3 years ago
Read 2 more answers
4. A student started with a 0.032 g sample of copper which he took through the series of reactions described in this experiment.
Elodia [21]

Answer:

Y=48.6\%

Explanation:

Hello,

In this case, we can consider the following chemical reaction for the oxidation of copper which only occurs at high temperatures:

2Cu+O_2\rightarrow 2CuO

In such a way, for 0.032 grams of copper, the following grams of copper (II) oxide (black product) are yielded:

m_{CuO}=0.032gCu*\frac{1molCu}{63.546gCu} *\frac{2molCuO}{2molCu}*\frac{79.546gCuO}{1molCuO}  =0.078gCuO

Therefore, the percent yield is:

Y=\frac{0.038g}{0.078g}*100\%\\ \\Y=48.6\%

Best regards.

6 0
3 years ago
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