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LUCKY_DIMON [66]
3 years ago
11

Gerry is looking at salt under a powerful microscope and notices a crystalline structure. What can be known about the salt sampl

e that Gerry is looking at?
Chemistry
2 answers:
Phoenix [80]3 years ago
4 0
Mined salt has sharp edges v s sea salt with rounded edged
mezya [45]3 years ago
3 0

Answer:

It shows that the atoms are vibrating in place.

Explanation:

In the crystal, the molecules exhibit a definite position, and they are combined with each other with the help of electrostatic forces. The molecules exhibit certain kind of energy due to which they vibrate in their locations. However, their energy is not that much high to result in overcoming the strong bonding, unless the atoms in the crystals are irradiated or are heated.

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1,4-Pentadiene has a AHhydro = -254 kJ/mol while trans-1,3-pentadiene has a AHhydra = -226 kJ/mol. Explain this difference in he
julsineya [31]

Answer:

trans-1,3-pentadiene is more stable than 1,4-pentadiene due to presence of a conjugated double bond.

Explanation:

Here, \Delta H_{hydro}=H(hydrogenated pdt.)-H(diene)

H(hydrogenated pdt.) is same for both 1,4-pentadiene and 1,3-pentadiene as they both produce pentane after hydrogenation

H(diene) depends on stability of diene.

More stable a diene, lesser will be it's H(diene) value (more neagtive).

trans-1,3-pentadiene is more stable than 1,4-pentadiene due to presence of a conjugated double bond.

Hence, \Delta H_{hydro} is higher (less negative) for trans-1,3-pentadiene

5 0
3 years ago
How many bromine atoms are present in 39.4 g of CH2Br2
strojnjashka [21]
37.8 g CH2Br2 X (1 mol CH2Br2 / 173.83 g) = 4.60X10^-3 mol CH2Br2 

<span>4.60X10^-3 mol CH2Br2 X (2 mol Br / 1 mol CH2Br2) X 6.02X10^23 atoms/mol = 5.54X10^21 bromine atoms</span>
7 0
3 years ago
A 151.5-g sample of a metal at 75.0°C is added to 151.5 g at 15.1°C. The temperature of the water rises to 18.7°C. Calculate the
Kryger [21]

Answer:

The specific heat capacity of the metal is 0.268 J/g°C

Explanation:

Step 1: Data given

Mass of the metal = 151.5 grams

The temperature of the metal = 75.0 °C

Temperature of water = 15.1 °C

The temperature of the water rises to 18.7°C.

The specific heat capacity of water is 4.18 J/°C*g

Step 2: Calculate the specific heat capacity of the metal

heat lost = heat gained

Q = m*c*ΔT

Qmetal = - Qwater

m(metal) * c(metal) * ΔT(metal) = m(water) * c(water) * ΔT(water)

⇒ mass of the metal = 151.5 grams

⇒ c(metal) = TO BE DETERMINED

⇒ΔT( metal) = T2 - T1 = 18.7 °C - 75.0 °C = -56.3 °C

⇒ mass of the water = 151.5 grams

⇒ c(water) = 4.184 J/g°C

⇒ ΔT(water) = 18.7° - 15.1 = 3.6 °C

151.5g * c(metal) * -56.3°C = 151.5g * 4.184 J/g°C * 3.6 °C

c(metal) = 0.268 J/g°C

The specific heat capacity of the metal is 0.268 J/g°C

5 0
3 years ago
A 1.2 kg sample of Th-228 has a half-life of 1.9 years. How many grams of Th is left after 13.3 years?
Bumek [7]
.009375 kg or 9.375 grams will remain
7 0
3 years ago
What coefficient should be used to make the following equation balanced? N2+O2--&gt; _NO
IgorLugansk [536]
<h2>                  ║∧║······→Hello.←······║∧║</h2>

                                          Your answer should be:

                                                N2 + O2 → 2NO

<h2>····················································································</h2>

I hope this answered your question!

Thanks for spending time reading this :)

(<em>A brainliest would be appreciated!)</em>

⊕If this answer doesn't answer your question or you are displeased by it please tell me in the comments I would like to know. And if I can, I will remove it.⊕

<h2>Have a wonderful day</h2>
7 0
2 years ago
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