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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?

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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A sample of ammonium nitrate, weighing 25.2 g, was added to a 150.0 mL container of H2O at 20.0 °C, and then is thoroughly disso

OLga [1]

Explanation:

It is known that the relation between heat energy, specific heat and change in temperature is as follows.

q = $m \times C \times \Delta T$

where, q = heat released or absorbed

m = mass of substance

C = specific heat

$\Delta T$ = change in temperature

It is given that mass is 25.2 g and it is added to 150 ml container so total mass will be (25.2 g + 150 g) = 175.2 g, $T_{1}$ is $20^{o}C$ , $T_{2}$ is $8.5^{o}C$ , and specific heat of water is 4.184 $J/g ^{o}C$ .

Hence, putting the given values into the above formula as follows.

q = $m \times C \times \Delta T$

= $175.2 g \times 4.184 J/g^{o}C \times (8.5 - 20)^{o}C$

= $175.2 g \times 4.184 J/g^{o}C \times -11.5^{o}C$

= -8430 J

As, 1 J = 0.001 kJ. Hence, -8430 J will also be equal to -8.43 kJ. The negative sign indicates that heat is being lost.

Thus, we can conclude that heat (kJ) was lost by the solution (surroundings) is 8.43 kJ.

6 0

3 years ago

Calculate the no. of each atoms presents in 10 grams of calcium carbonate.

raketka [301]

Molar Mass of Calcium carbonate:-

$\\ \sf\longmapsto CaCO_3$

$\\ \sf\longmapsto 40u+12u+3(16u)$

$\\ \sf\longmapsto 52u+48u$

$\\ \sf\longmapsto 100u$

$\\ \sf\longmapsto 100g/mol$

Given Mass=10g

$\boxed{\sf No\:of\;moles=\dfrac{Given\:Mass}{Molar\:Mass}}$

$\\ \sf\longmapsto No\:of\;moles=\dfrac{10}{100}$

$\\ \sf\longmapsto No\:of\:moles=0.1mol$

Now

$\boxed{\sf No\:of\:molecules=No\;of\:moles\times Avagadro\: No}$

$\\ \sf\longmapsto 0.1\times 6.023\times 10^{23}$

$\\ \sf\longmapsto 6.023\times 10^{22}molecules$

8 0

3 years ago

Nickel carbonyl decomposes to form nickel and carbon monoxide, like this:

Schach [20]

Answer: The value of the equilibrium constant for this reaction is $2.8\times 10^{-5}$

Explanation:

To calculate the moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar mass}}$

Moles of $Ni(CO)_4$ = $\frac{0.597g}{170.7g/mol}=0.0035moles$

Moles of $Ni$ = $\frac{12.7g}{58.7g/mol}=0.216moles$

Moles of $CO$ = $\frac{1.98g}{28.01g/mol}=0.071moles$

Volume of solution = 2.7 L

Equilibrium concentration of $Ni(CO)_4$ = $\frac{0.0035mol}{2.7L}=1.29\times 10^{-3}M$

Equilibrium concentration of $Ni$ = $\frac{0.216mol}{2.7L}=0.08M$

Equilibrium concentration of $CO$ = $\frac{0.071mol}{2.7L}=0.026M$

The given balanced equilibrium reaction is,

$Ni(CO)_4\rightleftharpoons Ni+4CO$

The expression for equilibrium constant for this reaction will be,

$K_c=\frac{[Ni]^1\times [CO]^4}{[Ni(CO_4]^1}$

Now put all the given values in this expression, we get :

$K_c=\frac{(0.08)^1\times (0.026)^4}{(1.29\times 10^{-3})^1}$

$K_c=2.8\times 10^{-5}$

Thus the value of the equilibrium constant for this reaction is $2.8\times 10^{-5}$

5 0

3 years ago

HELP ASAP PLEASE!!!

serg [7]

The answer is B. Neutrons

Hope this helps

7 0

4 years ago

I need correct answer cuz this is exam so don't write anything just for point plzzz :)

Pie

Answer: B

Explanation:

7 0

3 years ago