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Leto [7]
3 years ago
6

Explain the role that gravitation played in the formation of the Moon after the large planetesimal hit Earth.

Chemistry
2 answers:
oksano4ka [1.4K]3 years ago
4 0

Answer:

The gravity pulled the bits back and formed a ball which is the moon

Explanation:

hope this helps :)

Misha Larkins [42]3 years ago
4 0

Answer:

While some of the debris from the collision merged back into Earth, the gravity caused the accretion of much of the debris into a separate ball, forming the Moon.

Explanation:

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Which element, if combined with chlorine, will have the greatest attraction for the chlorine electrons?
german
Phosphorus is the element among the elements given in the question that if <span>combined with chlorine, will have the greatest attraction for the chlorine electrons. The correct option among all the options given in the question is the second option or option "b" and the fact is that you have chosen correctly. </span>
6 0
3 years ago
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The forensic technician at a crime scene has just prepared a luminol stock solution by adding 16.0 g of luminol to H2O creating
san4es73 [151]

Answer : The concentration of luminol in the solution is, 1.20 M.

Explanation : Given,

Mass of luminol = 16.0 g

Volume of solution = 75.0 mL

Molar mass of luminol = 177.16 g/mole

Molarity : It is defined as the number of moles of solute present in one liter of volume of solution.

Formula used :

\text{Molarity}=\frac{\text{Mass of luminol}\times 1000}{\text{Molar mass of luminol}\times \text{Volume of solution (in mL)}}

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

\text{Molarity}=\frac{16.0g\times 1000}{177.16g/mole\times 75.0mL}=1.20mole/L=1.20M

Therefore, the concentration of luminol in the solution is, 1.20 M.

4 0
3 years ago
Here are some notes if anybody needs em, you just gotta mirror the image!
san4es73 [151]

Answer:

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4 0
3 years ago
Which substance has a coefficient of 2 when you balance the equation for this redox reaction? cu(s) + hno3(aq) → cu(no3)2(aq) +
Mama L [17]
In this redox reaction, the Cu goes from oxidation state of (0) to (+2), therefore it oxidises. N in HNO₃ goes from oxidation state of (+5) to N in NO with oxidation state of (+2) and becomes reduced. 
Cu acts as the reducing reagent and HNO₃ is the oxidising agent.
oxidation half reaction
Cu ---> Cu²⁺ + 2e  --1)
reduction half reaction 
4H⁺ + 3e + NO₃⁻ ---> NO + 2H₂O --2)
to balance the number of electrons , 1) x3 and 2) x2
3Cu ---> 3Cu²⁺ + 6e
8H⁺ + 6e + 2NO₃⁻ ---> 2NO + 4H₂O 
add the 2 equations 
3Cu + 8H⁺ + 2NO₃⁻ --->  3Cu²⁺ + 2NO + 4H₂O 
add 6 nitrate ions to both sides to add up to 8 and form acid with 8H⁺ ions
3Cu + 8HNO₃ --->  3Cu(NO₃)₂ + 2NO + 4H₂O 
Balanced equation for the redox reaction is as follows;
3Cu(s) + 8HNO₃(aq) → 3Cu(NO₃)₂(aq) + 2NO(g) + 4H₂O<span>(l)

NO has a coefficient of 2

</span>
8 0
3 years ago
When you combine 50.0 mL of 0.100 M AgNO3 with 50.0 mL of 0.100 M HCl in a coffee-cup calorimeter, the temperature changes from
RideAnS [48]

Answer : The enthalpy of reaction (\Delta H_{rxn}) is, 67.716 KJ/mole

Explanation :

First we have to calculate the moles of AgNO_3 and HCl.

\text{Moles of }AgNO_3=\text{Molarity of }AgNO_3\times \text{Volume}=(0.100mole/L)\times (0.05L)=0.005mole

\text{Moles of }HCl=\text{Molarity of }HCl\times \text{Volume}=(0.100mole/L)\times (0.05L)=0.005mole

Now we have to calculate the moles of AgCl formed.

The balanced chemical reaction will be,

AgNO_3(aq)+HCl(aq)\rightarrow AgCl(s)+HNO_3(aq)

As, 1 mole of AgNO_3 react with 1 mole of HCl to give 1 mole of AgCl

So, 0.005 mole of AgNO_3 react with 0.005 mole of HCl to give 1 mole of AgCl

The moles of AgCl formed  = 0.005 mole

Total volume of the solution = 50.0 ml + 50.0 ml = 100.0 ml

Now we have to calculate the mass of solution.

Mass of the solution = Density of the solution × Volume of the solution

Mass of the solution = 1.00 g/ml × 100.0 ml = 100 g

Now we have to calculate the heat.

q=m\times C\Delta T=m\times C \times (T_2-T_1)

where,

q = heat

C = specific heat capacity = 4.18J/g^oC

m = mass = 100 g

T_2 = final temperature = 24.21^oC

T_1 = initial temperature = 23.40^oC

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

q=100g\times (4.18J/g^oC)\times (24.21-23.40)^oC

q=338.58J

Now  we have to calculate the enthalpy of the reaction.

\Delta H_{rxn}=\frac{q}{n}

where,

\Delta H_{rxn} = enthalpy of reaction = ?

q = heat of reaction = 338.58 J

n = moles of reaction = 0.005 mole

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

\Delta H_{rxn}=\frac{338.58J}{0.005mole}=6771.6J/mole=67.716KJ/mole

Conversion used : (1 KJ = 1000 J)

Therefore, the enthalpy of reaction (\Delta H_{rxn}) is, 67.716 KJ/mole

4 0
3 years ago
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