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siniylev [52]
3 years ago
3

What is the difference between Claire’s test of the collision scene where Vehicle 2 fell off the cliff and the film, Iceworld Re

venge, where it did not?*
Which claim is more convincing?

Claim 1: The vehicles in Iceworld Revenge had different masses; in Claire’s test, the vehicles had the same mass.

Claim 2: The friction of the surface that was used in Iceworld Revenge was different from the friction of the surface in Claire’s test.

Did the Science Seminar cause your thinking about the claims to change? Explain.
Chemistry
1 answer:
RUDIKE [14]3 years ago
7 0

Complete Question:

Check the file attached to get the complete question

Answer:

In the film Ice word Revenge, vehicle 2 did not fall of the cliff because,  but in Claire's test, vehicle 2 off the cliff because

Explanation:

In Claire's test, the weight of vehicle 1 is either equal to or greater than the weight of vehicle 2, so it was sufficient to push it down the cliff.  In the film Ice word revenge, the weight of vehicle 1 is less than the weight of vehicle 2, it is not sufficient to make it fall off the cliff ( Note: Looking exactly the same in the movie, as Claire claimed, does not mean they have the same mass). Therefore if Claire wants a collision that will not make the vehicle 2 fall off the cliff, he should collide it with a vehicle of lesser mass/weight.

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Which of these are true for the reaction above?
agasfer [191]

Answer:

a, d, f

Explanation:

ΔHrxn = ΔH(CCl4) -ΔH(CH4) = - 106.7 -(-74.8) = - 31.9 kJ/mol

6 0
4 years ago
Each step in the following process has a yield of 90.0 %. CH 4 + 4 Cl 2 ⟶ CCl 4 + 4 HCl CCl 4 + 2 HF ⟶ CCl 2 F 2 + 2 HCl The CCl
Rainbow [258]

Answer:

4.86 moles of HCl

Explanation:

1. First write the balanced chemical equations involved in the process:

CH_{4}+_4Cl_{2}=CCl_{4}+_4HCl

CCl_{4}+_2HF=CCl_{2}+F_{2}+_2HCl

2. Calculate what amount of CCl_{4} is formed in the first reaction.

3.00molesCH_{4}*\frac{1molCCl_{4}}{1molCH_{4}}=3.00molesCCl_{4}

As the yield of each reaction is 90.0%, the amount of CCl_{4} produced is the following:

3.00molesCCl_{4}*0.90=2.7molesCCl_{4}

3. Calculate the amount of HCl produced.

2.7molesCCl_{4}*\frac{2molHCl}{1molCCl_{4}}=5.4molesHCl

The total amount of HCl produced with a 90.0% yield is:

5.4molesHCl*0.90=4.86molesHCl

4 0
3 years ago
When a radioactive atom emits an alpha particle, the original atom's atomic number decreases by four.
solniwko [45]
Ya it is true, alpha particle is actually helium atom with no electrons !
He2+ and it has mass of 4 u ....So that is correct !
5 0
4 years ago
Which statement is true of a
soldier1979 [14.2K]
I think the awnser to your question is C
6 0
3 years ago
Pentaborane−9 (B5H9) is a colorless, highly reactive liquid that will burst into flames when exposed to oxygen. The reaction is
mote1985 [20]

Answer : The heat released per gram of the compound reacted with oxygen is 71.915 kJ/g

Explanation :

Enthalpy change : It is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as \Delta H^o

The equation used to calculate enthalpy change is of a reaction is:  

\Delta H^o_{rxn}=\sum [n\times \Delta H^o_f(product)]-\sum [n\times \Delta H^o_f(reactant)]

The equilibrium reaction follows:

2B_5H_9(l)+12O_2(g)\rightleftharpoons 5B_2O_3(s)+9H_2O(l)

The equation for the enthalpy change of the above reaction is:

\Delta H^o_{rxn}=[(n_{(B_2O_3)}\times \Delta H^o_f_{(B_2O_3)})+(n_{(H_2O)}\times \Delta H^o_f_{(H_2O)})]-[(n_{(B_5H_9)}\times \Delta H^o_f_{(B_5H_9)})+(n_{(O_2)}\times \Delta H^o_f_{(O_2)})]

We are given:

\Delta H^o_f_{(B_5H_9(l))}=73.2kJ/mol\\\Delta H^o_f_{(O_2(g))}=0kJ/mol\\\Delta H^o_f_{(B_2O_3(s))}=-1271.94kJ/mol\\\Delta H^o_f_{(H_2O(l))}=-285.83kJ/mol

Putting values in above equation, we get:

\Delta H^o_{rxn}=[(5\times -1271.94)+(9\times -285.83)]-[(2\times 73.2)+(12\times 0)]=-9078.57kJ/mol

Now we have to calculate the heat released per gram of the compound reacted with oxygen.

From the reaction we conclude that,

As, 2 moles of compound released heat = -9078.57 kJ

So, 1 moles of compound released heat = \frac{-9078.57}{2}=-4539.28kJ

For per gram of compound:

Molar mass of B_5H_9 = 63.12 g/mole

\Delta H^o_{rxn}=\frac{-4539.28}{63.12}=-71.915kJ/g

Therefore, the heat released per gram of the compound reacted with oxygen is 71.915 kJ/g

7 0
3 years ago
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