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3 years ago

Why can't a bullet with a tiny mass have the same momentum as a moving truck?

1 answer:

3 0

A bullet with a tiny mass cant have the same momentum as a moving truck because the trucks mass and size makes it a lot harder to slow down. There for the wait of a 10 ton truck does a lot more damage then that of a tiny bullet. Due to the difference in the mass.

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Which statement best describes an array of 10 elements where the elements are of a class type?

ladessa [460]

The array will be created with the elements equal to null.

3 0

3 years ago

Determine the free energy(ΔG) from the standard cell potential (Ecell0 ) for the reaction:2ClO2-(aq)+Cl2(g)→2ClO2(g)+ 2Cl-(aq)wh

Dima020 [189]

Answer: The $\Delta G^o$ for the given reaction is $-7.84\times 10^4J$

Explanation:

For the given chemical reaction:

$2ClO_2^-(aq.)+Cl_2(g)\rightarrow 2ClO_2(g)+2Cl^-(aq.)$

Half reactions for the given cell follows:

Oxidation half reaction: $ClO_2^-\rightarrow ClO_2+e^-;E^o_{ClO_2^-/ClO_2}=0.954V$ ( × 2)

Reduction half reaction: $Cl_2+2e^-\rightarrow 2Cl(g);E^o_{Cl_2/2Cl^-}=1.36V$

Oxidation reaction occurs at anode and reduction reaction occurs at cathode.

To calculate the $E^o_{cell}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$

Putting values in above equation, we get:

$E^o_{cell}=1.36-(0.954)=0.406V$

To calculate standard Gibbs free energy, we use the equation:

$\Delta G^o=-nFE^o_{cell}$

Where,

n = number of electrons transferred = 2

F = Faradays constant = 96500 C

$E^o_{cell}$ = standard cell potential = 0.406 V

Putting values in above equation, we get:

$\Delta G^o=-2\times 96500\times 0.406=-78358J=-7.84\times 10^4J$

Hence, the $\Delta G^o$ for the given reaction is $-7.84\times 10^4J$

4 0

3 years ago

What does the position of an element in the periodic table tell you?

loris [4]

They are in order according to their atomic number, so the position should tell you the atomic number.

3 0

3 years ago

A sample of 3.62 moles of diphosphorous trioxide is

Sindrei [870]

$H_3PO_3$ $2H_3PO_3$ $2H_3PO_3$ $P_2O_3$ Answer:

Explanation:

This question is about stoichiometry. From the balanced equation $P_2O_3 + 3H_2O$ ⇒ $2H_3PO_3$ , we see that 3 moles of water is needed to react with 1 mole of $P_2O_3$ .

This means that, to fully react 3.62 moles of $P_2O_3$ , we would need 3*3.62 or 10.86 moles of water. However, we only have 6.31 moles, so water is the limiting reactant.

Since 3 moles of water react with 1 mole of $P_2O_3$ , 6.31 moles of water can fully react with 6.31÷3 or 2.1033 moles of $P_2O_3$ .

From the balanced equation, we see that every mole of $P_2O_3$ reacted gets you 2 moles of $2H_3PO_3$ . Therefore, 2.1033 moles of $P_2O_3$ would give you approximately 4.21 moles of $H_3PO_3$ .

5 0

2 years ago

What is Magnitude? ⚫ No Spam

Rainbow [258]

the great size or extent of something

4 0

3 years ago

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