All categories

3 years ago

Predict: Will a 1500 kg car moving at the same speed as the tennis ball have

1 answer:

4 0

The car will have more momentum because it has a greater mass. Momentum= mass x velocity therefore if both the objects have the same velocity the object with a greater mass will have more momentum

You might be interested in

Which of the following hydrocarbons has a double bond in its carbon skeleton?1) C3H82) C2H63) CH44) C2H45) C2H2

Kipish [7]

Answer:

C₂H₄ (option number 4)

Explanation:

A hydrocarbon with a double bond in its carbon skeleton is an alkene and has the general form:

$C_nH_{2n}$ .

This is, the number of hydrogen atoms is twice the number of carbon atoms.

On the other hand, alkanes have only single bonds, and the compounds with a triple bond in its carbon skeleton are alkynes.

Review each choice:

1) C₃H₈:

In this case, the number of hydrogen atoms is 2×3 + 2 = 6 + 2 = 8, which is corresponds to an alkane, not an alkene.

2) C₂H₆

For this, the number of hydrogen atoms is 2 × 2 + 2 = 4 + 2 = 6. Again an alkane, not alkene.

3) CH₄

Hydrogen atoms: 1 × 2 + 2 = 4 ⇒ an alkane

4) C₂H₄

Hydrogen atoms: 2 × 2 = 4. This is precisely the relation for an alkene, so this is the hydrocarbon that has a double bond in its carbon skeleton.

The chemical formula may be writen as CH₂ = CH₂, to show the double bond.

So, this is the correct answer.

5) C₂H₂

Hydrogen atoms: 2 × 2 - 2 = 4 - 2 = 2. This relation of carbon and hydrogen atoms corresponds to a compound with triple bond, i.e an alkyne: CH≡CH.

8 0

3 years ago

The equilibrium constant for the reaction AgBr(s) Picture Ag+(aq) + Br− (aq) is the solubility product constant, Ksp = 7.7 × 10−

barxatty [35]

Answer:

The reaction will be non spontaneous at these concentrations.

Explanation:

$AgBr(s)\rightarrow Ag^+(aq) + Br^- (aq)$

Expression for an equilibrium constant $K_c$ :

$K_c=\frac{[Ag^+][Br^-]}{[AgCl]}=\frac{[Ag^+][Br^-]}{1}=[Ag^+][Br^-]$

Solubility product of the reaction:

$K_{sp}=[Ag^+][Br^-]=K_c=7.7\times 10^{-13}$

Reaction between Gibb's free energy and equilibrium constant if given as:

$\Delta G^o=-2.303\times R\times T\times \log K_c$

$\Delta G^o=-2.303\times R\times T\times \log K_{sp}$

$\Delta G^o=-2.303\times 8.314 J/K mol\times 298 K\times \log[7.7\times 10^{-13}]$

$\Delta G^o=69,117.84 J/mol=69.117 kJ/mol$

Gibb's free energy when concentration $[Ag^+] = 1.0\times 10^{-2} M$ and $[Br^-] = 1.0\times 10^{-3} M$

Reaction quotient of an equilibrium = Q

$Q=[Ag^+][Br^-]=1.0\times 10^{-2} M\times 1.0\times 10^{-3} M=1.0\times 10^{-5}$

$\Delta G=\Delta G^o+(2.303\times R\times T\times \log Q)$

$\Delta G=69.117 kJ/mol+(2.303\times 8.314 Joule/mol K\times 298 K\times \log[1.0\times 10^{-5}])$

$\Delta G=40.588 kJ/mol$

For reaction to spontaneous reaction: $\Delta G$ .
For reaction to non spontaneous reaction: $\Delta G>0$ .

Since ,the value of Gibbs free energy is greater than zero which means reaction will be non spontaneous at these concentrations

5 0

3 years ago

How many molecules of he are contained in a 10.0 l tank at 7.53 atm and 485 k?

ankoles [38]

Answer:

1.126 x 10^22

Explanation:

pV = nRT

7.53 x 10 = n x 8.31 x 485

n = (7.53 x 10) / (8.31 x 485) = 0.0187 moles

M = n x Avogadros number

0.0187 x 6.02 x 10^23 = 1.126 x 10^22

4 0

3 years ago

A potassium atom (atomic number 19) and a bromine atom (atomic number 35) can form a chemical bond through a transfer of one ele

liubo4ka [24]

Answer:

A potassium atom (atomic number 19) and a bromine atom (atomic number 35) can form a chemical bond through a transfer of one electron. The potassium ion that forms has 18 electrons. What best describes the bromide ion that forms? It is a negative ion that has one more valence electron than a neutral bromine atom.

Explanation:

7 0

2 years ago

The total bonding energy for the products of a reaction is 2535 kJ/mol and the bonding energy of the reactants is 1375 kJ/mol. C

Dmitry [639]

Answer:

-1160kj/mol

Explanation:

the reaction is exothermic because heat is released to the environment

7 0

3 years ago