All categories

3 years ago

Explain how friction and mass affect the motion of an object.

Chemistry

1 answer:

agasfer [191]3 years ago

8 0

Answer: Ithink that the mass is the one that affects the friction.

Explanation:

You might be interested in

What is the difference between primary scale and extension scale?

elena-14-01-66 [18.8K]

Answer:

8txirzrz8rzrirzriz£=,£=,,-8/-9/-¥/

Explanation:

otcitxfi

8 0

3 years ago

If the mass of a material is 118 grams and the volume of the material is 12 cm3, what would the density of the material be?

Sliva [168]

Density is equal to mass divided by volume.

So D=118/12

D=9.83

5 0

3 years ago

Calculate ΔGrxn at 298 K under the conditions shown below for the following reaction.

algol13

Answer : The value of $\Delta G_{rxn}$ is -24.9 kJ/mol

Explanation :

First we have to calculate the value of 'Q'.

The given balanced chemical reaction is,

$Fe_2O_3(s)+3CO(g)\rightarrow 2Fe(s)+3CO_2(g)$

The expression for reaction quotient will be :

$Q=\frac{(p_{CO_2})^3}{(p_{CO})^3}$

In this expression, only gaseous or aqueous states are includes and pure liquid or solid states are omitted.

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

$Q=\frac{(2.1)^3}{(1.4)^2}$

$Q=3.375$

Now we have to calculate the value of $\Delta G_{rxn}$ .

The formula used for $\Delta G_{rxn}$ is:

$\Delta G_{rxn}=\Delta G^o+RT\ln Q$ ............(1)

where,

$\Delta G_{rxn}$ = Gibbs free energy for the reaction = ?

$\Delta G_^o$ = standard Gibbs free energy = -28.0 kJ/mol

R = gas constant = $8.314\times 10^{-3}kJ/mole.K$

T = temperature = 298 K

Q = reaction quotient = 3.375

Now put all the given values in the above formula 1, we get:

$\Delta G_{rxn}=(-28.0kJ/mol)+[(8.314\times 10^{-3}kJ/mole.K)\times (298K)\times \ln (3.375)$

$\Delta G_{rxn}=-24.9kJ/mol$

Therefore, the value of $\Delta G_{rxn}$ is -24.9 kJ/mol

5 0

3 years ago

Upon adding solid potassium hydroxide pellets to water the following reaction takes place: KOH(s) → KOH(aq) + 43 kJ/mol Answer t

nekit [7.7K]

Answer:

a) Warmer

b) Exothermic

c) -10.71 kJ

Explanation:

The reaction:

KOH(s) → KOH(aq) + 43 kJ/mol

It is an exothermic reaction since the reaction liberates 43 kJ per mol of KOH dissolved.

Hence, the dissolution of potassium hydroxide pellets to water provokes that the beaker gets warmer for being an exothermic reaction.

The enthalpy change for the dissolution of 14 g of KOH is:

$n = \frac{m}{M}$

Where:

m: is the mass of KOH = 14 g

M: is the molar mass = 56.1056 g/mol

$n = \frac{m}{M} = \frac{14 g}{56.1056 g/mol} = 0.249 mol$

The enthalpy change is:

$\Delta H = -43 \frac{kJ}{mol}*0.249 mol = -10.71 kJ$

The minus sign of 43 is because the reaction is exothermic.

I hope it helps you!

5 0

3 years ago

You need 1.50 moles of ammonia (nh3) for a reaction at stp. what volume of ammonia must you measure? also,how many molecules of

professor190 [17]

what volume of ammonia must you measure?

1.50 moles NH3 ( 22.4 L / 1 mole ) = 33.6 L NH3 should be measured

how many molecules of nacl are in 23.40 grams nacl

23.40 g NaCl ( 1 mol / 58.44 g) (6.022 x 10^23 molecules / 1 mol ) = 2.4113 x 10^23 molecules

Hope these answer the questions. Have a nice day.

4 0

4 years ago