Luca made the following chart for Earth’s four basic movements. What needs to be correct on his chart

How are mass and weight different? mass is the space an object takes up and weight is the effect of gravity on mass. mass is the

Solnce55 [7]

Answer:

if you know what they have in common please comment what here

Explanation:

7 0

3 years ago

Read 2 more answers

What does X represent for this transmutation?

Dimas [21]

Answer:

24He

Explanation:

When a nuclide is bombarded with alpha particles, we notice an increase in its mass and charge since an alpha particle has a mass of four and a charge of +2. recall that nuclear equations must be balanced. That is, the masses and charges on the left hand six and the right hand side of the nuclear reaction equation must be exactly the same.

Looking at the nuclear reaction at hand, the total mass on the left hand side has to be 246 while the total charge on the left hand side of the equation will be 98. Coming over to the right hand side, the total masses are also found to be the same as a neutron with no charge and a mass of 1 is produced. Hence the nuclear reaction equation is balanced when an alpha particle (helium nucleus) is substituted for X in the equation.

8 0

3 years ago

A reaction A(aq)+B(aq)↽−−⇀C(aq) has a standard free‑energy change of −4.20 kJ/mol at 25 °C. What are the concentrations of A, B,

Dafna1 [17]

Answer : The concentration of $A,B\text{ and }C$ at equilibrium are 0.132 M, 0.232 M and 0.168 M respectively.

Explanation :

The given chemical reaction is,

$A(aq)+B(aq)\rightleftharpoons C(aq)$

First we have to calculate the equilibrium constant for the reaction.

The relation between the equilibrium constant and standard free‑energy is:

$\Delta G^o=-RT \ln k$

where,

$\Delta G^o$ = standard free‑energy change = -4.20 kJ/mole

R = universal gas constant = 8.314 J/mole.K

k = equilibrium constant = ?

T = temperature = $25^oC=273+25=298K$

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

$-4.20kJ/mole=-(8.314J/mole.K)\times (298K) \ln k$

$k=5.45$

Now we have to calculate the concentrations of A, B, and C at equilibrium.

The given equilibrium reaction is,

$A(aq)+B(aq)\rightleftharpoons C(aq)$

Initially 0.30 0.40 0

At equilibrium (0.30-x) (0.40-x) x

The expression of equilibrium constant will be,

$k=\frac{[C]}{[A][B]}$

$5.45=\frac{x}{(0.30-x)\times (0.40-x)}$

By solving the term x, we get

$x=0.168\text{ and }0.716$

From the values of 'x' we conclude that, x = 0.716 can not more than initial concentration. So, the value of 'x' which is equal to 0.716 is not consider.

The value of x will be, 0.168 M

The concentration of $A$ at equilibrium = (0.30-x) = 0.30 - 0.168 = 0.132 M