What kind of bond would a metal and a nonmetal typically make?

If earth did not rotate what would happen to the hlobal wind

Readme [11.4K]

It would probably stop moving. Earth has motion and we do do. without the world moving, there would be No wind at all.

8 0

3 years ago

quizlet a basement is successfully sealed so that more radon may not enter the space, but 5.7×107 radon atoms are already trappe

12345 [234]

After 25 days, it remains radon 5.9x10^5 atoms.

Half-life is the time required for a quantity (in this example number of radioactive radon) to reduce to half its initial value.

N(Ra) = 5.7×10^7; initial number of radon atoms

t1/2(Ra) = 3.8 days; the half-life of the radon is 3.8 days

n = 25 days / 3.8 days

n = 6.58; number of half-lifes of radon

N1(Ra) = N(Ra) x (1/2)^n

N1(Ra) = 5.7×10^7 x (1/2)^6.58

N1(Ra) = 5.9x10^5; number of radon atoms after 25 days

The half-life is independent of initial concentration (size of the sample).

More about half-life: brainly.com/question/1160651

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6 0

2 years ago

Calculate the volume of one mole of a gas at 1.00 atm pressure and 0 °C.

N76 [4]

Answer:

Solution:-

The gas is in the standard temperature and pressure condition i.e. at S.T.P

Therefore,

V

i

=22.4dm

3

V

f

=?

As given that the expansion is isothermal and reversible

∴ΔU=0

Now from first law of thermodynamics,

ΔU=q+w

∵ΔU=0

∴q=–w

Given that the heat is absorbed.

∴q=1000cal

⇒w=−q=−1000cal

Now,

Work done in a reversible isothermal expansion is given by-

w=−nRTln(

V

i

V

f

)

Given:-

T=0℃=273K

n=1 mol

∴1000=−nRTln(

V

i

V

f

)

⇒1000=−1×2.303×2×273×log(

22.4

V

f

)

Explanation:

6 0

2 years ago

Just as one dozen eggs always has 12 eggs in it, one mole of a

zzz [600]

Answer:

$6.022x10^{23}atoms \ Al$

Explanation:

Hello,

In this case, given the described concept regarding the Avogadro's number, we can easily notice that 27.0 g of aluminium foil has 6.022x10²³ atoms as shown below based on the mass-mole-particles relationship:

$27.0gAl*\frac{1molAl}{27.0gAl} *\frac{6.022x10^{23}atoms \ Al}{1molAl} \\\\=6.022x10^{23}atoms \ Al$

Notice this is backed up by the fact that aluminium molar mass if 27.0 g/mol.

Best regards.

8 0

3 years ago

Three kilograms of steam is contained in a horizontal, frictionless piston and the cylinder is heated at a constant pressure of

lakkis [162]

Answer:

Final temperature: 659.8ºC

Expansion work: 3*75=225 kJ

Internal energy change: 275 kJ

Explanation:

First, considering both initial and final states, write the energy balance:

$U_{2}-U_{1}=Q-W$

Q is the only variable known. To determine the work, it is possible to consider the reversible process; the work done on a expansion reversible process may be calculated as:

$dw=Pdv$