Scientists must be careful not to use inductive reasoning, because it can lead to faulty conclusions true or false?

If an unknown element didplays many extremly strong metallic properties, where is it most likely to be located on the periodic t

Ipatiy [6.2K]

This is late but for anyone else who needs it...It's D. Far left

4 0

3 years ago

Identify three reasons why

BARSIC [14]

The radioactive decay of unstable isotopes continually generates new energy within Earth's crust and mantle, providing the primary source of the heat that drives mantle convection. Plate tectonics can be viewed as the surface expression of mantle convection.

6 0

2 years ago

For the reaction C2H4(g) + H2O(g) --> CH3CH2OH(g)

Dominik [7]

Answer : The value of equilibrium constant for this reaction at 262.0 K is $3.35\times 10^{2}$

Explanation :

As we know that,

$\Delta G^o=\Delta H^o-T\Delta S^o$

where,

$\Delta G^o$ = standard Gibbs free energy = ?

$\Delta H^o$ = standard enthalpy = -45.6 kJ = -45600 J

$\Delta S^o$ = standard entropy = -125.7 J/K

T = temperature of reaction = 262.0 K

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

$\Delta G^o=(-45600J)-(262.0K\times -125.7J/K)$

$\Delta G^o=-12666.6J=-12.7kJ$

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

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

where,

$\Delta G^o$ = standard Gibbs free energy = -12666.6 J

R = gas constant = 8.314 J/K.mol

T = temperature = 262.0 K

K = equilibrium constant = ?

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

$-12666.6J=-(8.314J/K.mol)\times (262.0K)\times \ln k$

$k=3.35\times 10^{2}$

Therefore, the value of equilibrium constant for this reaction at 262.0 K is $3.35\times 10^{2}$

3 0

3 years ago

Legend says that archimedes, in determining whether or not the king's crown was made of pure gold, measured its volume by the wa

amm1812

Mass=volume x density
if we have mass and density we can calculate volume using the formula: volume=mass/density
volume of the displaced water = 600g/19.3g/cm3
volume = 31.09cm3

6 0

3 years ago

In an acid-base titration, a student uses 21.35 mL of 0.150 M NaOH to neutralize 25.00 mL of H2SO4. How many moles of acid are i

GalinKa [24]

Answer: There are 0.006 moles of acid in the flask.

Explanation:

Given: $V_{1}$ = 21.35 mL, $M_{1}$ = 0.150 M