All categories

3 years ago

Which of the following does not describe a characteristic of carbon?

2 answers:

7 0

The characteristic that is not descriptive of C, would be D. It forms bonds with only itself. It forms bonds with almost all other nonmetals, present, like H,O,N, etc.

BaLLatris [955]3 years ago

3 0

Carbon can bond with other atoms, such a oxygen, to produce carbon monoxide, carbon dioxide etc.

You might be interested in

If volume remains the same while the mass of a substance ________, the density of the substance will______________

victus00 [196]

Black and blank sorry idk

4 0

4 years ago

A certain weak acid, HA, has a Ka value of 2.6×10−7. Calculate the percent ionization of HA in a 0.10 M solution.

Andrej [43]

Answer:

The percent ionization is 0,16%

Explanation:

The percent ionization is defined as the number of ions that exist in a substance.

$PI=\frac{[A-]}{[HA]} x100$

First, we find the [A-] using the ka equation

HA ⇄ $H^{+} + A^{-}$

[H+] = [A-]

$Ka=\frac{[H+][A-]}{[HA]}\\ \\$

since the ionization constant is very small we can assume that the final concentration of [HA] is the same

$Ka=\frac{[H+]^{2} }{[HA]} \\\\$

$[H+]=\sqrt[2]{Ka.[HA]} \\\\$

$[H+] =\sqrt{(2,610^{-7} )(0,1)} = 1,61210^{-4}$

Now we calculate the percent ionization using these values

$PI=\frac{1,61210^{-4} }{0,1} X100$

PI=0,16%

4 0

4 years ago

A radioactive isotope, 14C decays to become 14N. After a time period of about 6,000 years, only about 12.5% of an original sampl

satela [25.4K]

Answer:

2000 years

Explanation:

A radioactive molecule will continuously decay and turn into another molecule. This nature of the radioactive molecule makes them can be used to estimate the age of an object. Half-life is the unit of time needed for radioactive molecules to decay to half of its mass. The formula for the mass remaining will be:

$N(t)= N_{0} (\frac{1}{2})^{\frac{t}{t_{1/2} } }$

Where

N(t)= number of the molecule remains

N0= number of molecule initially

t= time elapsed

t1/2= half time

We have all variable besides the half time, the calculation will be:

$N(t)= N_{0} (\frac{1}{2})^{\frac{t}{t_{1/2} } }$

$0.125= 1 (\frac{1}{2})^{\frac{6000}{t_{1/2} } }$

$(\frac{1}{8})= (\frac{1}{2})^{\frac{6000}{t_{1/2} } }$

$(\frac{1}{2})^3= (\frac{1}{2})^{\frac{6000}{t_{1/2} } }$

3= 6000/ (t1/2)

t1/2= 6000/3= 2000

The half-life is 2000 years

8 0

3 years ago

1. What is the difference between charging by conduction and charging by induction?

telo118 [61]

Answer:a

Explanation:

7 0

3 years ago

One reaction involved in the conversion of iron ore to the metal is FeO(s) + CO(g) → Fe(s) + CO2(g) Use Hess’s Law to calculate

Ugo [173]

Answer:

$\delta H_{rxn} = -66.0 \ kJ/mole$

Explanation:

Given that:

$3FeO_3_{(s)}+CO_{(g)} \to 2Fe_3O_4_{(s)} +CO_{2(g)} \ \ \delta H = -47.0 \ kJ/mole -- equation (1) \\ \\ \\ Fe_2O_3_{(s)} +3CO_{(g)} \to 2FE_{(s)} + 3CO_{2(g)} \ \ \delta H = -25.0 \ kJ/mole -- equation (2) \\ \\ \\ Fe_3O_4_{(s)} + CO_{(g)} \to 3FeO_{(s)} + CO_{2(g)} \ \delta H = 19.0 \ kJ/mole -- equation (3)$

From equation (3) , multiplying (-1) with equation (3) and interchanging reactant with the product side; we have:

$3FeO_{(s)} + CO_{2(g)} \to Fe_3O_4_{(s)} + CO_{(g)} \ \delta H = -19.0 \ kJ/mole -- equation (4)$

Multiplying (2) with equation (4) ; we have:

$6FeO_{(s)} + 2CO_{2(g)} \to 2Fe_3O_4_{(s)} + 2CO_{(g)} \ \delta H = -38.0 \ kJ/mole -- equation (5)$

From equation (1) ; multiplying (-1) with equation (1); we have:

$2Fe_3O_4_{(s)} +CO_{2(g)} \to 3FeO_3_{(s)}+CO_{(g)} \ \ \delta H = 47.0 \ kJ/mole -- equation (6)$

From equation (2); multiplying (3) with equation (2); we have:

$3 Fe_2O_3_{(s)} +9CO_{(g)} \to 6FE_{(s)} + 9CO_{2(g)} \ \ \delta H = -75.0 \ kJ/mole -- equation (7)$

Now; Adding up equation (5), (6) & (7) ; we get:

$6FeO_{(s)} + 2CO_{2(g)} \to 2Fe_3O_4_{(s)} + 2CO_{(g)} \ \delta H = -38.0 \ kJ/mole -- equation (5)$

$2Fe_3O_4_{(s)} +CO_{2(g)} \to 3FeO_3_{(s)}+CO_{(g)} \ \ \delta H = 47.0 \ kJ/mole -- equation (6)$

$3 Fe_2O_3_{(s)} +9CO_{(g)} \to 6FE_{(s)} + 9CO_{2(g)} \ \ \delta H = -75.0 \ kJ/mole -- equation (7)$

$FeO \ \ \ + \ \ \ CO \ \ \to \ \ \ \ Fe_{(s)} + \ \ CO_{2(g)} \ \ \ \delta H = - 66.0 \ kJ/mole$

$\delta H_{rxn} = \delta H_1 + \delta H_2 + \delta H_3$ (According to Hess Law)

$\delta H_{rxn} = (-38.0 + 47.0 + (-75.0)) \ kJ/mole$

$\delta H_{rxn} = -66.0 \ kJ/mole$

8 0

3 years ago