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3 years ago

Which periodic table groups have elements that cannot be found unbounded in nature?

2 answers:

7 0

Hello,

The answer is titanium because although it is not found unbound to other elements in nature, Titanium is the ninth most abundant element in the Earth's crust

Hope this Helps
Plz mark me as Brainist or verified answer Thanks!!!!!

miss Akunina [59]3 years ago

3 0

Lanthanides and actinides cannot be found in nature

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If 12.5 grams of the original sample of cesium-137 remained after 90.6 years, what was the mass of the original sample?

myrzilka [38]

Answer:

Mass of original sample = 100 g

Explanation:

Half life of cesium-137 = 30.17 years

$t_{1/2}=\frac {ln\ 2}{k}$

Where, k is rate constant

So,

$k=\frac{\ln2}{t_{1/2}}$

$k=\frac{\ln2}{30.17}\ year^{-1}$

The rate constant, k = 0.02297 year⁻¹

Time = 90.6 years

Using integrated rate law for first order kinetics as:

$[A_t]=[A_0]e^{-kt}$

Where,

$[A_t]$ is the concentration at time t

$[A_0]$ is the initial concentration

Initial concentration $[A_0]$ = ?

Final concentration $[A_t]$ = 12.5 grams

Applying in the above equation, we get that:-

$12.5\ g=[A_0]e^{-0.02297\times 90.6}$

$[A_0]=\frac{12.5}{e^{-0.02297\times 90.6}}\ g=100\ g$

Mass of original sample = 100 g

8 0

3 years ago

Carbon atoms can be identified based on the # of which subatomic particles

34kurt

Answer: protons and neutrons.

The nucleus is made up of 3 subatomic particles that are protons,neutrons and electrons.

General notation of an element is $_{Z}^{A}\textrm{X}$

where, X is the Element, A is the Atomic Mass and Z is the Atomic Number

If we know the number of protons we can easily find out the atomic number of any element because Atomic Number = Number of protons in an element.

And in addition if we know the number of neutrons we can easily find out the atomic mass of an element because

Atomic Mass = (Number of protons) + (Number of neutrons)

If we get to know the atomic number and atomic mass, we can easily tell what element is it by looking from the periodic table.

6 0

2 years ago

PLZZZZZ I WILL MARK BRAINLEST

kenny6666 [7]

Answer:

hold on...

Explanation:

5 0

2 years ago

In a titration, 4.7 g of an acid (HX) requires 32.6 mL of 0.54 M NaOH(aq) for complete reaction. What is the molar mass of the a

katrin2010 [14]

Answer : The molar mass of an acid is 266.985 g/mole

Explanation : Given,

Mass of an acid (HX) = 4.7 g

Volume of NaOH = 32.6 ml = 0.0326 L

Molarity of NaOH = 0.54 M = 0.54 mole/L

First we have to calculate the moles of NaOH.

$\text{Moles of }NaOH=\text{Molarity of }NaOH\times \text{Volume of solution}=0.54mole/L\times 0.0326L=0.017604mole$

Now we have to calculate the moles of an acid.

In the titration, the moles of an acid will be equal to the moles of NaOH.

Moles of an acid = Moles of NaOH = 0.017604 mole

Now we have to calculate the molar mass of and acid.

$\text{Moles of an acid}=\frac{\text{Mass of an acid}}{\text{Molar mass of an acid}}$

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

$0.017604mole=\frac{4.7g}{\text{Molar mass of an acid}}$

$\text{Molar mass of an acid}=266.985g/mole$

Therefore, the molar mass of an acid is 266.985 g/mole

3 0

3 years ago

If 200 ml of 0.15 M propionic acid (PA) is added to 300 ml of 0.02 M NaOH, what is the resulting pH of the solution? Round the a

vodomira [7]

Answer:

pH = 4.543

Explanation:

CH3CH2COOH + H2O ↔ CH3CH2COO- + H3O+
pKa = - Log Ka

∴ Ka = [H3O+][CH3CH2COO-]/[CH3CH2COOH]

∴ pKa = 4.87

⇒ Ka = 1.349 E-5 = [H3O+][CH3CH2COO-]/[CH3CH2COOH]

added 300 mL 0f 0.02 M NaOH:

⇒ C CH3CH2COOH = ((0.200 L)(0.15 M)) - ((0.300 L)(0.02 M))/(0.3 + 0.2)

⇒ C CH3CH2COOH = 0.048 M

⇒ C NaOH = (0.300 L)(0.02 M) / (0.3 +0.2) = 0.012 M

mass balance:

⇒ 0.048 + 0.012 = 0.06 M = [CH3CH2COO-] + [CH3CH2COOH].......(1)

charge balance:

⇒ [H3O+] + [Na+] = [CH3CH2COO-]

∴ [Na+] = 0.02 M

⇒ [CH3CH2COO-] = [H3O+] + 0.02 M.............(2)

(2) in (1):

⇒ [CH3CH2COOH] = 0.06 M - 0.02 M - [H3O+] = 0.04 M - [H3O+]

replacing in Ka:

⇒ 1.349 E-5 = [H3O+][([H3O+] + 0.02) / (0.04 - [H3O+])

⇒ (1.349 E-5)(0.04 - [H3O+]) = [H3O+]² + 0.02[H3O+]

⇒ 5.396 E-7 - 1.349 E-5[H3O+] = [H3O+]² + 0.02[H3O+]

⇒ [H3O+]² + 0.02001[H3O+] - 5.396 E-7 = 0

⇒ [H3O+ ] = 2.867 E-5 M

∴ pH = - Log [H3O+]

⇒ pH = 4.543

3 0

3 years ago