How many grams of the parent isotope are left in the sample after three half lives?

A spectator ion is (Select all that apply.) a. an ionic component of a reactant that is unchanged by the reaction your eye, b. c

DENIUS [597]

Correct Question:

A spectator ion is (Select all that apply.)

- a piece of french fry contaminating the reaction mixture

- an ionic component of a reactant that is unchanged by the reaction

-in this experiment, nitrate ion

- your eye, carefully watching the progress of the reaction

Answer:

- an ionic component of a reactant that is unchanged by the reaction

Explanation:

A spectator ion is an ion that exists as a reactant and a product in a chemical equation. A spectator ion is one that exists in the same form on both the reactant and product sides of a chemical reaction.

Spectator ions are ions that are present in a solution but don't take part in the reaction. When reactants dissociate into ions, some of the ions may combine to form a new compound. The other ions don't take part in this chemical reaction and are therefore called spectator ions.

The correct option is therefore the option;

- an ionic component of a reactant that is unchanged by the reaction

5 0

3 years ago

Which term is described as a long, narrow depression in the ocean floor?

ahrayia [7]

The term which is described as a long, narrow depression in the ocean floor would be ocean trench. They are hemispheric-scale long but narrow topographic depressions of the sea floor. They are also the deepest parts of the ocean floor. Hope this answers the question.

8 0

2 years ago

Read 2 more answers

Methane CH4 (g) reacts with oxygen gas to produce carbon dioxide and water. ____________________________________________________

nordsb [41]

Answer: The chemical equations are written below.

Explanation:

For a: Methane reacts with oxygen gas to produce carbon dioxide and water.

Combustion reaction is defined as the reaction in which a hydrocarbon reacts with oxygen gas to produce carbon dioxide and water

The chemical equation for the combustion of methane follows:

$CH_4(g)+2O_2(g)\rightarrow CO_2(g)+2H_2O(g)$

For b: Butane reacts with oxygen gas to produce carbon dioxide and water.

This is also an example of combustion reaction.

The chemical equation for the combustion of butane follows:

$2C_4H_{10}(g)+13O_2(g)\rightarrow 8CO_2(g)+10H_2O(g)$

For c: An aqueous solution of sulfuric acid reacts with aqueous potassium hydroxide to produce potassium sulfate and water.

When an acid reacts with a base, it leads to the formation of salt and water. This reaction is known as neutralization reaction

The chemical equation for the reaction of potassium hydroxide and sulfuric acid follows:

$H_2SO_4+2KOH\rightarrow K_2SO_4+2H_2O$

Hence, the chemical equations are written above.

4 0

3 years ago

NEED HELP NOW 100 POINTS 5 pictures to answer WILL MARK BRAINLIEST ANSWER!!!!!!!!! on question 9 and 10 you have to write a writ

irga5000 [103]

Ok number one is A number 2 is C number 3 is A number 4 is B hope it helps

3 0

3 years ago

Read 2 more answers

Phosphoric acid is a triprotic acid with the following pKa values:

lisabon 2012 [21]

Answer:

Mass NaH₂PO₄ = 1.920 g

Mass Na₂HPO₄ = 4.827 g

Explanation:

For a buffer solution we know its pH can be calculated from the Henderson-Hasselbach formula:

pH = pKa + log [A⁻]/[HA]

where [A⁻] and [HA] are the concentrations of the weak acid and its conjugate base in the buffer.

We want to prepare a buffer at pH 7.540 so we have chosen salts NaH₂PO₄ and Na₂HPO₄ as the weak acid and conjugate base respectively.

To calculate the mass of these salts we need to compute their ratio in the Henderson- Hasselbach equation .

Now since we are asked to determine the masses of NaH₂PO₄ and Na₂HPO₄ and we know we want to prepare 1.000 L of a 0.05 M phosphate buffer, we can setup a system of 2 equations with two unknowns from the ratio mentioned above:

pH = pKa + log [A⁻]/[HA]

7.540 = 7.198 + log[HPO₄²⁻] / [H₂PO₄ ⁻]

0.342 = log[HPO₄²⁻] / [H₂PO₄ ⁻]

taking inverse log function to both sides of this equation:

2.198 = [HPO₄²⁻] / [H₂PO₄ ⁻]

but this is also equivalent to

2.198 = mol HPO₄²⁻ / mol H₂PO₄⁻ (M = mol/V)

We also know that in 1 liter of 0.05 M phosphate, we have 0.05 total mol HPO₄²⁻ and H₂PO₄⁻ , thus

mol HPO₄²⁻ + mol H₂PO₄⁻ = 0.05 mol

2.198 = mol HPO₄²⁻ / mol H₂PO₄⁻

solving this system of equations calling x = mol HPO₄²⁻ and y = mol H₂PO₄⁻ , we have:

2.198 = x /y ⇒ x = 2.198y

x + y = 0.05

2.198y + y = 0.05

3.198 y = 0.05 ⇒ y = 0.05 / 3.198 = 0.016

x = 0.05 - 0.016 = 0.034

and the masses can be calculated from the molar masses ( 141.96 g/mol Na₂HPO₄ and 119.98 g/mol NaH₂PO₄

mol HPO₄²⁻ = 0.034 mol x 141.96 g/mol = 4.827 g

mol H₂PO₄⁻ = 0.016 mol x 119.98 g/mol = 1.920 g

6 0

3 years ago