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3 years ago

What is the difference between carbon 12 and carbon 14 in terms of abundance, stability, and structure?

1 answer:

Pie3 years ago

7 0

<span>Carbon-12 and carbon-14 are two isotopes of the element carbon. The difference between carbon-12 and carbon-14 is the number of neutrons in each atom. The number given after the atom name (carbon) indicates the number of protons plus neutrons in an atom or ion. Atoms of both isotopes of carbon contain 6 protons.</span>

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A chemist must prepare 900.0mL of sodium hydroxide solution with a pH of 13.90 at 25°C. She will do this in three steps: Fill a

Ierofanga [76]

Answer:

28.58 g of NaOH

Explanation:

The question is incomplete. The missing part is:

<em>"Calculate the mass of sodium hydroxide that the chemist must weigh out in the second step"</em>

To do this, we need to know how much of the base we have to weight to prepare this solution.

First we know that is a sodium hydroxide aqueous solution so, this will dissociate in the ions:

NaOH -------> Na⁺ + OH⁻

As NaOH is a strong base, it will dissociate completely in solution, so, starting with the pH we need to calculate the concentration of OH⁻.

This can be done with the following expression:

14 = pH + pOH

and pOH = -log[OH⁻]

So all we have to do is solve for pOH and then, [OH⁻]. To get the pOH:

pOH = 14 - 13.9 = 0.10

[OH⁻] = 10⁽⁻⁰°¹⁰⁾

[OH⁻] = 0.794 M

Now that we have the concentration, let's calculate the moles that needs to be in the 900 mL:

n = M * V

n = 0.794 * 0.9

n = 0.7146 moles

Finally, to get the mass that need to be weighted, we need to molecular mass of NaOH which is 39.997 g/mol so the mass:

m = 39.997 * 0.7146

3 0

4 years ago

This chemical equation represents a chemical reaction. 2HNO3+Na2SO3-2NaNO3+H2O+SO2

muminat

Answer:

D. Na₂SO₃

Explanation:

The given equation is:

2HNO₃ + Na₂SO₃ → 2NaNO₃ + H₂O + SO₂

The reaction equation is shown above. In chemical reactions, the left hand side is made up of the reactants and the right hand side is made up of the products.

The reactants in the equation are:

HNO₃ and Na₂SO₃

6 0

3 years ago

Read 2 more answers

What are three components that make a complete description of a location?

vodka [1.7K]

Answer:

Distance, direction and symbol.

Explanation:

Distance, direction and symbol are the three components of the map which make a complete description of a location. On the map, these three components are present which provides information about a specific location on the map. Map is a drawing on the paper that shows the geography of the whole world and provides information in detail.

4 0

3 years ago

An element, X, can form a chloride (XCl3) and an iodide (XI3). The chloride can be converted quantitatively into the iodide when

Arlecino [84]

Answer : The chemical symbol for this element is, (La)

Explanation : Given,

Mass of $XCl_3$ = 0.760 g

Mass of $XI_3$ = 1.610 g

The given chemical reaction is:

$2XCl_3+3I_2\rightarrow 2XI_3+3Cl_2$

First we have to calculate the moles of $XCl_3$ and $XI_3$ .

$\text{Moles of }XCl_3=\frac{\text{Mass of }XCl_3}{\text{Molar mass of }XCl_3}$

Molar mass of Cl = 35.5 g/mole

Let the molar mass of element 'X' be, M

$\text{Moles of }XCl_3=\frac{0.760}{[M+3(35.5)]}$

and,

$\text{Moles of }XI_3=\frac{\text{Mass of }XI_3}{\text{Molar mass of }XI_3}$

Molar mass of I = 126.9 g/mole

Let the molar mass of element 'X' be, M

$\text{Moles of }XI_3=\frac{1.610}{[M+3(126.9)]}$

From the balanced chemical reaction we conclude that,

The moles of ratio of $XCl_3$ and $XI_3$ is, 1 : 1

That means,

$\frac{0.760}{[M+3(35.5)]}=\frac{1.610}{[M+3(126.9)]}$

$\frac{0.760}{[M+106.5]}=\frac{1.610}{[M+380.7]}$

$M=138.667g/mol$

From this we conclude that the element (X) is lanthanum (La) that has molecular weight, 138.667 g/mol.

Hence, the chemical symbol for this element is, (La)

6 0

4 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:

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

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

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

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

<u />

$\delta H_{rxn} = \delta H_1 + \delta H_2 + \delta H_3$ (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