What is the density of ammonia (NH3) at 273 K and 1.00 atm?

A - 0.7601 g/L

B - 0.982 g/L

C - 1.23 g/L

D - 2.43 g/L

Answer:

The answer is HSO4- is a weak acid which will be C

The acid dissociation constant, ka, of HSO4- indicates that it is a strong acid. Details about acid dissociation constant can be found below.

The acid dissociation constant denoted by Ka is the measure of the strength of an acid.

The dissociation constant of an acid is used to identify strong acids from their weak counterparts. Strong acids have high Ka values while weak acids have low Ka values.

According to this question, the Ka of HSO4- is 1.2 × 10-², which is quite high.

Therefore, the dissociation constant, ka, of HSO4- suggests that it is a strong acid.

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Answer:

a. B

b. C

c. A

d. A

e. B

f. -266

g. B and D

Answer:

ii and iv

Explanation:

atomic mass is the sum of protons and neutrons

protons ( postively charged) usually have the same number like electrons( negatively charged)

Answer:

answer of your question is 4th

Explanation:

P205

Answer:

conjugate base

Explanation:

The following equation is given in this question:

HCI(g) + H2O(l) → H30+(aq) + Cl-(aq)

In accordance to Brønsted–Lowry acid–base theory, an acid is a substance that loses an hydrogen ion (H+) or proton in an aqueous solution. This theory further indicated that the substance formed after an acid loses its hydrogen ion is called a CONJUGATE BASE.

In the above equation, HCl is the acid because it loses H+ while the substance formed after the loss of H+, which is Cl- is a conjugate base.

Answer:

A balanced chemical equation must always include coefficients on every reactant and product.

Explanation:

A balanced chemical equation does not need to include coefficients on every reactant and product.

For example, below is a balanced chemical equation in which the reactants and the products have no coefficients whatsoever:

NaOH(aq) + HCl (aq) -----> NaCl (s) + H2O (l)

Of course, a properly written chemical equation must include the states of matter of all the substances in the reaction and the number of atoms of each element must balance both in the reactant and product sides of the equation. Generally, a balanced chemical equation must obey the law of conservation of matter which opines that matter can neither be created nor destroyed but can only be converted from one form to another.

Hence, that a balanced chemical equation must always include coefficients on every reactant and product is not true.

The concentration of hydronium ions (H₃O⁺) is generally greater than 1 × 10⁻⁷ for all acidic solutions: True.

An acidic solution can be defined as a type of aqeous solution that has a high concentration of hydrogen ions (H⁺), which is typically greater than that of pure water.

As a result of the fact that an acidic solution has a power of hydrogen ions (pH) below 7, the concentration of hydronium ions (H₃O⁺) derived from the protonation of water would be greater than 1 × 10⁻⁷ for all acidic solutions.

In conclusion, we can deduce that a property of all acidic solution is a concentration of hydronium ions (H₃O⁺) that is greater than 1 × 10⁻⁷.

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Answer:

Phosphorus atoms can bond with oxygen atoms to form ester groups. These can bond with carbon atoms, yielding a large number of organic phosphorus chemicals. These are found in many important biological processes

Answer:

Ba

Explanation:

Answer:

40.68 % C

23.73 % N

8.47% H

27.12% O

Explanation:

The mass percent of an element X in a compound is calculated as the molar mass (MM) of X multiplied by the number of atoms of X in the compound, divided into the molecular weight (MW) of the compound, as follows:

mass percent of X = (MM(X) x number of atoms of X)/MW compound x 100

Thus, we first calculate the MW of acetamide (C₂H₅NO) by using the molar mass of the chemical elements C, H, N and O:

MW(C₂H₅NO) = (12 g/mol C x 2) + (1 g/mol H x 5) + 14 g/mol N + 16 g/mol O = 59 g/mol

Now, we can calculate the mass percent of each element (C, H, N, O) in C₂H₅NO:

          % mass C = (12 g/mol x 2)/(59 g/mol) x 100 = 40.68 %

       % mass N = (14 g/mol x 1)/(59 g/mol) x 100 = 23.73 %

       % mass H = (1 g/mol x 5)/(59 g/mol) x 100 = 8.47 %

       % mass O = (16 g/mol x 1)/(59 g/mol) x 100 = 27.12 %

The sum of the mass percents has to be equal to 100%:

40.68 % C + 23.73 % N + 8.47% H + 27.12% O = 100%

Answer:

Mg

Explanation:

Mg forms 2+ ions and mgcl2 has a melting point over 700°C I believe.

Answer:

No

Explanation:

Napthalene cannot conduct electricity

Respuesta:

NaF: Enlace iónico

SO₂: Enlace covalente polar

I₂: Enlace covalente apolar

Explicación:

El tipo de enlace entre 2 elementos depende de la diferencia de electronegatividad.

NaF

ΔEN = |EN(Na)-EN(F)| = |0.9-4.0| = 3.1

El enlace es iónico.

SO₂

ΔEN = |EN(S)-EN(O)| = |2.6-3.5| = 0.9

El enlace es covalente polar.

I₂

ΔEN = |EN(I)-EN(I)| = |2.7-2.7| = 0

El enlace es covalente apolar.

Answer:

C. Enlargement

Explanation:

Answer:

answer is in the picture.

Answer:

Too much steam expands within the can, causing it to explode

Explanation:

Beans contain water, and when water is heated, steam is created. As the steam increases, there's less and less room for it to go. The built up pressure causes the can to explode.

Answer:

This answer is 108g mass of aluminum is produced from 100g of aluminum oxide

Answer:

A supersaturated solution contains more dissolved solute than required for preparing a saturated solution and can be prepared by heating a saturated solution, adding more solute, and then cooling it gently. Excess dissolved solute crystallizes by seeding supersaturated solution with a few crystals of the solute.

Explanation:

Answer:

see explanation

Explanation:

Solutions are prepared from one of three solute sources ...

- solid solute from manufacturer,

-liquid solute from manufacturer,

-stock concentrate of solute for dilution to lower concentrations.

In this problem, sucrose is a stock solid with a formula mass of 342.3 grams/mol. The amount of solid needed can be determined using the formula:

mass of solute needed (grams) = (Molarity needed x Volume needed in Liters x formula weight) / (decimal fraction purity of stock solid)

Molarity needed = 0.3M

Volume needed = 500 ml = 0.500 liters

Formula weight = 342.3 grams/mole

Purity factor (assumed) = 100% = 1.00 (should be posted on stock bottle label) For example, a 95% pure stock solid => purity factor of 0.95.

∴ grams of sucrose needed = (0.3M*)(0.500L)(342.3g/mole)/(1.00) =51.345 grams sucrose.

Measure 51.345 grams of sucrose into mixing vessel and add solvent water up to, but not to exceed 500 ml total volume. Mix until homogeneous.

Answer:

1 Ca(OH)2 + 2 HCl ---> 1 CaCl2 + 2 H2O

Explanation:

When chlorine bonds and becomes stable, (had a full outer electron shell), because we know that chlorine has 7 electrons in it’s outer shell, it requires 1 to gain noble gas structure. So, when if bonds with an element and then becomes stable, the gaining of an electron means it becomes a negatively charged ion. (an ion being a charged particle.)

Answer:

Explanation:

2,5M

Answer:

2,5m

Explanation:

c=n:v

Answer:

23°C + 273.15 = 296.15K

HAVE A GREAT DAY :)

Answer:

1302 nm

Explanation:

Del formulario de Rydberg para longitud de onda

1 / λ = R (1 / nf ^ 2 - 1 / ni ^ 2)

1 / λ = 1.097 × 10 ^ 7 (1/3 ^ 2 - 1/5 ^ 2)

1 / λ = 1.097 × 10 ^ 7 (0.11 - 0.04)

1 / λ = 1.097 × 10 ^ 7 (0.07)

= 1,302 × 10 ^ -6 m

o 1302 nm

Answer:

1s2 2s2 2p6 3s2 3p6

Explanation:

Chlorine is a groups 17 element. The halogens for ions by accepting one electron to form univalent negative ions.

Since chlorine normally contains seventeen electrons, the chloride ion consists of eighteen electrons.

Hence the electronic configuration of chlorine ion is; 1s2 2s2 2p6 3s2 3p6.

Cu2+ is reduced to Cu.

Change in oxidation state is from +2 to 0

Answer:

Option A:

Zn(s) + Cu^(2+) (aq) → Cu(s) + Zn^(2+)(aq)

Explanation:

The half reactions given are:

Zn(s) → Zn^(2+)(aq) + 2e^(-)

Cu^(2+) (aq) + 2e^(-) → Cu(s)

From the given half reactions, we can see that in the first one, Zn undergoes oxidation to produce Zn^(2+).

While in the second half reaction, Cu^(2+) is reduced to Cu.

Thus, for the overall reaction, we will add both half reactions to get;

Zn(s) + Cu^(2+) (aq) + 2e^(-) → Cu(s) + Zn^(2+)(aq) + 2e^(-)

2e^(-) will cancel out to give us;

Zn(s) + Cu^(2+) (aq) → Cu(s) + Zn^(2+)(aq)