What is the bond character of this molecule? A.) strongly covalent
B.) Positively charged
C.) Strongly ionic
D.) Negatively charged

Answers

Answer 1

Answer: There are two types of atomic bonds - ionic bonds and covalent bonds. They differ in their structure and properties. Covalent bonds consist of pairs of electrons shared by two atoms, and bind the atoms in a fixed orientation. ... This results in a positively charged ion (cation) and negatively charged ion (anion)

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Two identical containers, one red and one yellow, are inflated with different gases at the same volume and pressure. Both containers have an identically sized hole that allows the gas to leak out. It takes four times as long for the yellow container to leak out compared to the red container. If the red container is twice as hot as the yellow container, what is the ratio of the molar masses of the gases (Myellow / Mred)

Answers

Answer:

Explanation:

Here we're dealing with the root mean square velocity of gases. We'll provide the formula in order to calculate the root mean square velocity of a gas:

$v_(rms)=\sqrt{(3RT)/(M)}$

Here:

$R = 8.314 (J)/(K mol)$ is the ideal gas law constant;

$T$ is the absolute temperature in K;

$M$ is the molar mass of a compound in kg/mol.

We know that the gas from the red container is 4 times faster, as it takes 4 times as long for the yellow container to leak out, this means:

$(v_(rms, red))/(v_(rms, yellow)) = 4$

We also know that the temperature of the red container is twice as large:

$(T_(red))/(T_(yellow)) = 2$

Write the ratio of the velocities and substitute the variables:

$(v_(rms, red))/(v_(rms, yellow))=\frac{\sqrt{(3RT_(red))/(M_(red))}}{\sqrt{(3RT_(yellow))/(M_(yellow))}}=4$

Then:

$\frac{\sqrt{(3RT_(red))/(M_(red))}}{\sqrt{(3RT_(yellow))/(M_(yellow))}}=\sqrt{(3RT_(red))/(M_(red))\cdot (M_(yellow))/(3RT_(yellow))}=\sqrt{(T_(red))/(T_(yellow))\cdot (M_(yellow))/(M_(red))}=4$

From here:

$16 = (T_(red))/(T_(yellow))\cdot (M_(yellow))/(M_(red))$

Then:

$(M_(yellow))/(M_(red)) = (16)/((T_(red))/(T_(yellow))) = (16)/(2) = 8$

Final answer:

Considering Graham's Law of Effusion, and given that the temperature in the red container is twice that in yellow, the molar mass of the gas in the yellow container is 16 times that of the gas in the red container.

Explanation:

The question is about comparing the molar masses of the gases based on the rate at which they escape or effuse from two different containers. The key to this problem lies in understanding Graham's Law of Effusion, which states that the rate at which a gas effuses is inversely proportional to the square root of its molar mass.

Firstly, note that it is given that the red container takes 1/4th the time as yellow to effuse completely, meaning the gas in the red container effuses 4 times faster than the gas in the yellow container. Hence, the ratio of rates of effusion is 4:1

It is also given that the temperature in the red container is twice that in the yellow. Given the gases are in the same volume and pressure, by Graham's law, the ratio of molar masses (Myellow / Mred) would be the square of the ratio of their effusion rates, however when different temperatures are considered, it's the square of [ratio of their effusion rates x (Tred / Tyellow)].

So the ratio of the molar mass of the yellow container to the red would be (4*22)2 = 16, implying that the molar mass of the gas in the yellow container is 16 times that of the gas in the red container.

Learn more about Graham's Law here:

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When radioactive uranium decays to produce thorium, it also emits a particle. As seen in the balanced nuclear equation, this particle can BEST be described asA) a helium atom.
B) an alpha particle or a helium atom.
C) a beta particle or a hydrogen nucleus.
D) an alpha particle or a helium nucleus.

Answers

The radioactive uranium decays to produce thorium and it emits an alpha particle or helium atom. Thus, option A is correct.

What is radioactive decay?

Unstable heavy isotopes of elements undergo nuclear decay to produce stable atoms by the emission of charged particle such as alpha or beta particles.

Based on the emitted particle, there are two types of decay process namely alpha decay and beta decay. In alpha decay atoms emits alpha particles which are helium nuclei and the atom losses its mass number by 4 units and atomic number by two units,

In beta decay, electrons are emitted by the atom, where no change occurs in mass number and atomic number increases by one unit. Uranium undergo alpha decay by emitting alpha particle or helium nuclei.

To find more on alpha decay, refer here:

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You’re answer would be D love!

9. The first step in industrial nitric acid production is the catalyzed oxidation of ammonia. Without a catalyst, a different reaction predominates: 4NH3(g) + 3O2(g) ⇔ 2N2(g) + 6H2O(g) When 0.0120 mol gaseous NH3 and 0.0170 mol gaseous O2 are placed in a 1.00 L container at a certain temperature, the N2 concentration at equilibrium is 2.20×10-3 M. Calculate Keq for the reaction at this temperature.

Answers

Answer: The value of $K_(eq)$ is $4.66* 10^(-5)$

Explanation:

We are given:

Initial moles of ammonia = 0.0120 moles

Initial moles of oxygen gas = 0.0170 moles

Volume of the container = 1.00 L

Concentration of a substance is calculated by:

$\text{Concentration}=\frac{\text{Number of moles}}{\text{Volume}}$

So, concentration of ammonia = $(0.0120)/(1.00)=0.0120M$

Concentration of oxygen gas = $(0.0170)/(1.00)=0.0170M$

The given chemical equation follows:

$4NH_3(g)+3O_2(g)\rightleftharpoons 2N_2(g)+6H_2O(g)$

Initial: 0.0120 0.0170

At eqllm: 0.0120-4x 0.0170-3x 2x 6x

We are given:

Equilibrium concentration of nitrogen gas = $2.20* 10^(-3)M=0.00220M$

Evaluating the value of 'x', we get:

$\Rightarrow 2x=0.00220\n\n\Rightarrow x=(0.00220)/(2)=0.00110M$

Now, equilibrium concentration of ammonia = $0.0120-4x=[0.0120-(4* 0.00110)]=0.00760M$

Equilibrium concentration of oxygen gas = $0.0170-3x=[0.0170-(3* 0.00110)]=0.0137M$

Equilibrium concentration of water = $6x=[6* 0.00110]=0.00660M$

The expression of $K_(eq)$ for the above reaction follows:

$K_(eq)=([N_2]^2* [H_2O]^6)/([NH_3]^4* [O_2]^3)$

Putting values in above expression, we get:

$K_(eq)=((0.00220)^2* (0.00660)^6)/((0.00760)^4* (0.0137)^3)\n\nK_(eq)=4.66* 10^(-5)$

Hence, the value of $K_(eq)$ is $4.66* 10^(-5)$

Four beakers containing potassium nitrate dissolved in water are allowed to evaporate to dryness. Beakers 1 through 4 contain 2.3, 1.91, 5.985, and 0.52 g of dry potassium nitrate respectively. How many moles of potassium nitrate were recovered after the water evaporated?

Answers

Explanation:

Molar mass of potassium nitrate will be calculated as follows.

Molar mass $KNO_(3)$ = molar mass of K + molar mass of N + 3 × molar mass of O

= 39.098 g/mol + 14.006 g/mol + 3 × 15.999 g/mol

= 102.102 g/mol

Now, adding the given amount of potassium nitrate present in each beaker as follows.

(2.3 + 1.91 + 5.985 + 0.52) g

= 10.715 g

Therefore, calculate number of moles as follows.

No. of moles = $\frac{mass}{\text{molar mass}}$

= $(10.715 g)/(102.102 g/mol)$

= 0.105 mol

Thus, we can conclude that 0.105 moles of potassium nitrate were recovered after the water evaporated.

A piece of charcoal used for cooking is found at the remains of an ancient campsite. a 0.94 kg sample of carbon from the wood has an activity of 1580 decays per minute. find the age of the charcoal. living material has an activity of 15 decays/minute per gram of carbon present and the half-life of 14c is 5730 y. answer

Answers

Mass of sample of charcoal = 0.94 kg = 0.00094

∴, activity = decay rate / mass = 1580/0.00094
= 1.681 X 10^6 decays per min per gram

Using the half-life formula, we have:
activity of sample / activity of modern carbon = (1/2)^(age / half-life)
∴, Age = half-life x log (base 2) (modern activity / coal activity)
= 5730 x log(base 2)(1.681X10^6/ 15)
= 96115 years.

Answer: Age of the charcoal = 96115 years

Final answer:

Using the radiocarbon dating technique and applying the decay formula, it is calculated that the age of the charcoal from the an ancient campsite is approximately 9,500 years.

Explanation:

The age of the charcoal can be found using the technique of radiocarbon dating, which capitalizes on the process of radioactive decay. The isotope carbon-14 (¹4C) is used in this method as it has a known half-life of 5730 years. The number of decays per minute per gram of carbon in a live organism is known as its activity.

Initially, the activity was given as 15 decays per minute per gram. The present activity of the carbon in the charcoal is provided at 1580 decays per minute for a 0.94 kg or 940 gram sample. Thus, the current activity per gram is 1580/940 equals approximately 1.68 decays per minute per gram.

Given that the half-life of ¹4C is 5730 years, we can apply the formula for calculating the time passed using the rate of decay, which is given as T = (t1/2 / ln(2)) * ln(N0/N), where 'ln' is the natural logarithm, 'N0' is the initial quantity (15 decays/minute per gram), 'N' is the remaining quantity (1.68 decays/minute per gram).

Plugging in the given values, we get T = (5730 / ln(2)) * ln(15/1.68), which gives us approximately 9,500 years. Therefore, the age of the charcoal is around 9,500 years.