How much energy is produced when 13.1 g of tin reacts with 2.715 g of N2 ?
I.
3 Sn + 2 N2-----------Sn3N4 + 632 KJ
Hint change grams to moles first.
1 mole Sn= 119g
1 mole N2= 28 g

Answer:

sending heat waves and vibrations

Answer:

7505.19 J

Explanation:

We'll begin by calculating the change in the temperature of copper. This can be obtained as follow:

Initial temperature (T₁) = 20 °C

Final temperature (T₂) = 875 °C

Change in temperature (ΔT) =?

ΔT = T₂ – T₁

ΔT = 875 – 20

ΔT = 855 °C

Finally, we shall determine the heat required. This can be obtained as follow:

Specific heat capacity (C) = 0.385 J/gºC

Change in temperature (ΔT) = 855 °C

Mass (M) = 22.8 g

Heat (Q) required =?

Q = MCΔT

Q = 22.8 × 0.385 × 855

Q = 7505.19 J

Thus, 7505.19 J of heat energy is required.

Answer:

A

Explanation:

frequency is the product of time and wave length

Answer:

(B) 0.2Hz

Explanation:

took the test and it for sure was not 2.0Hz

Answer:

40.4 kJ

Explanation:

Step 1: Given data

Step 2: Calculate the moles corresponding to 55.0 g of CO₂

The molar mass of CO₂ is 44.01 g/mol.

n = 55.0 g × 1 mol/44.01 g = 1.25 mol

Step 3: Calculate the heat (Q) required to sublimate 1.25 moles of CO₂

We will use the following expression.

Q = n × ΔH°sub

Q = 1.25 mol × 32.3 kJ/mol = 40.4 kJ

[tex]P_{1} = \text{2.50 atm}[/tex]

[tex]T_{1} = 15^{\circ}\text{C + 273 = 288 K}[/tex]

[tex]V_{1} = \text{4.5 L}[/tex]

[tex]P_{2} = \text{0.85 atm}[/tex]

[tex]V_{2} = \text{2.5 L}[/tex]

[tex]T_{2}[/tex]

[tex]\dfrac{P_{1}V_{1}}{T_{1}} = \dfrac{P_{2}V_{2}}{T_{2}}[/tex]

[tex]T_{2} = T_{1} \times \dfrac{P_{2}}{P_{1}} \times \dfrac{V_{2}}{V_{1}}[/tex]

[tex]T_{2} = \text{288 K} \times \dfrac{\text{0.85 atm}}{\text{2.50 atm}} \times \dfrac{\text{2.5 L}}{\text{4.5 L}}[/tex]

[tex]\boxed{T_{2} = \text{54.4 K}}[/tex]

[tex]\\[/tex]

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Answer:

Explanation:

55

Answer:

C2H4I2

Explanation:

Viscosity of a fluid has to do with the internal friction between the internal layers of the fluid.

Molecular weight is found to be related to the viscosity of a fluid even though the relationship may not be strictly linear.

However, the greater the molecular weight of a substance, the greater the viscosity of the material.

Since C2H4I2 has the greatest molecular weight (281.86 g/mol), it is also expected to display the greatest viscosity among all the compounds listed in the question.

Answer:

The three freezing points will all be slightly different. It is given that a water solution has a freezing point of zero degrees Celsius, so water would have a freezing temperature below that. Salt will lower the freezing point, the more that is added.

Explanation:

Answer:

[tex]m_{H_2O}=73.0gH_2O[/tex]

Explanation:

Hello there!

In this case, since the formation of water from hydrogen and oxygen is:

[tex]2H_2+O_2\rightarrow 2H_2O[/tex]

Whereas we find a 2:2 mole ratio of hydrogen to water. In such a way, by using the Avogadro's number, the aforementioned mole ratio and the molar mass of water (18.02 g/mol), we obtain the following grams of water product:

[tex]m_{H_2O}=2.44x10^{24}molec*\frac{1molH_2}{6.022x10^{23}molec}*\frac{2molH_2O}{2molH_2}*\frac{18.02gH_2O}{1molH_2O}\\\\ m_{H_2O}=73.0gH_2O[/tex]

Regards!

Answer:

A. Mass of KMnO₄ = 316.08 g

B. Mass of KMnO₄ = 41.49 g

C. Mass of KMnO₄ = 0.13 g.

Explanation:

A. Determination of the mass of KMnO₄

We'll begin by determining the number of mole of KMnO₄ in the solution. This can be obtained as follow:

Volume = 1 L

Molarity = 2 M

Mole of KMnO₄ =?

Mole = Molarity × Volume

Mole of KMnO₄ = 2 × 1

Mole of KMnO₄ = 2 moles

Finally, we shall determine the mass of KMnO₄. This can be obtained as follow:

Mole of KMnO₄ = 2 moles

Molar mass of KMnO₄ = 158.04 g/mole

Mass of KMnO₄ =?

Mass = mole × molar mass

Mass of KMnO₄ = 2 × 158.04

Mass of KMnO₄ = 316.08 g

B. Determination of the mass of KMnO₄

We'll begin by determining the number of mole of KMnO₄ in the solution. This can be obtained as follow:

Volume = 350 mL = 350/1000 = 0.35 L

Molarity = 0.75 M

Mole of KMnO₄ =?

Mole = Molarity × Volume

Mole of KMnO₄ = 0.75 × 0.35

Mole of KMnO₄ = 0.2625 mole

Finally, we shall determine the mass of KMnO₄. This can be obtained as follow:

Mole of KMnO₄ = 0.2625 mole

Molar mass of KMnO₄ = 158.04 g/mole

Mass of KMnO₄ =?

Mass = mole × molar mass

Mass of KMnO₄ = 0.2625 × 158.04

Mass of KMnO₄ = 41.49 g

C. Determination of the mass of KMnO₄

We'll begin by determining the number of mole of KMnO₄ in the solution. This can be obtained as follow:

Volume = 80 mL = 80/1000 = 0.08 L

Molarity = 0.01 M

Mole of KMnO₄ =?

Mole = Molarity × Volume

Mole of KMnO₄ = 0.01 × 0.08

Mole of KMnO₄ = 0.0008 mole

Finally, we shall determine the mass of KMnO₄. This can be obtained as follow:

Mole of KMnO₄ = 0.0008 mole

Molar mass of KMnO₄ = 158.04 g/mole

Mass of KMnO₄ =?

Mass = mole × molar mass

Mass of KMnO₄ = 0.0008 × 158.04

Mass of KMnO₄ = 0.13 g

Answer:

D. length × width

Explanation:

eg. 30 cm(length) × 20cm(width) = 600cm²

Answer:

the value of H° is below -6535 kj. +6H2O

Explanation:

6H2O answer solved

For the given reaction, 2 moles of C₆H₆ the heat energy released is - 6535 KJ. Then, for 16 g of the compound or 0.205 moles needs 669.83 KJ of heat released in combustion.

Combustion is a chemical reaction that occurs between a fuel and an oxidizing agent, typically oxygen, resulting in the release of heat, light, and various combustion products, such as carbon dioxide and water vapor.

The process of combustion involves a rapid and exothermic (heat-releasing) oxidation reaction that produces a flame, which is visible in many cases.

Here, 2 moles of the hydrocarbon releases - 6535 KJ of energy.

molar mass of C₆H₆ = 78 g/mol

then no.of moles in 16 g = 16 /78 = 0.205 moles.

then energy released by 0.205 moles  = 0.205 moles × 6535 KJ /2 moles = 669.83 kJ

Therefore, the heat energy released by 16 g of the compound in combustion is 669.83 kJ.

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Answer:

potentiol or kenetic

Explanation:

Answer:

The electrons will be added by the hydrogen.

Explanation:

If we fire a single electron towards the hydrogen atom, the hydrogen atoms added the electron to its shell by applying force of attraction and becomes stable as well as non reactive in nature because the hydrogen attains the electronic configuration of helium which is a noble gas and have completed its outermost shell. The proton that is present in the nucleus attracts this electron and compel it to add in the electron.

Answer:

180 amu

C₆H₁₂O₆

Explanation:

Step 1: Determine the molecular mass of the compound

The sample has a mass (m) of 3.06 g and it contains (n) 0.0170 moles. The molar mass M is:

M = m/n = 3.06/0.0170 mol = 180 g/mol

Then, the molecular mass is 180 amu.

Step 2: Determine the molar mass of the empirical formula.

M(CH₂O) = 1 × M(C) + 2 × M(H) + 1 × M(O)

M(CH₂O) = 1 × 12 g/mol + 2 × 1 g/mol + 1 × 16 g/mol = 30 g/mol

Step 3: Determine the molecular formula

First, we will determine "n" according to the following expression.

n = molar mass molecular formula / molar mass empirical formula

n = 180 g/mol / 30 g/mol = 6

The molecular formula is:

n × CH₂O = 6 × CH₂O = C₆H₁₂O₆

Answer:

d. 4 Al(s) + 3 O₂(g) → 2 Al₂O₃(s)

Explanation:

Aluminum metal reacts with oxygen gas in a combination reaction that forms a product that coats the metal preventing it from further oxidation: aluminum oxide. Aluminum is a cation with charge 3+ (Al³⁻) and oxide is an anion with charge 2- (O²⁻). Thus, the neutral compound aluminum oxide has the chemical formula Al₂O₃. The unbalanced chemical equation is:

Al(s) + O₂(g) → Al₂O₃(s)

We can balance using the trial and error method. First, we will balance O atoms by multiplying Al₂O₃ by 2 and O₂ by 3.

Al(s) + 3 O₂(g) → 2 Al₂O₃(s)

Finally, we get the balanced equation by multiplying Al by 4.

4 Al(s) + 3 O₂(g) → 2 Al₂O₃(s)

Answer:

7.16 × 10⁻³ M

Explanation:

Let's consider the reduction reaction of copper during the electroplating.

Cu²⁺(aq) + 2 e⁻ ⇒ Cu(s)

We can calculate the moles of Cu²⁺ present in the solution using the following relations.

The moles of Cu²⁺ reduced are:

[tex]2.30 min \times \frac{60s}{1min} \times \frac{3.00C}{s} \times \frac{1mole^{-} }{96486C} \times \frac{1molCu^{2+} }{2mole^{-} } = 2.15 \times 10^{-3} molCu^{2+}[/tex]

[tex]2.15 \times 10^{-3} moles[/tex] of Cu²⁺ are in 0.300 L of solution.

[Cu²⁺] = 2.15 × 10⁻³ mol/0.300 L = 7.16 × 10⁻³ M

The element nitrogen exists as a gas  and is  obtained from air on a large scale​ by fractional distillation of air.

An element is defined as a substance which cannot be broken down further into any other substance. Each element is made up of its own type of atom. Due to this reason all elements are different from one another.

Elements can be classified as metals and non-metals. Metals are shiny and conduct electricity and are all solids at room temperature except mercury. Non-metals do not conduct electricity and are mostly gases at room temperature except carbon and sulfur.

The number of protons in the nucleus is the defining property of an element and is related to the atomic number.All atoms with same atomic number are atoms of same element.

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Answer:

The equation: (NH₄)₂SO₄ = 2NH4(+) + SO4(-2)

The number of moles = 5 g / 132.14 g/mol = 0.038 mol

The number of molecules = 0.038 X 6.022x10^23 = 2.29x10^23

the number of positive ions present in the ammonium sulphate solution:

2 positive ions for every 1 molecule of (NH₄)₂SO₄

so 2 x 2.29x10^23 = 4.58x10^23

the number of negative ions present in the ammonium sulphate solution

1 negative ion for every 1 molecule of (NH₄)₂SO₄

so 1 x 2.29x10^23 = 2.29x10^23

the total number of ions present in the ammonium sulphate solution​

4.58x10^23 + 2.29x10^23 = 6.87x10^23

my answer

Explanation:

Answer. Filtration is the method of separating a solid from a liquid. A sieve sets a threshold calibration through which all undersized materials pass through. Filtration differs from sieving, where separation occurs at a single perforated layer (a sieve).

Answer: hello your question is poorly written below is the complete question

answer:

For N1 :  sp³  orbital

For N2:  p orbital

For N3 : p orbital

For N4 : sp² orbital

For N5 : sp² orbital

Explanation:

Determining the type of orbital in which the lone pair on each N atom will reside.

From the configuration attached below we can determine the type of orbital and they are ;

For N1 :  sp³  orbital

For N2:  p orbital

For N3 : p orbital

For N4 : sp² orbital

For N5 : sp² orbital

Answer:

36704 $

Explanation:

First we calculate how much the worker gets paid in one week:

Then we calculate how many weeks does the worker work in 1.7 years:

Finally we calculate how much does the worker make in 1.7 years:

Answer:

drought

Explanation:

droughts are long periods without water

Answer:

The more concentrated acetic acid buffer has a better buffer capacity because requires more moles of acid or base to change the pH than a more diluted acetic acid buffer.

Explanation:

Buffer capacity is defined as the moles of an acid or base that are needed to change the pH of a buffer in 1 unit.

A more concentrated solution of acetic buffer contains more moles of the acid per liter of solution. A solution that contains more moles of the acetic ion or the acetic acid requires more moles of base or acid to change the pH, that means:

The more concentrated acetic acid buffer has a better buffer capacity because requires more moles of acid or base to change the pH than a more diluted acetic acid buffer.

Answer:

d . The volume of the solvent used was less than 5 liters.

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to calculate the volume of the stock (initial) solution by using the following equation:

[tex]M_1V_1=M_2V_2[/tex]

Thus, we solve for, V1, which stands for the aforementioned volume of stock solution:

[tex]V_1=\frac{M_2V_2}{M_1}[/tex]

Then, we plug in to obtain:

[tex]V_1=\frac{5L*1M}{10M}\\\\V_1=0.5L[/tex]

Now, since the final volume was 5 L, we can infer that the volume of solvent is 4.5 L and that of the stock solution 0.5 L for a total of 5 L of diluted solution; therefore, the correct reasoning is d . The volume of the solvent used was less than 5 liters.

Regards!

Answer:

The volume of the solvent is less than 5

Explanation: