What is the oxidation state for bh3

Answer:

[tex]\boxed {\boxed {\sf 0.14 \ mol \ CaCl_2}}[/tex]

Explanation:

We are asked to calculate the number of moles of 15 grams of calcium chloride (CaCl₂).

To convert from grams to moles, we use the molar mass, or the mass of 1 mole of a substance. Molar masses are found on the Periodic Table because they are equivalent to the atomic masses, but the units are grams per mole instead of atomic mass units.

Look up the individual elements in the compound: calcium and chloride.

Notice the chemical formula has a subscript of 2 after Cl or chlorine. There are 2 moles of chlorine in every 1 mole of calcium chloride. We must multiply chlorine's molar mass by 2 before adding calcium's molar mass.

We will convert using dimensional analysis, so we must create a ratio using the molar mass.

[tex]\frac {110.98 \ g \ CaCl_2}{ 1 \ mol \ CaCl_2}[/tex]

We are converting 15 grams of calcium chloride to moles, so we must multiply the ratio by this value.

[tex]15 \ g \ CaCl_2 *\frac {110.98 \ g \ CaCl_2}{ 1 \ mol \ CaCl_2}[/tex]

Flip the ratio so the units of grams of calcium chloride cancel.

[tex]15 \ g \ CaCl_2 *\frac { 1 \ mol \ CaCl_2}{110.98 \ g \ CaCl_2}[/tex]

[tex]15 *\frac { 1 \ mol \ CaCl_2}{110.98}[/tex]

[tex]\frac { 15}{110.98} \ mol \ CaCl_2[/tex]

[tex]0.1351594882\ mol \ CaCl_2[/tex]

The original measurement of grams (15) has 2 significant figures, so our answer must have the same. For the number we calculated, that is the hundredth place. The 5 in the thousandth place tells us to round the 3 up to a 4.

[tex]0.14 \ mol \ CaCl_2[/tex]

15 grams of calcium chloride is approximately 0.14 moles of calcium chloride.

Answer:

6.4×10¯³ g of O₂.

Explanation:

We'll begin by writing the balanced equation for the reaction. This is given below:

CH₄ + 2O₂ —> CO₂ + 2H₂O

Next, we shall determine the masses of CH₄ and O₂ that reacted from the balanced equation. This can be obtained as follow:

Molar mass of CH₄ = 12 + (4×1)

= 12 + 4

= 16 g/mol

Mass of CH₄ from the balanced equation = 1 × 16 = 16 g

Molar mass of O₂ = 2 × 16 = 32 g/mol

Mass of O₂ from the balanced equation = 2 × 32 = 64 g

SUMMARY:

From the balanced equation above,

16 g of CH₄ reacted with 64 g of O₂.

Finally, we shall determine the mass of O₂ needed to react with 1.6×10¯³ g of CH₄. This can be obtained as illustrated below:

From the balanced equation above,

16 g of CH₄ reacted with 64 g of O₂.

Therefore, 1.6×10¯³ g of CH₄ will react with = (1.6×10¯³ × 64) / 16 = 6.4×10¯³ g of O₂

Thus, 6.4×10¯³ g of O₂ is needed for the reaction.

Answer:

Explanation:

1. A solution is made by dissolving 5.84g of NaCl is enough distilled water to a give a final volume of 1.00L. What is the molarity of the solution? a. 0.100 M b. 1.00 M c. 0.0250 M d. 0.400 M 2. A 0.9% NaCl (w/w) solution in water is a. is made by mixing 0.9 moles of NaCl in a 100 moles of water b. made and has the same final volume as 0.9% solution in ethyl alcohol c. a solution that boils at or above 100°C d. All the above (don't choose this one) 3. In an exergonic process, the system a. gains energy b. loses energy c. either gains or loses energy d. no energy change at all

Answer:

[tex]\boxed {\boxed {\sf 0.100 \ M }}[/tex]

Explanation:

Molarity is a measure of concentration in moles per liter.

[tex]molarity = \frac{moles \ of \ solute}{liters \ of \ solution}}[/tex]

The solution has 5.84 grams of sodium chloride or NaCl and a volume of 1.00 liters.

We are given the mass of solute in grams, so we must convert to moles. This requires the molar mass, or the mass of 1 mole of a substance. These values are found on the Periodic Table as the atomic masses, but the units are grams per mole, not atomic mass units.

We have the compound sodium chloride, so look up the molar masses of the individual elements: sodium and chlorine.

The chemical formula (NaCl) contains no subscripts, so there is 1 mole of each element in 1 mole of the compound. Add the 2 molar masses to find the compound's molar mass.

There are 58.4397693 grams of sodium chloride in 1 mole. We will use dimensional analysis and create a ratio using this information.

[tex]\frac {58.4397693 \ g\ \ NaCl} {1 \ mol \ NaCl}[/tex]

We are converting 5.84 grams to moles, so we multiply by that value.

[tex]5.84 \ g \ NaCl *\frac {58.4397693 \ g\ NaCl} {1 \ mol \ NaCl}[/tex]

Flip the ratio. It remains equivalent and the units of grams of sodium chloride cancel.

[tex]5.84 \ g \ NaCl *\frac {1 \ mol \ NaCl}{58.4397693 \ g\ NaCl}[/tex]

[tex]5.84 *\frac {1 \ mol \ NaCl}{58.4397693 }[/tex]

[tex]0.09993194823 \ mol \ NaCl[/tex]

We can use the number of moles we just calculated to find the molarity. Remember there is 1 liter of solution.

[tex]molarity= \frac{moles \ of \ solute}{liters \ of \ solution}[/tex]

[tex]molarity= \frac{ 0.09993194823 \ mol \ NaCl}{1 \ L}[/tex]

[tex]molarity= 0.09993194823 \ mol \ NaCl/L[/tex]

The original measurements of mass and volume have 3 significant figures, so our answer must have the same. For the number we calculated, that is the thousandths place. The 9 in the ten-thousandths place tells us to round the 9 to a 0, but then we must also the next 9 to a 0, and the 0 to a 1.

[tex]molarity \approx 0.100 \ mol \ NaCl/L[/tex]

1 mole per liter is 1 molar or M. We can convert the units.

[tex]molarity \approx 0.100 \ M \ NaCl[/tex]

The molarity of the solution is 0.100 M.

Answer:

5 Br₂ + S₂O₃²⁻ + 5 H₂O ⇒ 10 Br⁻ + 2 SO₄²⁻ + 10 H⁺

Explanation:

We will balance the redox reaction through the ion-electron method.

Step 1: Identify both half-reactions

Reduction: Br₂ ⇒ Br⁻

Oxidation: S₂O₃²⁻ ⇒ SO₄²⁻

Step 2: Perform the mass balance, adding H⁺ and H₂O where appropriate

Br₂ ⇒ 2 Br⁻

5 H₂O + S₂O₃²⁻ ⇒ 2 SO₄²⁻ + 10 H⁺

Step 3: Perform the charge balance, adding electrons where appropriate

2 e⁻ + Br₂ ⇒ 2 Br⁻

5 H₂O + S₂O₃²⁻ ⇒ 2 SO₄²⁻ + 10 H⁺ + 10 e⁻

Step 4: Make the number of electrons gained and lost equal

5 × (2 e⁻ + Br₂ ⇒ 2 Br⁻)

1 × (5 H₂O + S₂O₃²⁻ ⇒ 2 SO₄²⁻ + 10 H⁺ + 10 e⁻)

Step 5: Add both half-reactions

5 Br₂ + S₂O₃²⁻ + 5 H₂O ⇒ 10 Br⁻ + 2 SO₄²⁻ + 10 H⁺

385 x 42.13 x 0.079 = 1281.38395

Answer:

The mole ratio of AgNO3 to Ag2SO4 IS 2:1 .0.657 g Ag2SO4 x 1 mol / 312 g = 0.00211 mol Ag2SO4.

0.00211 mol Ag2SO4 x 2 mol AgNO3 / 1 mol Ag2SO4 = 0.00421 mol AgNO3

0.00421 mol AgNO3 x 1 L / 0.123 mol AgNO3 = 0.0342 L = 34.2 mL of AgNO3 solution.Therefore,34.2ml of 0.123M AgNO3 will be required.

Answer:

endet nach selam nw

4gh7

Answer:

AgNO3 + NaOH = AgOH + NaNO3.

Explanation:

Balancing Strategies: In this reaction, the products are initially NaNO3 + AgOH. However the AgOH would break down into Ag2O and H2O. This would give us NaNO3 + Ag2O + H2O as our products for the overall reaction.

Balancing Strategies: In this reaction, the products are initially NaNO3 + AgOH. However the AgOH would break down into Ag2O and H2O. This would give us NaNO3 + Ag2O + H2O as our products for the overall reaction.However, the equation balanced here is the initial reaction which produces AgOH and NaNO3.

Answer:

495 J

Explanation:

When the hot pot was set on the counter to cool, heat energy was lost from the pot. Note that according to the first law of thermodynamics, heat is neither created nor destroyed.

This implies that, the heat energy lost from the pot must be gained by the surrounding air. Therefore, if 495 J of energy is lost from the pot, then 495 J of energy is gained by the surrounding air.

Answer:

lots more of the carbon 12 than the others

havent calculated it percentage-wise but you can see its very close to 12 meaning it is of far greater abundance that carbon 13 and 14

Explanation:

Answer:

0.1 mol

Explanation:

6/(15*3+15)

0.1 mol moles of (CH3)3NH+ are in 6.0 g of (CH3)3NH+

The mole, symbol mol, exists as the SI base unit of the amount of substance. The quantity amount of substance exists as a measure of how many elementary entities of a provided substance exist in an object or sample.A mole corresponds to the mass of a substance that includes 6.023 x 1023 particles of the substance. The mole exists the SI unit for the amount of a substance. Its symbol stands mol.

The compound trimethylamine, (CH3 )3N, exists as a  weak base when dissolved in water.

A mole exist expressed as 6.02214076 × 1023 of some chemical unit, be it atoms, molecules, ions, or others. The mole exists as a convenient unit to utilize because of the great number of atoms, molecules, or others in any substance.

To find the amount of the substance (CH3)3NH+ to calculate its molar mass:

M((CH3)3NH+) = (12+3)*3 + 14+1 = 60 g/mol

n((CH3)3NH+) = m/M

m((CH3)3NH+) = 6g

Thus,

n((CH3)3NH+) = 6g/60 g/mol = 0.1 mol

Hence,

n((CH3)3NH+) = 0.1 mol

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Answer:

the mas is .291 g

Explanation:

the mass of a object does not change. so when added the substance the beaker. you had the mass of both objects together. you know the mass of the beaker and you know the mass of both. since mass does not change. the beakers mass is still 74.605g. the mass of both objects is 74.896. all you have to do is subtract the mass of the beaker from the total mass. 74.896 - 74.605 equals .291g. so the mass of the unknown substance Is .291g

Answer:

a. 2NO₂ (g) + 2OH⁻ (aq) → NO₃⁻ (aq) + NO₂⁻  (aq) + H₂O (l)

b. i. NO₂⁻ is the oxidizing agent

ii. NO₃⁻ is the reducing agent.

Explanation:

a. Balance the following skeleton reaction

The reaction is

NO₂ (g) → NO₃⁻ (aq) + NO₂⁻ (aq)

The half reactions are

NO₂ (g) → NO₃⁻ (aq)  (1) and

NO₂ (g) → NO₂⁻  (aq) (2)

We balance the number of oxygen atoms in equation(1) by adding one H₂O molecule to the left side.

So, NO₂ (g) + H₂O (l) → NO₃⁻ (aq)

We now add two hydrogen ions 2H⁺ on the right hand side to balance the number of hydrogen atoms

NO₂ (g) + H₂O (l) → NO₃⁻ (aq) + 2H⁺ (aq)

The charge on the left hand side is zero while the total charge on the right hand side is -1 + 2 = +1. To balance the charge on both sides, we add one electron to the right hand side.

So, NO₂ (g) + H₂O (l) → NO₃⁻ (aq) + 2H⁺ (aq) + e⁻  (4)

Since the number of atoms in equation two are balanced, we balance the charge since the charge on the left hand side is zero and that on the right hand side is -1. So, we add one electron to the left hand side.

So, NO₂ (g) + e⁻ → NO₂⁻  (aq) (5)

We now add equation (4) and (5)

So, NO₂ (g) + H₂O (l) → NO₃⁻ (aq) + 2H⁺ (aq) + e⁻  (4)

+  NO₂ (g) + e⁻ → NO₂⁻  (aq) (5)

2NO₂ (g) + H₂O (l) + e⁻ → NO₃⁻ (aq) + NO₂⁻  (aq) + 2H⁺ (aq) + e⁻  (4)

2NO₂ (g) + H₂O (l)  → NO₃⁻ (aq) + NO₂⁻  (aq) + 2H⁺ (aq)  

We now add two hydroxide ions to both sides of the equation.

So, 2NO₂ (g) + H₂O (l) + 2OH⁻ (aq) → NO₃⁻ (aq) + NO₂⁻  (aq) + 2H⁺ (aq) + 2OH⁻ (aq)

The hydrogen ion and the hydroxide ion become a water molecule

2NO₂ (g) + H₂O (l) + 2OH⁻ (aq) → NO₃⁻ (aq) + NO₂⁻  (aq) + 2H₂O (l)

2NO₂ (g) + 2OH⁻ (aq) → NO₃⁻ (aq) + NO₂⁻  (aq) + H₂O (l)

So, the required reaction is

2NO₂ (g) + 2OH⁻ (aq) → NO₃⁻ (aq) + NO₂⁻  (aq) + H₂O (l)

b. Identify the oxidizing agent and reducing agent

Since the oxidation number of oxygen in NO₂ is -2. Since the oxidation number of NO₂ is zero, we let x be the oxidation number of N.

So, x + 2 × (oxidation number of oxygen) = 0

x + 2(-2) = 0

x - 4 = 0

x = 4

Since the oxidation number of oxygen in NO₂⁻ is -1. Since the oxidation number of NO₂⁻ is -1, we let x be the oxidation number of N.

So, x + 2 × (oxidation number of oxygen) = 0

x + 2(-2) = -1

x - 4 = -1

x = 4 - 1

x = 3

Also, the oxidation number of oxygen in NO₃⁻ is -1. Since the oxidation number of NO₃⁻ is -1, we let x be the oxidation number of N.

So, x + 2 × (oxidation number of oxygen) = -1

x + 3(-2) = -1

x - 6 = -1

x = 6 - 1

x = 5

i. The oxidizing agent

The oxidation number of N changes from +4 in NO₂ to +3 in NO₂⁻. So, Nitrogen is reduced and thus  NO₂⁻ is the oxidizing agent

ii. The reducing agent

The oxidation number of N changes from +4 in NO₂ to +5 in NO₃⁻. So, Nitrogen is oxidized and thus and  NO₃⁻ is the reducing agent.

Answer:

1087.84 J

Explanation:

From the question given above, the following data were obtained:

Mass of metal (Mₘ) = 70 g

Temperature of metal (Tₘ) = 80 °C

Mass of water (Mᵥᵥ) = 100 g

Temperature of water (Tᵥᵥ) = 22 °C

Equilibrium temperature (Tₑ) = 24.6 °C

Heat lost by metal (Qₘ) =?

NOTE: Specific heat capacity of water (Cᵥᵥ) = 4.184 J/gºC

Heat lost by metal (Qₘ) = Heat gained by water (Qᵥᵥ)

Qₘ = Qᵥᵥ

Thus, we shall determine the heat gained by water. This can be obtained as follow:

Qᵥᵥ = MᵥᵥCᵥᵥ(Tₑ – Tᵥᵥ)

Qᵥᵥ = 100 × 4.184 (24.6 – 22)

Qᵥᵥ = 418.4 × 2.6

Qᵥᵥ = 1087.84 J

Thus, the heat gained by water is 1087.84 J.

Heat lost by metal (Qₘ) = Heat gained by water (Qᵥᵥ)

Qₘ = Qᵥᵥ

Qᵥᵥ = 1087.84 J

Qₘ = 1087.84 J

Therefore, the heat lost by the metal is 1087.84 J

A 70.0‑g piece of metal at 80.0 °C is placed in 100 g of water at 22.0 °C contained in a calorimeter. After reaching a temperature of 24.6 °C, the heat given up by the metal to the water is -1.08 kJ.

A calorimeter is an object used for calorimetry, or the process of measuring the heat of chemical reactions or physical changes as well as heat capacity.

A 70.0‑g piece of metal at 80.0 °C is placed in 100 g of water at 22.0 °C contained in a calorimeter. The final temperature of the system is 24.6 °C.

Let's use the following expression to calculate the heat absorbed by the water.

Qw = c × m × ΔT

Qw = (4.184 J/g.°C) × 100 g × (24.6 °C - 22.0 °C) = 1.08 kJ

where,

According to the law of conservation of energy, the sum of the heat absorbed by the water and the heat released by the metal (Qm) is zero.

Qw + Qm = 0

Qm = -Qw = -10.8 kJ

A 70.0‑g piece of metal at 80.0 °C is placed in 100 g of water at 22.0 °C contained in a calorimeter. After reaching a temperature of 24.6 °C, the heat given up by the metal to the water is -1.08 kJ.

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Answer:

46.2%

Explanation:

Number of moles benzoic acid reacts = 3.3g/122.12 g/mol = 0.027 moles

Since the reaction is 1:1, 0.027 moles of methyl benzoate is formed.

Hence;

Theoretical yield of methyl benzoate = 0.027 moles × 136.15 g/mol = 3.68 g

% yield = actual yield/theoretical yield × 100

% yield = 1.7 g/3.68 g × 100

% yield = 46.2%

Answer:

ag and au are sure not to react. but hg and sn might or might not

The concentration of CO₃²⁻ ions will be higher

To explain, I want you to imagine H₂CO₃ in water.

we know that it will lose 2 of it's protons, and form 2 ions

The ion which is more stable will have a higher concentration because that ion will refuse to react with anything else, so once something turns into that specific ion, it's not going back... unless there's a more stabler ion possible

In this case, the 2 ions formed are: HCO₃⁻ and CO₃⁽²⁻⁾, drawing the structures of both the ions tells us that both of them have resonance, but the CO₃⁽²⁻⁾ ion has more resonance structures and hence is more stable

The equilibrium constant for the reaction is K = 0.137

We obtain the equilibrium constant considering the following equilibria and their constants:

COO(s) + H₂(g) → Co(s) + H₂O(g)    K₁ = 67

COO(s) + CO(g) → Co(s) + CO₂(g)   K₂ = 490

We write the first reaction in the forward direction because we need H₂(g) in the reactants side:

(1)     COO(s) + H₂(g) → Co(s) + H₂O(g)    K₁ = 67

Then, we write the second reaction in the reverse direction because we need CO₂(g) in the reactants side. Thus, the equilibrium constant for the reaction in the reverse direction is the reciprocal of the constant for the reaction in the forward direction (K₂):

(2)   Co(s) + CO₂(g) → COO(s) + CO(g)   K₂ = 1/490

From the addition of (1) and (2), we obtain:

COO(s) + H₂(g) → Co(s) + H₂O(g)    K₁ = 67

+

Co(s) + CO₂(g) → COO(s) + CO(g)   K₂ = 1/490

-------------------------------------------------

H₂(g) +  CO₂(g) → CO(g) + H₂O(g)

Notice that Co(s) and COO(s) are removed that appear in the same amount at both sides of the chemical equation.

Now, the equilibrium constant K for the reaction that is the sum of other two reactions is calculated as the product of the equilibrium constants, as follows:

K = K₁ x K₂ = 67 x 1/490 = 67/490 = 0.137

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Answer:

we don't understand why humans have only 46 chromosomes

Answer:

46 chromosomes is what we don't understand

Answer:

Fe(NO3)3, Cr(NO3)3, Co(NO3)3

Explanation:

According to the question, no precipitate is observed when HCl was added. This means that we must rule out AgNO3.

Again, the sulphides of Cu^2+, Bi^3+ are soluble in acidic medium but according to the question, the sulphides do not precipitate at low pH hence Cu(NO3)2 and Bi(NO3)3 are both ruled out.

The sulphides of Fe^3+, Cr^3+ and Co^3+ all form precipitate in basic solution hence Fe(NO3)3, Cr(NO3)3, Co(NO3)3 may be present.

The presence of Ca(NO3)2 and KNO3 may be confirmed by flame tests.

Answer:

c

Explanation:

it takes 10,000 years to just reduce down the decay

Answer:

Half life is the time taken by a radio active isotope to reduce by half of its original amount. Radium-226 has a half life of 1602 years meaning that it would take 1602 years for a mass of radium to reduce by half.

Number of half lives in 9612 years = 9612/1602 = 6 half lives

New mass = Original mass x (1/2)n where n is the number of half lives.

Therefore, New mass= 500 x (1/2)∧6

                                 = 500 x 0.015625

                                 = 7.8125 g

Hence the mass of radium after 9612 years will be 7.8125 grams.

Explanation:

Answer:

[tex]\boxed {\boxed {\sf 6.25 \ grams}}[/tex]

Explanation:

We are asked to find the mass of a sample of radium-226 after half-life decay. We will use the following formula:

[tex]A= A_o *\frac{1}{2}^{\frac{t}{h}}[/tex]

In this formula, [tex]A_o[/tex] is the initial amount, t is the time, and h is the half-life.

For this problem, the initial amount is 200 grams of radium-226, the time is 8,000 years, and the half-life is 1,600 years.

[tex]\bullet \ A_o= 200 \ g \\\\bullet \ t= 8,000 \ \\\bullet \ h= 1,600[/tex]

Substitute the values into the formula.

[tex]A= 200 \ g * \frac{1}{2} ^{\frac{8.000}{1,600}[/tex]

Solve the fraction in the exponent.

[tex]A= 200 \ g * \frac{1}{2}^{5}[/tex]

Solve the exponent.

[tex]A= 200 \ g *0.03125[/tex]

[tex]A= 6.25 \ g[/tex]

In addition, we can solve this another way. First, we find the number of half-lives by dividing the total time by the half-life.

Every half-life, 1/2 of the mass decays. Divide the initial mass in half, then that result in half, and so on 5 times.

After 8,000 years, 6.25 grams of radium-226 remains.

Answer:

Explanation:

418.4kj is the correct answer

Thermochemistry has to do with  heat evolved or absorbed in a chemical reactions. Thermochemical equations are equations in which the heat of reaction is included in the reaction equation. The reaction of moles and heat of reaction is important here.

This question has to do with thermochemistry and thermochemical equations.

The answers to each of the questions are shown below;

a) 300.52 KJ

b) 11.39 g

c) 5.78 g

The equation of the thermochemical reaction is;

2C12H26 + 37O2-------> 24CO2 + 15026KJ

Number of moles of CO2 released = 21.3g/44g/mol = 0.48 moles

From the reaction equation;

15026KJ is released when 24 moles of CO2 is released

x KJ is released when  0.48 moles of CO2 is released

x = 15026KJ  * 0.48 moles/24 moles

x = 300.52 KJ

b) If 2 moles of C12H26 released 15026KJ of heat

     x moles of C12H26  released 500.00KJ

x = 2 * 500.00KJ/15026KJ

x = 0.067 moles

Mass of C12H26 consumed =  0.067 moles * 170 g/mol = 11.39 g

c) Heat gained by water = heat released by combustion of kerosene

Heat gained by water = 0.75 Kg * 4200  * (90 -10)

Heat gained by water = 252 KJ

If 2 moles of C12H26  produced 15026KJ

x moles of C12H26  produces 252 KJ

x = 2 * 252/15026

x = 0.034 moles

Mass of C12H26   = 0.034 moles *  170 g/mol = 5.78 g

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Answer:

This question is asking to find the new temperature

The answer for the final temperature is 429.73K

Explanation:

Using Charles law equation as follows:

V1/T1 = V2/T2

Where;

V1 = initial volume (L)

V2 = final volume (L)

T1 = initial temperature (K)

T2 = final temperature (K)

According to this question;

V1 = 3.0L

V2 = 4.4L

T1 = 20°C = 20 +273 = 293K

T2 = ?

Using V1/T1 = V2/T2

3/293 = 4.4/T2

Cross multiply

293 × 4.4 = 3 × T2

1289.2 = 3T2

T2 = 1289.2 ÷ 3

T2 = 429.73K

Answer:

main asteroid belt separates the inner planets from the outer planets

Answer:

Dung dịch nào sau đây chỉ chứa các ion (bỏ qua sự điện li của nước, các chất điện li mạnh phân li hoàn toàn)?

A. HBr, Na2S, Mg(OH)2, Na2CO3.

B. H2SO4, NaOH, NaCl, HF.

C. HNO3, H2SO4, KOH, K2SiO3.

D. Ca(OH)2, KOH, CH3COOH, NaCl.