You are given mixture made of 290 grams of water and 14.2 grams of salt. Determine the % by mass of salt in the salt solution.

Answer:

1. The coefficients are: 1, 3, 2

2. From the balanced equation, we obtained the following:

3 moles oxygen, O2 reacted.

2 moles of Hydrogen sulfide, H2S reacted.

2 moles of water were produced.

2 moles of sulphur dioxide, SO2 were produced.

Explanation:

1. Determination of the coefficients of the equation.

This is illustrated below:

P2O5(s) + H2O(l) <==> H3PO4(aq)

There are 2 atoms of P on the left side and 1 atom on the right side. It can be balance by putting 2 in front of H3PO4 as shown below:

P2O5(s) + H2O(l) <==> 2H3PO4(aq)

There are 2 atoms of H on the left side and 6 atoms on the right side. It can be balance by putting 3 in front of H2O as shown below:

P2O5(s) + 3H2O(l) <==> 2H3PO4(aq)

Now the equation is balanced.

The coefficients are: 1, 3, 2.

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

2H2S(g) + 3O2(g) => 2H2O(l) + 2SO2(g)

From the balanced equation above,

3 moles of oxygen, O2 reacted with 2 moles of Hydrogen sulfide, H2S to produce 2 moles of water, H2O and 2 moles of sulphur dioxide, SO2.

Final answer:

In the balanced chemical equation provided, 3 moles of oxygen react with 2 moles of hydrogen sulfide to produce 2 moles of water and 2 moles of sulfur dioxide.

Explanation:

When the balanced chemical equation 2H2S(g) + 3O2(g) =2H2O(l) + 2SO2(g) is considered, we can deduce that 3 moles of oxygen and 2 moles of hydrogen sulfide react together in this reaction. The products of this chemical reaction are 2 moles of water and 2 moles of sulfur dioxide. Each of these quantities is directly inferred from the coefficients in front of each compound in the balanced chemical equation.

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Answer:

The specific heat of the alloy is 2.324 J/g°C

Explanation:

Step 1: Data given

Mass of water = 0.3 kg = 300 grams

Temperature of water = 20°C

Mass of alloy = 0.090 kg

Initial temperature of alloy = 55 °C

The final temperature = 25°C

The specific heat of water = 4.184 J/g°C

Step 2: Calculate the specific heat of alloy

Qlost = -Qwater

Qmetal = -Qwater

Q = m*c*ΔT

m(alloy) * c(alloy) * ΔT(alloy) = -m(water)*c(water)*ΔT(water)

⇒ mass of alloy = 90 grams

⇒ c(alloy) = the specific heat of alloy = TO BE DETERMINED

⇒ ΔT(alloy) = The change of temperature = T2 - T1 = 25-55 = -30°C

⇒ mass of water = 300 grams

⇒ c(water) = the specific heat of water = 4.184 J/g°C

⇒ ΔT(water) = The change of temperature = T2 - T1 = 25 - 20 = 5 °C

90 * c(alloy) * -30°C = -300 * 4.184 J/g°C * 5°C

c(alloy) = 2.324 J/g°C

The specific heat of the  alloy is 2.324 J/g°C

Final answer:

To neutralize the KOH solution, we need 61.4 mL of 1.33 mol L−1 H2SO4(aq).

Explanation:

To find the volume of the H2SO4 solution needed to neutralize the KOH solution, we can use the equation:

Mole of H2SO4 = Molarity of KOH x Volume of KOH

First, calculate the moles of KOH:
Moles of KOH = Molarity of KOH x Volume of KOH = 0.830 mol/L x (49.3 mL / 1000 mL) = 0.04089 mol

Since H2SO4 is a diprotic acid and KOH is a strong base, the reaction will be:
H2SO4 + 2 KOH -> K2SO4 + 2 H2O

Therefore, the ratio between the moles of H2SO4 and KOH is 1:2. This means that twice the moles of KOH will be needed to neutralize the H2SO4. Calculate the moles of H2SO4 needed:
Moles of H2SO4 needed = 2 x Moles of KOH

= 2 x 0.04089 mol

= 0.08178 mol

Finally, calculate the volume of the H2SO4 solution needed:
Volume of H2SO4 = Moles of H2SO4 / Molarity of H2SO4 = 0.08178 mol / 1.33 mol/L

= 0.0614 L

= 61.4 mL

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Answer:

Explanation:

Hello,

In this case, considering that 1 m equals 10 dm, the required diameter of the atom in decimetres is:

Best regards.

Final answer:

The diameter of an atom is approximately 1 * 10^-10 * m, which translates to 1 * 10^-11 decimeters by multiplying the meter measurement by 10.

Explanation:

The subject of this question is a unit conversion. We are trying to convert the diameter of an atom from meters to decimeters.

The process is as follows:

• First, we need to understand that 1 meter equals 10 decimeters because the prefix 'deci-' means one-tenth.
• So, if the diameter of an atom is approximately 1 * 10^-10 * m, then in decimeters the diameter would be 1 * 10^-11 * dm

In essence, to convert meters to decimeters, we need to multiply the meter measurement by 10. But because the meter measurement of the atom is in the power of -10, the decimeter equivalent would be in the power of -11.

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Answer:

the equilibrium partial pressure of BrCl is pBC = 784.52 torr

Explanation:

Since

Br₂(g) + Cl₂(g) ⇌ 2BrCl(g) , Kp=1.112 at 150 K

denoting BC as BrCl , B as Br₂ , C as Cl₂, p as partial pressure , then

Kp = pBC²/[pB*pC]

solving for pBC

pBC = √(Kp*pB*pC)

replacing values

pBC = √(Kp*pB*pC) = √(1.112*751 torr*737 torr) = 784.52 torr

pBC = 784.52 torr

then the equilibrium partial pressure of BrCl is pBC = 784.52 torr

Final answer:

To calculate the equilibrium partial pressure of BrCl, use the equilibrium constant expression and substitute the given partial pressures of Br2 and Cl2. The equilibrium partial pressure of BrCl is approximately 0.0375 atm.

Explanation:

To calculate the equilibrium partial pressure of BrCl, we need to use the equilibrium constant expression:

Kp = ([BrCl]^2) / ([Br2] * [Cl2])

Given that the equilibrium partial pressures of Br2 and Cl2 are 0.450 atm and 0.115 atm, respectively, we can substitute these values into the expression:

1.112 = ([BrCl]^2) / (0.450 * 0.115)

Simplifying the expression, we find that the equilibrium partial pressure of BrCl is approximately 0.0375 atm.

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