Platinum is a transition metal and forms Pt+2 and Pt+4 ions. Write the formulas for the compounds for each platinum ion with:

Which element is in the same family as Chlorine (Cl) and Fluorine (F)?

ValentinkaMS [17]

Answer/Explanation:

Chlorine and Fluorine are in the Halogen family. The elements in the Halogen family are:

Fluorine (F)

Chlorine (Cl)

Bromine (Br)

Iodine (I)

Astatine (At)

Tennessine (Ts)

Hydrogen (H) is a nonmetal

Oxygen (O) is a nonmetal

Lithium (Li) is an alkaline metal.

8 0

3 years ago

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12.70 L of a gas has a pressure of 0.63 atm. What is the volume of the gas at 105kPa?

Zielflug [23.3K]

Answer: 7.693 L

Explanation:

To calculate the new volume, we use the equation given by Boyle's law. This law states that pressure is directly proportional to the volume of the gas at constant temperature.

The equation given by this law is:

$P_1V_1=P_2V_2$

where,

$P_1\text{ and }V_1$ are initial pressure and volume.

$P_2\text{ and }V_2$ are final pressure and volume.

We are given:

$P_1=0.63atm\\V_1=12.70L\\P_2=105kPa=1.04atm(1kPa=0.009atm)\\V_2=?$

Putting values in above equation, we get:

$0.63\times 12.70mL=1.04\times V_2\\\\V_2=7.693L$

Thus new volume of the gas is 7.693 L

6 0

4 years ago

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You make an experiment to determine the percent mass of carbon in sugar. You find out that there is 46.63g of carbon in 75.00g o

Aleksandr [31]

Taking into account the definition of percentage composition, the percent composition of carbon in this sample is 62.17%.

<h3>Percentage composition</h3>

The Percentage Composition is a measure of the amount of mass that an element occupies in a compound and indicates the percentage by mass of each element that is part of a compound.

Then, the percentage by mass expresses the concentration and indicates the amount of mass of solute present in 100 grams of solution.

In other words, the percentage by mass of a component of the solution is defined as the ratio of the mass of the solute to the mass of the solution, expressed as a percentage.

The percentage by mass is calculated as the mass of the solute divided by the mass of the solution, the result of which is multiplied by 100 to give a percentage. This is:

$percentageby mass= \frac{mass of solute}{mass of solution}x100$

In this case, you know:

mass of solute= 46.63 g
mass of solution= 75 g

Replacing:

$percentageby mass= \frac{46.63 g}{75 g}x100$

Solving:

percentage by mass= 62.17 %

Finally, the percent composition of carbon in this sample is 62.17%.

Learn more about percent composition:

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3 0

3 years ago

Part A

Elden [556K]

Answer:

ΔG° = -5.4 kJ/mol

ΔG = 873.2 J/mol = 0.873 kJ /mol

Explanation:

Step 1: Data given

ΔG (NO2) = 51.84 kJ/mol

ΔG (N2O4) = 98.28 kJ/mol

Step 2:

ΔG = ΔG° + RT ln Q

⇒with Q = the reaction quatient

⇒with T = the temperature = 298 K

⇒with R = 8.314 J / mol*K

⇒with ΔG° = ΔG° (N2O4) - 2*ΔG°(NO2 )

⇒ ΔG° = 98.28 kJ/mol - 2* 51.84 kJ/mol

⇒ ΔG° = -5.4 kJ/mol

Part B

ΔG = ΔG° =RT ln Q

⇒with G° = -5.4 kj/mol = -5400 j/mol

⇒ with R = 8.314 J/K*mol

⇒with T = 298 K

⇒with Q = p(N2O4)/ [ p(NO2) ]² = 1.63/0.36² = 12.577

ΔG = -5400 + 8.314 * 298 * ln(12.577)

ΔG = -5400 + 8.314 * 298 * 2.532

ΔG = 873.2 J/mol = 0.873 kJ/mol

3 0

4 years ago

An analytical chemist weighs out 0.093g of an unknown monoprotic acid into a 250mL volumetric flask and dilutes to the mark with

Kamila [148]

Answer:

The molar mass of the unknown acid is 89 g/mol

Explanation:

Step 1: The balanced equation

HA(aq) + NaOH(aq) → NaA(aq) + H2O(l)

It takes 1 mole of NaOH to neutralize 1 mole of the triprotic acid. This is called the reaction stoichiometry.

Step 2: Data given

Mass of the acid = 0.093 grams

volume = 250 mL

titrates with 0.16 M NaOH

adds 6.5 mL NaOH

Step 3: Calculate moles of NaOH

We know the concentration and volume of NaOH needed to neutralize the acid.

By determining the moles of NaOH in that volume in liters (95.9mL=0.0959L), the moles of acid in the original sample can be determined from the reaction stoichiometry.

Moles = Molarity * Volume

Moles = 0.16 M * 0.0065 L

Moles = 0.00104 moles NaOH

Step 4: Calculate moles of the unknown acid:

It takes 1 mole of NaOH to neutralize 1 mole of the triprotic acid. This is called the reaction stoichiometry.

For 0.00104 moles NaOH we have 0.00104 moles of HA

Step 5: Calculate the molar mass of the acid

Molar mass Ha = Mass Ha / moles Ha

Molar mass Ha = 0.093 grams / 0.00104 moles

Molar mass Ha = 89.42 g/mol ≈89 g/mol

The molar mass of the unknown acid is 89 g/mol

3 0

3 years ago