What is the answer to 1 and 2?

What are the states of matter for the following equation?

White raven [17]

Answer:

Explanation:

On the reactant side, XeO3 is a strong oxidizing agent. It can be a solid or gas. Most likely it is in a solid state.

H+ and Br- combine to form HBr which is a weak acid. So H+ and Br- are in an aqueous state.

On the product side, both Br2 and Xe are not very soluble in water. They will be in gaseous state and H2O is water i.e. liquid state.

7 0

2 years ago

A solution is prepared by dissolving 0.5636 g oxalic acid (H2C2O4) in enough water to make 100.0 mL of solution. A 10.00-mL aliq

zepelin [54]

Answer:

The final molarity of the diluted oxalic acid solution is 0.002504 M

Explanation:

Step 1: Data given

Mass of oxalic acid (H2C2O4) = 0.5636 grams

Volume of the solution = 100.0 mL = 0.100 L

A 10 ml of this solution diluted in 250 ml of solution.

Molecular weight of H2C2O4 = 90.03 g/mol

Step 2: Calculate initial moles of H2C2O4

Moles H2C2O4 = mass / molar mass

Moles H2C2O4 = 0.5636 grams / 90.03 g/mol

Moles H2C2O4 = 0.00626 moles

Step 3: Calculate molarity of the solution

Molarity = moles / volume

Molarity = 0.00626 moles / 0.100 L

Molarity = 0.0626 M

Step 4: Calculate moles of a 10.00 mL aliquot

Moles = 0.0626 M * 0.010 L

Moles = 0.000626 moles

Step 5: Calculate the new molarity

Molarity = 0.000626 moles / 0.250 L

Molarity = 0.002504 M

The final molarity of the diluted oxalic acid solution is 0.002504 M

6 0

3 years ago

The equilibrium constant, Kc, is calculated using molar concentrations. For gaseous reactions another form of the equilibrium co

Anarel [89]

Answer:

Kp = 1.41 x 10⁻⁶

Explanation:

We have the chemical equation:

2 A(g) + 3 B(g)⇌ C(g)

In which A and B are the reactants and C is the product. We calculate first the change in the number of moles of gas (Δn or dn):

dn= (sum moles products - sum moles reactants)

= (moles C - (moles A + moles B))

= (1 - (2+3))

= 1 - 5

= -4

We have also the following data:

Kc = 63.2

T= 81∘C + 273 = 354 K

R = 0.082 L.atm/K.mol (it is a constant)

Thus, we introduce the data in the mathematical expression for the relation between Kp and Kc:

$Kc = (RT)^{dn}$ = (0.082 L.atm/K.mol x 354 K)⁻⁴ = 1.41 x 10⁻⁶

3 0

3 years ago

An ideal gas occupies a volume V at an absolute temperature T. If the volume is halved and the pressure kept constant, what will

Kruka [31]

Answer:

It will be halve of T

Explanation:

V1 = V

T1 = T

V2 = ½V

T2 = x

V1/T1 = V2/T2

V/T = ½V/x

Vx = ½VT

2Vx = VT

2x = T

x = ½T

6 0

3 years ago

A 25.0 g piece of aluminum (which has a molar heat capacity of 24.03 J/mol°C) is heated to 86.4°C and dropped into a calorimeter

Ira Lisetskai [31]

Answer:

m H2O = 56 g

Explanation:

Q = mCΔT

∴ The heat ceded (-) by the Aluminum part is equal to the heat received (+) by the water:

⇒ - (mCΔT)Al = (mCΔT)H2O

∴ m Al = 25.0 g

∴ Mw Al = 26.981 g/mol

⇒ n Al = (25.0g)×(mol/26.981gAl) = 0.927 mol Al

⇒ Q Al = - (0.927 mol)(24.03 J/mol°C)(26.8 - 86.4)°C

⇒ Q Al = 1327.64 J

∴ mH2O = Q Al / ( C×ΔT) = 1327.64 J / (4.18 J/g.°C)(26.8 - 21.1)°C

⇒ mH2O = 55.722 g ≅ 56 g

5 0

3 years ago