What is the acceleration of a jogger who moves in a straight line and increases her speed by 1 mile per hour every 0.5 hours?

A 13 g sample of P4010 contains how many

Alex777 [14]

Answer:

$\large \boxed{5.5 \times 10^{22}\text{ molecules of P$_{2}$O}_{5}}$

Explanation:

You must calculate the moles of P₄O₁₀, convert to moles of P₂O₅, then convert to molecules of P₂O₅.

1. Moles of P₄O₁₀

$\text{Moles of P$_{4}$O}_{10} = \text{13 g P$_{4}$O}_{10} \times \dfrac{\text{1 mol P$_{4}$O}_{10}}{\text{283.89 g P$_{4}$O}_{10}} = \text{0.0458 mol P$_{4}$O}_{10}$

2. Moles of P₂O₅

P₄O₁₀ ⟶ 2P₂O₅

The molar ratio is 2 mol P₂O₅:1 mol P₄O₁₀

$\text{Moles of P$_{2}$O}_{5} = \text{0.0458 mol P$_{4}$O}_{10} \times \dfrac{\text{2 mol P$_{2}$O}_{5}}{\text{1 mol P$_{4}$O}_{10}} = \text{0.0916 mol P$_{2}$O}_{5}$

3. Molecules of P₂O₅

$\text{No. of molecules} = \text{0.0916 mol P$_{2}$O}_{5} \times \dfrac{6.022 \times 10^{23}\text{ molecules P$_{2}$O}_{5}}{\text{1 mol P$_{2}$O}_{5}}\\\\= \mathbf{5.5 \times 10^{22}}\textbf{ molecules P$_{2}$O}_{5}\\\text{There are $\large \boxed{\mathbf{5.5 \times 10^{22}}\textbf{ molecules of P$_{2}$O}_{5}}$}$

6 0

3 years ago

Are enzyme-catalyzed reactions examples of homo- geneous or heterogeneous catalysis? Explain.

Tanzania [10]

Answer:Yes,enzymes are catalyzed reactions

Explanation:Enzymes are protein that speeds up chemical reactions. Enzyme catalyzed reaction are divided into two:

Homogeneous reaction

Heterogeneous reaction.

Homogeneous catalysts occupy the same phase as the reaction mixture, while heterogeneous catalysts occupy a different phase.

Acid catalysis, organometallic catalysis, and enzymatic catalysis are examples of homogeneous catalysis.

Vanadium oxide (V2 O5) is a brown/yellow solid on which the oxygen and sulfur dioxide can adsorb in order to react with each other to form sulfuric acid.

8 0

3 years ago

How many formula units are in 0.96 moles of H2O

never [62]

Answer:

18.02 amu

Explanation:

6 0

2 years ago

QUESTION 1 1.What would happen if producers were removed from the food chain?

gregori [183]

It would cause the whole food chain to collapse.

explanation: primary consumers or herbivores, which feed of producers would directly die off. Higher level consumers would suffer as organisms from lower tropic levels start to die off

5 0

2 years ago

Read 2 more answers

Please help!! ASAP Calculate the pH of the solution after the addition of the following amounts of 0.0574 M HNO3 to a 80.0 ml so

alexira [117]

Answer:

a) 10.457.

b) 9.32.

c) 8.04.

d) 6.58.

e) 4.76.

f) 2.87.

Explanation:

Aziridine is an organic compounds containing the aziridine functional group, a three-membered heterocycle with one amine group (-NH-) and two methylene bridges (-CH2-). The parent compound is aziridine (or ethylene imine), with molecular formula C2H5N.

Aziridine has a basic character.

It has pKa = 8.04.

So, pKb = 14 – 8.04 = 5.96

Kb = 1.1 x 10⁻⁶.

If we denote Aziridine the symbol (Az), It is dissociated in water as:

Az + H₂O → AzH⁺ + OH⁻

a) 0.00 ml of HNO₃:

There is only Az,

[OH⁻] = √(Kb.C)

Kb = 1.1 x 10⁻⁶. & C = 0.0750 M.

[OH⁻] = √(1.1 x 10⁻⁶)(0.075) = 2.867 x 10⁻⁴.

∵ pOH = - log[OH-] = - log (2.867 x 10⁻⁴) = 3.542.

∴ pH = 14 – pOH = 14 – 3.542 = 10.457.

b) 5.27 ml of HNO₃

To solve this point, we compare the no. of millimoles of acid (HNO₃) and the base (Az).

No. of millimoles of Az before addition of HNO₃ = (0.0750 mmol/ml) × (80.0 ml) = 6.00 mmol.

No. of millimoles of HNO₃, H⁺ = (MV) = (0.0574 mmol/ml) × (5.27 ml) = 0.302 mmol.

The no. of millimoles of the base Az (6.0 mmol) is higher than that of the acid HNO₃ (0.302).

This will form a basic buffer in the presence of weak base (Az).

pOH = pKb + log[salt]/[base]

[salt] = no. of millimoles of the limiting reactant HNO₃ / total volume = (0.302) / (85.27) = 3.54 x 10⁻³ M.

[base] = (no. of millimoles of Az – no. of millimoles of HNO₃) / total volume = (6.00 mmol - 0.302 mmol) / (85.27 ml) = 0.0668 M.

pOH = pKb + log[salt]/[base] = 5.96 + log[3.54 x 10⁻³]/[ 0.0668 M] = 4.68.

pH = 14 – pOH = 14 – 4.68 = 9.32.

c) Volume of HNO₃ equal to half the equivalence point volume :

At half equivalence point, the concentration of the salt formed is equal to the concentration of the remaining base (aziridine), [salt] = [base].

pOH = pKb + log[salt]/[base] = 5.96 + log[1.0] = 5.96.

pH = 14 – pOH = 14 – 5.96 = 8.04.

d) 101 ml of HNO₃:

To solve this point, we compare the no. of millimoles of acid (HNO₃) and the base (Az).

No. of millimoles of Az before addition of HNO₃ = (0.0750 mmol/ml) × (80.0 ml) = 6.00 mmol.

No. of millimoles of HNO₃, H⁺ = (MV) = (0.0574 mmol/ml) × (101.0 ml) = 5.7974 mmol.

The no. of millimoles of the base Az (6.0 mmol) is higher than that of the acid HNO₃ (5.7974).

This will form a basic buffer in the presence of weak base (Az).

pOH = pKb + log[salt]/[base]

[salt] = no. of millimoles of the limiting reactant HNO₃ / total volume = (5.7974) / (181.0) = 0.032 M.

[base] = (no. of millimoles of Az – no. of millimoles of HNO₃) / total volume = (6.00 mmol - 5.7974 mmol) / (181.0 ml) = 0.00112 M.

pOH = pKb + log[salt]/[base] = 5.96 + log[0.032]/[ 0.00112] = 7.416.

pH = 14 – pOH = 14 – 7.416 = 6.58.

e) Volume of HNO₃ equal to the equivalence point :

At the equivalence point the no. of millimoles of the base is equal to that of the acid.

Volume of HNO₃ needed for the equivalence point = (6.00 mmol) / (0.0574 mmol/ml) = 104.5 ml

At the equivalence point:

[Az] = 0.

[AzH⁺] = (6.00 mmol) / (80.0 + 104.5 ml) = 0.0325 M.

As Ka is very small, the dissociation of AzH⁺ can be negligible.

Hence, [AzH⁺] at eqm ≈ 0.0325 M.

[H+] = √(Ka.C) = √(10⁻⁸˙⁰⁴ x 0.0325) = 1.72 x 10⁻⁵.

pH = - log[H+] = - log(1.72 x 10⁻⁵) = 4.76.

f) 109 ml of HNO₃:

No. of milli-moles of H⁺ added from HNO₃ = (0.0574 mmol/ml) × (109 ml) = 6.257 mmol.

Which is higher than the no. of millimoles of the base (Az) = 6.0 mmol.

After the addition, [H⁺] = (6.257 - 6.00) / (80.0 + 109 mL) = 0.00136 M.

As Ka is small and due to the common ion effect in the presence of H⁺, the dissociation of Az is negligible.

pH = -log[H⁺] = -log(0.00136) = 2.87.

3 0

3 years ago