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3 years ago

45.0 g of Ca(NO3)2 are used to create a 1.3 M solution. What is the volume of the solution

1 answer:

6 0

As we know that Molarity is given as,

M = moles / V
Solving for V,
V = moles / M ------------------(1)
Also, moles is equal to,
moles = mass / M. mass -------------(2)
puting value of moles from eq. 2 into eq. 1,
V = (mass / M.mass) / M
Putting values,
V = (45 g / 164 g/mol) / 1.3 mol/dm³

V = 0.21 dm³

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How many molecules are in 2.570 moles of H2

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A lot of molecules will be in 2.570 moles of H2

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3 years ago

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Please help I need the answer now. Write a balanced equation for combustion of pentane in plenty supply of air and in limited su

Ratling [72]

Pentane burns in plenty of air: CO₂ and H₂O is produced.

C₅H₁₂ + 8 O₂ → 5 CO₂ + 6 H₂O

Pentane burns in limited amount of air: CO or even C is produced along with H₂O.

2 C₅H₁₂ + 11 O₂ → 10 CO + 12 H₂O

<h3>Explanation</h3>

Pentane is a hydrocarbon. There are five carbon atoms in each of its molecule. Its molecular formula will be C₅H₁₂.

Hydrocarbon fuels burn to produce CO₂ when there's plenty of air.

? C₅H₁₂ + ? O₂ → ? CO₂ + ? H₂O

Among all species in this reaction, C₅H₁₂ has the largest number of atoms per molecule. Assume that the coefficient of C₅H₁₂ is one.

1 C₅H₁₂ + ? O₂ → ? CO₂ + ? H₂O

C₅H₁₂ is the only reactant that contains C atoms. There are 5 C atoms in a C₅H₁₂ molecule. There should be the same number of C atoms in the products.
CO₂ is the only product that contains C atoms. There are one C atom in each CO₂ molecule. 5 C atoms correspond to 5 CO₂ molecules.

1 C₅H₁₂ + ? O₂ → 5 CO₂ + ? H₂O

Similarly, C₅H₁₂ is the only reactant that contains H atoms. H₂O is the only product that contains H atoms. There are 12 H atoms in one C₅H₁₂ molecule, which corresponds to 6 H₂O molecules.

1 C₅H₁₂ + ? O₂ → 5 CO₂ + 6 H₂O

Both CO₂ and H₂O are products that contains O atoms. There are 5 × 2 + 6 × 1 = 16 O atoms in total in 5 CO₂ molecules and 6 H₂O molecules. The 16 O atoms on the product side corresponds to 8 O₂ molecules on the reactant side.

1 C₅H₁₂ + 8 O₂ → 5 CO₂ + 6 H₂O

1 C₅H₁₂ + 8 O₂ → 5 CO₂ + 6 H₂O

All coefficients shall be whole numbers. If there's any fraction in this equation, multiply both sides by the least common multiple of their denominators.

Hydrocarbon fuels burn to produce H₂O and CO when there's a limited supply of air. C (soot) might also be produced. Assuming that only CO is produced. Try to balance the equation using the same method.

1 C₅H₁₂ + 11/2 O₂ → 5 CO + 6 H₂O

2 C₅H₁₂ + 11 O₂ → 10 CO + 12 H₂O

Less O₂ is consumed for each mole of C₅H₁₂.

Consider: What would be the balanced equation when only C is produced?

<h3>Reference</h3>

"Products and effects of combustion", GCSE Chemistry (Single Science), BBC Bitesize.

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3 years ago

An enzyme is discovered that catalyzes the chemical reaction SAD ↔ HAPPY A team of motivated researchers sets out to study the e

Pepsi [2]

Answer:

The $K_{m}$ of a substrate will be "10 μM".

Explanation:

The given values are:

$E_{t} = 20 \ nM$

$[Substract] = 40 \ \mu M$

$K_{cat}=600 \ s^{-1}$

Reaction velocity, $Vo=9.6 \ \mu M s^{-1}$

As we know,

⇒ $Vo=\frac{K_{cat}[E_{t}][S]}{K_{m}+[S]}$

On putting the estimated values, we get

⇒ $9.6=\frac{600\times 20\times 10^{-3}\times 40}{K_{m}+40}$

⇒ $K_{m}+40=\frac{600\times 20\times 10^{-3}\times 40}{9.6}$

⇒ $K_{m}+40=50$

On subtracting "40" from both sides, we get

⇒ $K_{m}+40-40=50-40$

⇒ $K_{m}=10 \ \mu M$

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3 years ago

Two moles of magnesium and five moles of oxygen are placed in a vessel. When magnesium is ignited ca two moles of magnesium and

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Limiting reactant in this experiment would be Magnesium since it will run out first

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#2 The Correct Answer is B

B) Police Siren

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3 years ago