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2 years ago

How do catalysts increase reaction rates? *

Chemistry

1 answer:

RSB [31]2 years ago

3 0

Answer:

By forming an activated complex with lower energy.

Explanation:

When catalyst is added to a reaction , it forms an activated catalyst which has lower activation energy . So initiation of reaction requires less energy and reaction becomes fast .

Hence third option is correct.

You might be interested in

Assuming complete dissociation of the solute, how many grams of KNO3 must be added to 275 mL of water to produce a solution that

iragen [17]

Answer:

108.43 grams KNO₃

Explanation:

To solve this problem we use the formula:

ΔT = Kf * b * i

Where

ΔT is the temperature difference (14.5 K)

Kf is the cryoscopic constant (1.86 K·m⁻¹)

b is the molality of the solution (moles KNO₃ per kg of water)

and i is the van't Hoff factor (2 for KNO₃)

We solve for b:

14.5 K = 1.86 K·m⁻¹ * b * 2

b = 3.90 m

Using the given volume of water and its density (aprx. 1 g/mL) we calculate the necessary moles of KNO₃:

275 mL water ≅ 275 g water

275 g /1000 = 0.275 kg

moles KNO₃ = molality * kg water = 3.90 * 0.275

moles KNO₃ = 1.0725 moles KNO₃

Finally we convert KNO₃ moles to grams, using its molecular weight:

1.0725 moles KNO₃ * 101.103 g/mol = 108.43 grams KNO₃

5 0

2 years ago

A student is given the question: "what is the mass of a gold bar that is 7.379 × 10–4 m3 in volume? the density of gold is 19.3

IRINA_888 [86]

The mass of a gold bar = 1.424 x 10⁴ gram

<h3>Further Explanation</h3>

Density is a quantity derived from the mass and volume

density is the ratio of mass per unit volume

With the same mass, the volume of objects that have a high density will be smaller than type

The unit of density can be expressed in g / cm3 or kg / m3

Density formula:

$\large{\boxed{{\bold{\rho~=~\frac{m}{V}}}}$

ρ = density

m = mass

v = volume

A common example is the water density of 1 gr / cm3

The mass itself is often equated with weight, even though it is different. Mass is the amount of matter in the matter while weight is related to the gravitational force

Known variable

volume gold bar 7.379 × 10⁻⁴ m³

a density of 19.3 g/cm³

Asked

the mass of a gold bar

Answer

volume is known to be 7.379 × 10⁻⁴ m³, then we change it first to units of cm³ adjusting to units of density

7.379 × 10⁻⁴ m³ = 7.379 × 10² cm³

then:

mass = volume x density

mass = 7.379 × 10² cm³ x 19.3 g/cm³

mass = 1.424 x 10⁴ gram