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3 years ago

A general term for a chemical which accelerates a reaction without becoming chemically involved is called a

Chemistry

2 answers:

Rudiy273 years ago

6 0

Answer:

D. Catalyst

Explanation:

A catalyst is a substance that speeds up a chemical reaction, but it is not consumed by the reaction. Hence a catalyst can be recovered chemically unchanged at the end of the reaction .

A catalyst lowers the activation energy and hence a large number of reactant particles collide and undergoes reaction. The particles have enough energy since the activation energy which is the minimum energy required for a reaction to occur becomes less.

Nastasia [14]3 years ago

3 0

Answer: Catalyst

Explanation:

The catalyst is a chemical which accelerates a reaction decreasing the activation energy, that is the energy that is necessary for a reaction occur. We can see in a standard graphic that the uncatalyzed reaction has a big curve while the catalyzed reaction has a minor curve. It shows the more facility that a catalyzed reaction has to occur.

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I just want to check to see if the answer my child has is right.

Law Incorporation [45]

Answer:

2= Moles of SO₂ produced from 0.356 moles of PbS = 0.356 mol

3= Mass of hydrogen = 0.45 g

3= Mass of oxygen = 3.616 g

4 = Theoretical yield of P₂O₅ = 14.2 g.

Explanation:

Q2= Given data:

Moles of PbS = 0.356 mol

Moles of PbO = ?

Moles of SO₂ = ?

Solution:

Chemical equation:

2PbS + 3O₂ → 2PbO + 2SO₂

Now we will compare the moles of PbS with PbO and SO₂ from balanced chemical equation.

PbS : PbO

2 : 2

0.356 : 0.356

Moles of PbO produced from 0.356 moles of PbS = 0.356 mol

PbS : SO₂

2 : 2

0.356 : 0.356

Moles of SO₂ produced from 0.356 moles of PbS = 0.356 mol

Q= 3

Given data:

Mass of water = 4.05 g

Mass of hydrogen = ?

Mass of oxygen = ?

Solution:

Chemical equation:

2H₂O → 2H₂ + O₂

Number of moles of water:

Number of moles of water = mass/ molar mass

Number of moles of water = 4.05 g/ 18 g/mol

Number of moles of water = 0.225 mol

Now we compare the moles of water with hydrogen and oxygen.

H₂O : H₂

2 : 2

0.225 : 0.225

Mass of hydrogen:

Mass of hydrogen = moles × molar mass

Mass of hydrogen = 0.225 mol × 2g/mol

Mass of hydrogen = 0.45 g

H₂O : O₂

2 : 1

0.225 : 1/2×0.225 = 0.113 mol

Mass of oxygen:

Mass of oxygen = moles × molar mass

Mass of oxygen = 0.113 mol × 32g/mol

Mass of oxygen = 3.616 g

Q 4

Given data:

Mass of phosphorus = 3.07 g

Mass of oxygen = 6.09 g

Theoretical yield of P₂O₅ = ?

Chemical equation:

4P + 5O₂ → 2P₂O₅

Number of moles of phosphorus = mass/ molar mass

Number of moles of phosphorus = 3.07 g/ 31 g/mol

Number of moles of phosphorus = 0.1 mol

Number of moles of oxygen = mass/ molar mass

Number of moles of oxygen = 6.09 g/ 32 g/mol

Number of moles of oxygen = 0.2 mol

Now we will compare the moles of P₂O₅ with oxygen and phosphorus.

O₂ : P₂O₅

5 : 2

0.2 : 2/5 ×0.2 = 0.08 mol

P : P₂O₅

4 : 2

0.1 : 2/4×0.1 = 0.05 mol

The number of moles of P₂O₅ produced from phosphorus are less that's why phosphorus will be limiting reactant.

Mass of P₂O₅ = moles × molar mass

Mass of P₂O₅ = 0.05 mol × 283.88 g/mol

Mass of P₂O₅ = 14.2 g

Theoretical yield of P₂O₅ = 14.2 g.

3 0

4 years ago

A 0.420 M Ca(OH)2 solution was prepared by dissolving 64.0 grams of Ca(OH)2 in enough water. What is the total volume of the sol

Juliette [100K]

Answer:
2.05 liters

Calculate the molar mass of Ca(OH)2 =74.09g/mol

Calculate how many mol of Ca(OH) are in 64g

n=m/MM
n=64/74.09
n=0.86 mol

Calculate the volume of solution
M=n/V => V=n/M

V= 0.86 mol/ 0.420M
V= 2.05 liters

4 0

4 years ago

Read 2 more answers

CO is a primary pollutant.

Anuta_ua [19.1K]

True

Carbon monoxide is a primary pollutant which no odor results from incomplete combustion of fuel. The man sources are gasoline and burning of biomass.

Depending on the source of emission, pollutants can be classified into two groups that is primary and secondary pollutants.

A primary pollutant is emitted in the atmosphere directly from a source. It can be either natural sch as volcanic eruptions, sandstorms or man-made that is due to industrial and vehicle emissions. Examples of primary pollutants are nitrogen oxides, carbon monoxide and particulate matter.

Secondary pollutant is due to interactions between primary and secondary pollutants. These can be chemical or physical interactions. Examples are photo-chemical oxidants and secondary particulate matter.

Therefore, carbon monoxide CO is a primary pollutant.

7 0

3 years ago

Read 2 more answers

What does the molar mass of an element represent?

zysi [14]

The molar mass of an element represents how many grams of that specific element are in one mole of that specific element

5 0

3 years ago

A sample of CaCO3 (molar mass 100. g) was reported as being 30. percent Ca. Assuming no calcium was present in any impurities, c

natka813 [3]

Answer:

Approximately 75%.

Explanation:

Look up the relative atomic mass of Ca on a modern periodic table:

Ca: 40.078.

There are one mole of Ca atoms in each mole of CaCO₃ formula unit.

The mass of one mole of CaCO₃ is the same as the molar mass of this compound: $\rm 100\; g$ .
The mass of one mole of Ca atoms is (numerically) the same as the relative atomic mass of this element: $\rm 40.078\; g$ .

Calculate the mass ratio of Ca in a pure sample of CaCO₃:

$\displaystyle \frac{m(\mathrm{Ca})}{m\left(\mathrm{CaCO_3}\right)} = \frac{40.078}{100} \approx \frac{2}{5}$ .

Let the mass of the sample be 100 g. This sample of CaCO₃ contains 30% Ca by mass. In that 100 grams of this sample, there would be $\rm 30 \% \times 100\; g = 30\; g$ of Ca atoms. Assuming that the impurity does not contain any Ca. In other words, all these Ca atoms belong to CaCO₃. Apply the ratio $\displaystyle \frac{m(\mathrm{Ca})}{m\left(\mathrm{CaCO_3}\right)} \approx \frac{2}{5}$ :

$\begin{aligned} m\left(\mathrm{CaCO_3}\right) &= m(\mathrm{Ca})\left/\frac{m(\mathrm{Ca})}{m\left(\mathrm{CaCO_3}\right)}\right. \cr &\approx 30\; \rm g \left/ \frac{2}{5}\right. \cr &= 75\; \rm g \end{aligned}$ .

In other words, by these assumptions, 100 grams of this sample would contain 75 grams of CaCO₃. The percentage mass of CaCO₃ in this sample would thus be equal to:

$\displaystyle 100\%\times \frac{m\left(\mathrm{CaCO_3}\right)}{m(\text{sample})} = \frac{75}{100} = 75\%$ .

3 0

3 years ago