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skad [1K]
3 years ago
15

Which of the following correctly applies to a catalyst? a. They do not actually participate in the chemical reaction and therefo

re are unchanged at the end of the reaction. b. They provide an alternate lower energy mechanism by which the reaction proceeds. c. They must be in the same phase as the reactants in the chemical reaction. d. Biological catalysts are proteins called substrates.
Chemistry
1 answer:
lawyer [7]3 years ago
6 0

Answer: option B) They provide an alternate lower energy mechanism by which the reaction proceeds

Explanation:

Catalyst are involved in chemical reaction increasing the rate they occur by lowering the activation energy found in the bound- reactants complex.

The reaction moves forward because the lower energy mechanism when catalyst are added allows for a more easily splitted bond, thus, allowing the formation of products.

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When was the idea of a atom first devloped
bija089 [108]

Answer:

Around 450 B.C.

Explanation:

The idea was forgotten until the 1800 when John Dalton re-introduced the atom.

6 0
2 years ago
1. Which statement describes the properties of both elements that have just one valence electron and elements that have seven va
FromTheMoon [43]
Q1. They are highly reactive. Q2. High reactivity, nonmetallic. Q3. Oxygen has an ion charge of -2. Q4. LiCl I believe. Q5. How electrons are shared. Q6 1. Q7. Share 2 valence electrons, I believe.
3 0
3 years ago
Read 2 more answers
How many liters of hydrogen gas will be produced at STP from the reaction of 7.179×10^23 atoms of magnesium with 54.219g of phos
Alexeev081 [22]

Answer: The volume of hydrogen gas produced will be, 12.4 L

Explanation : Given,

Mass of H_3PO_4 = 54.219 g

Number of atoms of Mg = 7.179\times 10^{23}

Molar mass of H_3PO_4 = 98 g/mol

First we have to calculate the moles of H_3PO_4 and Mg.

\text{Moles of }H_3PO_4=\frac{\text{Given mass }H_3PO_4}{\text{Molar mass }H_3PO_4}

\text{Moles of }H_3PO_4=\frac{54.219g}{98g/mol}=0.553mol

and,

\text{Moles of }Mg=\frac{7.179\times 10^{23}}{6.022\times 10^{23}}=1.19mol

Now we have to calculate the limiting and excess reagent.

The balanced chemical equation is:

3Mg+2H_3PO_4\rightarrow Mg(PO_4)_2+3H_2

From the balanced reaction we conclude that

As, 3 mole of Mg react with 2 mole of H_3PO_4

So, 0.553 moles of Mg react with \frac{2}{3}\times 0.553=0.369 moles of H_3PO_4

From this we conclude that, H_3PO_4 is an excess reagent because the given moles are greater than the required moles and Mg is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of H_2

From the reaction, we conclude that

As, 3 mole of Mg react to give 3 mole of H_2

So, 0.553 mole of Mg react to give 0.553 mole of H_2

Now we have to calculate the volume of H_2  gas at STP.

As we know that, 1 mole of substance occupies 22.4 L volume of gas.

As, 1 mole of hydrogen gas occupies 22.4 L volume of hydrogen gas

So, 0.553 mole of hydrogen gas occupies 0.553\times 22.4=12.4L volume of hydrogen gas

Therefore, the volume of hydrogen gas produced will be, 12.4 L

4 0
3 years ago
Find the density of an object that has a mass of 5 kg and a<br> volume of 50 cm3.
iren2701 [21]
  • Mass=5kg=5000g
  • Volume=50cm^3

\\ \rm\longmapsto Density=\dfrac{Mass}{Volume}

\\ \rm\longmapsto Density=\dfrac{5000}{50}

\\ \rm\longmapsto Density=100g/cm^3

4 0
2 years ago
If a lab requires each lab group to have 25ml of a solution and it takes 15 grams of CuNO3 to make 1 liter of solution how many
tino4ka555 [31]

We need to do some general algebra here.

We will find that you need 8.25 grams of CuNO₃ to make enough solution for the 22 labs.

<em>We know that:</em>

  • Each lab group needs 25 ml of solution.
  • it takes 15 g of CuNO₃ to make one L of that solution.
  • There are 22 labs.

Because each lab needs 25 ml of solution, 22 labs will need that amount 22 times, so the <u>total amount of solution needed</u> is:

22*25ml = 550 ml

Now we know that we need 15 grams to make one liter of solution, and:

1 L = 1000ml

Then you need 15g to make 1000ml

and x (we want to find this amount) to make 550ml

Then we can write two equations (not actual equations, as these are different units) like:

x = 550ml

15g = 1000ml

Now we can take the quotient between these two equations:

x/15 g = (550ml/1000ml)

And now we can solve this for x:

x = (550ml/1000ml)*15g = 8.25g

So you need 8.25 grams of CuNO₃ to make enough solution for the 22 labs.

If you want to learn more, you can read:

brainly.com/question/8743486

5 0
3 years ago
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