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2 years ago

Coach Scott hits a baseball with his bat to help his players with fielding

Chemistry

1 answer:

dangina [55]2 years ago

8 0

Answer:

The ball exerts an equal and opposite force on the bat. This is the reaction force. Such an interaction pair is another example of Newton's Third Law. The baseball forces the bat in one direction and the bat forces the ball in the opposite direction.

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Carbon tetrachloride, CCl4, once used as a cleaning fluid and as a fire extinguisher, is produced by heating methane and chlorin

Elena-2011 [213]

Answer:

1.882 g

Explanation:

Data Given

mass of Cl₂ = 33.4 g

mass of CH₄ = ?

Reaction Given:

CH₄+ 4Cl₂ --------→ CCl₄ + HCl

Solution:

First find the mass of CH₄ from the reaction that it combine with how many grams of Chlorin.

Look at the balanced reaction

CH₄ + 4Cl₂ --------→ CCl₄ + 4HCl

1 mol 4 mol

So 1 mole of CH₄ combine with 4 moles of Cl₂

Now

convert the moles into mass for which we have to know molar mass of CH₄ and Cl₂

Molar mass of Cl₂ = 2 (35.5)

Molar mass of Cl₂ = 71 g/mol

mass of Cl₂

mass in grams = no. of moles x molar mass

mass of Cl₂ = 4 mol x 71 g/mol

mass of Cl₂ = 284 g

Molar mass of CH₄= 12+ 4(1)

Molar mass of CH₄= 16 g/mol

mass of CH₄

mass in grams = no. of moles x molar mass

mass of CH₄= 1 mol x 16 g/mol

mass of CH₄ = 16 g

So,

284 g of Cl₂ combine with 16 g of methane ( CH₄ ) then how many grams of CH₄ is needed to combine with 33.4 g of Cl₂

Apply unity Formula

284 g of Cl₂ ≅ 16 g of methane ( CH₄ )

33.4 g of Cl₂ ≅ X g of methane ( CH₄ )

By cross multiplication

X g of methane ( CH₄ ) = 16 g x 33.4 g / 284 g

X g of methane ( CH₄ ) = 1.88 g

1.882 g of methane (CH₄) will needed to combine with 33.4 g of Cl₂

methane (CH₄) = 1.882 g

5 0

3 years ago

How many moles of HCl are present in 40.0 mL of a 0.035 M solution? 1. 0.0014 mol 2. 0.0060 mol 3. 0.25 mol?

Vinil7 [7]

40.0mL(1 L/1000 mL) = 0.040 L

then plug into the formula M = moles/liters 

0.035 M = moles/ 0.040L 

multipy both sides by 0.040L, and you get 0.0014 moles 

so the answer is 1

4 0

3 years ago

The product formation was decreased when a substance b was added to an enzyme reaction. more substrate being added did not incre

borishaifa [10]

When the product formation is decreased if a substance B is added to an enzyme reaction and more substrate being added would not increase the amount of produce formed, then we assume that substance b could be a noncompetitive inhibitor. This type of inhibitor would be one that would bind to the enzyme with or without the presence of a substrate in different sites at the same time. It would change the conformation of the enzyme and also the active sites. As a result, the substrate would not be able to bind to the enzyme more effectively than the usual. The overall efficiency would decrease.

7 0

3 years ago

When light rays appear to be coming from behind a mirror, what type of image has been formed?

Setler79 [48]

Plane is the right answer I thinkkkkk

4 0

2 years ago

Read 2 more answers

For the following electron-transfer reaction:

creativ13 [48]

Answer:

1. The oxidation half-reaction is: Mn(s) ⇄ Mn²⁺(aq) + 2e⁻

2. The reduction half-reaction is: Ag⁺(aq) + 1e⁻ ⇄ Ag(s)

Explanation:

Main reaction: 2Ag⁺(aq) + Mn(s) ⇄ 2Ag(s) + Mn²⁺(aq)

In the oxidation half reaction, the oxidation number increases:

Mn changes from 0, in the ground state to Mn²⁺.

The reduction half reaction occurs where the element decrease the oxidation number, because it is gaining electrons.

Silver changes from Ag⁺ to Ag.

1. The oxidation half-reaction is: Mn(s) ⇄ Mn²⁺(aq) + 2e⁻

2. The reduction half-reaction is: Ag⁺(aq) + 1e⁻ ⇄ Ag(s)

To balance the hole reaction, we need to multiply by 2, the second half reaction:

Mn(s) ⇄ Mn²⁺(aq) + 2e⁻

(Ag⁺(aq) + 1e⁻ ⇄ Ag(s)) . 2

2Ag⁺(aq) + 2e⁻ ⇄ 2Ag(s)

Now we sum, and we can cancel the electrons:

2Ag⁺(aq) + Mn(s) + 2e⁻ ⇄ 2Ag(s) + Mn²⁺(aq) + 2e⁻

4 0

3 years ago