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2 years ago

Newtons second law in your own words.

Chemistry

2 answers:

Misha Larkins [42]2 years ago

8 0

Answer:

behavior of objects for which all existing forces are not balanced.

Explanation:

andre [41]2 years ago

6 0

Answer:

F=ma. Force is equal to mass times acceleration. The accleration of an object depends on the net force.

Explanation:

I know its F=ma but I think this is right I hope it helps

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A first-order reaction is 81omplete in 264s. A. What is the half-life for this reaction? b. How long will it take for the reacti

Arlecino [84]

A first-order reaction is 81omplete in 264s.The half-life for this reaction (i) t 1/2 = =3.465×10 −3 s.to reach 95% Completion = 285 s.

To measure reaction rates, chemists initiate the reaction, measure the concentration of the reactant or product at different times as the reaction progresses,

For a 0-order response, the mathematical expression that may be employed to determine the half of life is: t1/2 = [R]0/2k. For a first-order reaction, the half of-existence is given by: t1/2 = zero.693/ok. For a 2d-order response, the method for the half-life of the response is: 1/okay[R]0

The 1/2-life of a response (t1/2), is the quantity of time needed for a reactant concentration to lower via half of compared to its initial awareness. Its software is used in chemistry and medicine to are expecting the awareness of a substance over time

Half of the lifestyles is the time required for exactly 1/2 of the entities to decay 50%.

Learn more about first order reaction here:-

#SPJ4

5 0

1 year ago

Scientific explanations

Sergio039 [100]

I think that it's cannot include supernatural explanations, or C.

If i got it wrong sorry

6 0

3 years ago

. What is a carbonyl group? Draw the carbonyl groups that are characteristic of aldehydes, ketones, carboxylic acids, and esters

Elan Coil [88]

Answer :

Carbonyl group : It is a functional group composed of a carbon atom that double bonded to oxygen atom. It is represented as $C=O$

Carboxylic group : It is the class of organic compound in which the carboxylic (-COOH) group is attached to a hydrocarbon is known as carboxylic.

The general formula of carboxylic is, $C_nH_{2n+1}COOH$ . According to the IUPAC naming, the carboxylic are named as alkanoic acids.

Aldehyde group : It is the class of organic compound in which the (-CHO) group is attached to a hydrocarbon is known as aldehyde.

The general representation of aldehyde is, $R-CHO$ . According to the IUPAC naming, the aldehyde are named as alkanals.

Ketone group : It is the class of organic compound in which the (-CO) group is directly attached to the two alkyl group of carbon is known as ketone.

The general representation of ketone is, $R-CO-R^'$ . According to the IUPAC naming, the ketone are named as alkanone.

Ester group : It is the class of organic compound in which the (-COO) group is directly attached to the two alkyl group of carbon is known as ester.

The general representation of ester is, $R-COO-R^'$ . According to the IUPAC naming, the ester are named as alkyl alkanoate.

6 0

3 years ago

Calculate how many times more soluble Mg(OH)2 is in pure water Based on the given value of the Ksp, 5.61×10−11, calculate the ra

maks197457 [2]

Answer:

molar solubility in water = 2.412 * 10^-4 mol/L

molar solubility of NaOH in 0.130M = 3.32 * 10^-9 mol/L

Mg(OH)2 is a factor 0.73*10^5 more soluble in pure water than in 0.130 M NaOH

Explanation:

The Ksp refers to the partial solubilization of a mostly insoluble salt. This is an equilibrium process.

The equation for the solubilization reaction of Mg(OH)2 can be given as:

Mg(OH)2 (s) → Mg2+ (aq) + 2OH– (aq)

Ksp can then be given as followed:

Ksp = [Mg^2+][OH^–]²

Step 2: Calculate the solubility in water

Mg(OH)2 (s) → Mg2+ (aq) + 2OH– (aq)

The mole ratio Mg^2+ with OH- is 1:2

So there will react X of Mg^2+ and 2X of OH-

The concentration at equilibrium will be XM Mg^2+ and 2X OH-

Ksp = [Mg^2+][OH^–]²

5.61*10^-11 = X * (2X)² = X *4X² = 4X³

X = 2.412 * 10^-4 mol/L = solubility in water

Step 3: Calculate solubility in 0.130 M NaOH

The initial concentration of Mg^2+ = 0 M

The initial concentration of OH- = 0.130 M

The mole ratio Mg^2+ with OH- is 1:2

So there will react X of Mg^2+ and 2X +0.130 for OH-

The concentration at equilibrium will be XM Mg^2+ and 0.130 + 2X OH-

The value of "[OH–] + 2X" is, because the very small value of X, equal to the value of [OH–] .

Let's consider:

[Mg+2] = X

[OH] = 0.130

Ksp = [Mg^2+][OH^–]²

5.61*10^-11 = X *(0.130)²

5.61*10^-11 = X * (0.130)^2

X = 3.32*10^-9 = solubility in 0.130 M NaOH

Step 4: Calculate how many times Mg(OH)2 is better soluble in pure water.

(2.412*10^-4)/ (3.32*10^-9) = 0.73 * 10^5

Mg(OH)2 is a factor 0.73*10^5 more soluble in pure water than in 0.130 M NaOH

4 0

3 years ago

the combustion of a sample butane, C4H10 (lighter fluid) produced 2.46 grams of water. how many moles of water formed

enyata [817]

$2C_4H_{10}+13O_2$ ⇒ $8CO_2 + 10H_2O$

$n= \frac{m}{M} = \frac{m}{2M(H)+M(O)}= \frac{2,46}{2*1,0+16,0} = 0,14mol$

So 0,14mol are formed.

6 0

3 years ago