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sineoko [7]
3 years ago
10

A car (mass = 1090 kg) is traveling at 30.4 m/s when it collides head-on with a sport utility vehicle (mass = 2880 kg) traveling

in the opposite direction. In the collision, the two vehicles come to a halt. At what speed was the sport utility vehicle traveling?
Physics
1 answer:
Thepotemich [5.8K]3 years ago
7 0

Answer:

The sport utility vehicle was traveling at V2= 11.5 m/s.

Explanation:

m1= 1090 kg

V1= 30.4 m/s

m2= 2880 kg

V2= ?

m1*V1 = m2*V2

V2= (m1*V1)/m2

V2= 11.5 m/s

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You treat 9.540 g of the mixture with the acid and isolate 9.355 g of nacl. what is the weight percent of each substance in the
Mila [183]

The weight percentage of sodium carbonate in the mixture is = 67.71%  

The weight percentage of sodium bicarbonate in the mixture is = 32.57%

<h3>What does "molar mass" ?</h3>

The total mass throughout grams of all the atoms needed to form a molecule per mole is what makes up the molar mass, also known as the molecular weight. Grams per mole is the unit measuring molar mass.

<h3>According to the given information:</h3>

The interaction between sodium carbonate and hydrochloric acid has the following equation:

Na₂CO₃ + HCl  --> NaCl  + CO₂ + H₂O

The reaction's balanced equation is,

Na₂CO₃ + 2HCl  --> 2NaCl  + CO₂ + H₂O -------(2).

When sodium hydrogen carbonate as well as hydrochloric acid react, the following equation results.

NaCO₃ + HCl  --> NaCl  + CO₂ + H₂O ----------------(3)

The Molar mass :

Na₂CO₃ = 106g/mol

NaCO₃ = 84g/mol

NaCl  = 58.5g/mol

The mass of the mixture is Na₂CO₃/NaCO₃ = 9.540g

The molar masses of the constituent elements that make up the compounds are added to determine the molar weight of the substances. In both chemical equations (2) and (3), the unitary technique is employed to calculate the mass of the mixture based on the number of moles (3).

The mass of the sodium carbonate and sodium bicarbonate is calculated with the help of the mass of NaCl formed by the mixture of  Na₂CO₃/NaCO₃ with HCl.

The mixture has the following percentage of sodium carbonate:

        % of Na₂CO₃ =  \frac{x \times 106}{9.540} \times 100\\

                             =\frac{0.061 \times 106}{9.540} \times 100

                             = (6.46/9.540)*100

                             = 67.71%        

The weight percentage of sodium carbonate in the mixture is = 67.71%      

The mixture has the following percentage of NaHCO3:

        % of NaHCO₃ =  \frac{y \times 84}{9.540} \times 100

                             = =\frac{0.037 \times 84}{9.540} \times 100

                              = (3.108/9.540)*100

                             = 32.57%

The weight percentage of sodium bicarbonate in the mixture is = 32.57%

To know more about molar mass visit:

brainly.com/question/12127540

#SPJ4

I understand that the question you are looking for is:

A mixture of sodium carbonate and sodium hydrogen carbonate is treated with aqueous hydrochloric acid. the unbalanced equations for the resulting's reactions are:

Na2CO3 (s) + HCl (aq) --> NaCl (aq) + CO2(g) + H2O (l)

NaHCO3(s) + HCl (aq) --> NaCl (aq) + CO2(g) + H2O (l)

You treat 9.540g of Na2CO3/NaHCO3 mixture with an excess of aqueous HCl and isolate 9.355g of NaCl. What is the weight percent of each substance in the mixture?

5 0
1 year ago
In a closed system, the loss of momentum of one object ________ the gain in momentum of another object.
Virty [35]

In a closed system, the loss of momentum of one object is same as________ the gain in momentum of another object

according to law of conservation of momentum, total momentum before and after collision in a closed system in absence of any net external force, remains conserved . that is

total momentum before collision = total momentum after collision

P₁ + P₂ = P'₁ + P'₂

where P₁ and P₂ are momentum before collision for object 1 and object 2 respectively.

P'₁ - P₁  = - (P'₂ -  P₂)

so clearly gain in momentum of one object is same as the loss of momentum of other object

8 0
2 years ago
The bond enthalpy value for a carbon to hydrogen bond is 413 kJ. What does this mean?
AlladinOne [14]
Breaking bond requires energy. The bond between the carbon and hydrogen is broken when the energy is absorbed. The enthalpy is defined to be the energy taken to break the one mole of the stated carbon and hydrogen bond. Thus a should be the correct answer
6 0
3 years ago
Read 2 more answers
The rms (root-mean-square) speed of a diatomic hydrogen molecule at 50∘C is 2000 m/s. Note that 1.0 mol of diatomic hydrogen at
denis-greek [22]

Answer:

A) d. (1/4)(2000m/s) = 500 m/s

B) c. 4000 J

C) f. None of the above (2149.24 m/s)

Explanation:

A)

The translational kinetic energy of a gas molecule is given as:

K.E = (3/2)KT

where,

K = Boltzman's Constant = 1.38 x 1^-23 J/K

T = Absolute Temperature

but,

K.E = (1/2) mv²

where,

v = root mean square velocity

m = mass of one mole of a gas

Comparing both equations:

(3/2)KT = (1/2) mv²

v = √(3KT)/m  _____ eqn (1)

<u>FOR HYDROGEN:</u>

v = √(3KT)/m = 2000 m/s  _____ eqn (2)

<u>FOR OXYGEN:</u>

velocity of oxygen = √(3KT)/(mass of oxygen)  

Here,

mass of 1 mole of oxygen = 16 m

velocity of oxygen = √(3KT)/(16 m)

velocity of oxygen = (1/4) √(3KT)/m

using eqn (2)

<u>velocity of oxygen = (1/4)(2000 m/s) = 500 m/s</u>

B)

K.E = (3/2)KT

Since, the temperature is constant for both gases and K is also a constant. Therefore, the K.E of both the gases will remain same.

K.E of Oxygen = K.E of Hydrogen

<u>K.E of Oxygen = 4000 J</u>

C)

using eqn (2)

At, T = 50°C = 323 k

v = √(3KT)/m = 2000 m/s

m = 3(1.38^-23 J/k)(323 k)/(2000 m/s)²

m = 3.343 x 10^-27 kg

So, now for this value of m and T = 100°C = 373 k

v = √(3)(1.38^-23 J/k)(373 k)/(3.343 x 10^-27 kg)

<u>v = 2149.24 m/s</u>

<u></u>

8 0
3 years ago
Differences between Pressure and upthrust​
Angelina_Jolie [31]

Answer:

Pressure is equal to the ratio of thrust to the area in contact. Upthrust is a force exerted by the fluids on an object placed in the fluid . Upthrust acts in upward direction.

4 0
3 years ago
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