In any chemical or physical process energy is neither created nor destroyed is a statement of the

A deck of cards has dimensions 8.9cm x 5.72cm x 1.82 cm. What is the volume of the deck in cubic centimeters?

arsen [322]

Answer:

92.65256 cm^3

Explanation:

To find this, we can simply multiply all three dimensions to get the answer in cubic centimeters, and we get the answer above. If you want to be more specific, we can go by the sigfig rule and the answer would be rounded to 93 cm^3.

5 0

2 years ago

Which of the following solution is more dilute and explain why?a)1M b)2M c)0.1M or d)0.009M

Taya2010 [7]

I can’t access the pictures. Sorry!

6 0

2 years ago

Before landing, the brakes and the tires of an airliner have a temperature of 15.0∘C. Upon landing, the 90.7 kg carbon fiber bra

Goryan [66]

Answer:

0.921 J/g degrees C

Explanation:

Recall that the First Law of Thermodynamics demands that the total internal energy of an isolated system must remain constant. Any amount of energy lost by the brakes must be gained by the tires (in the form of heat in this situation). Therefore, heat given off by the brakes = −heat taken in by tires, or:

−qbrakes=qtires

The equation used to calculate the quantity of heat energy exchanged in this process is:

−qbrakes=−cbrakes mbrakes ΔTbrakes=ctires mtires ΔTtires=qtires

First we must convert the mass of the tires and the brakes from kg to g.

massbrakes=90.7 kg×1,000. g1 kg=9.07×104 g

masstires=123 kg×1,000. g1 kg=1.23×105 g

Next, substitute in known values and rearrange to solve for ctires. Note that the final temperature for both the tires and the brakes is 172∘C, the initial temperature of the brakes is 312∘C and the initial temperature of the tires is 15∘C.

−(1.400Jg∘C)(9.07×104 g)(172∘C−312∘C)=(ctires)(1.23×105 g)(172∘C−15∘C)

ctires=−(1.400 Jg∘C)(9.07×104 g)(−140∘C)(1.23×105 g)(157∘C)=17,777,200 J19311000 g∘C=0.9206Jg∘C

The answer should have three significant figures, so round to 0.921Jg∘C.

6 0

2 years ago

If 3.491 grams of the precipitate was formed, how many moles of strontium bromide were reacted

AVprozaik [17]

The balanced chemical equation that represents the reaction is as follows:
SrBr2(aq) + 2AgNO3(aq) → Sr(NO3)2(aq) + 2AgBr(s)

From the periodic table:
mass of silver = 108 grams
mass of bromine = 80 grams

molar mass of silver bromide = 108 + 80 = 188 grams

number of moles = mass / molar mass
number of moles of produced precipitate = 3.491/188 = 0.018 moles

From the balanced equation:
1 mole of strontium bromide produces 2 moles of silver bromide. Therefore, to calculate the number of moles of strontium bromide that produces 0.018 moles of silver bromide, you will just do a cross multiplication as follows:
amount of strontium bromide = (0.018x1) / 2 = 9.28 x 10^-3 moles

4 0

2 years ago

Explain the effects of nh3 and hcl on the cuso4 solution in terms of le chatelier's principle

Fittoniya [83]

The Principle of Le Chatelier states that if a system in equilibrium is subjected to a disturbance, the system will react in such a way that it will diminish the effect of that disturbance. Thus, when the concentration of one of the substances in an equilibrium system is changed, the equilibrium varies in such a way that it can compensate for this change.

For example, if the concentration of one of the reactants is increased, the equilibrium shifts to the right or to the side of the products. Also, if you add more reagents, the reaction will move even more to the right until the balance is re-established again, increasing the quantity of products.

In this way, adding HCl to a solution of CuSO4 will produce the following reaction:

CuSO4 (aq) + 2HCl (aq) ⇔ CuCl2 (aq) + H2SO4 (aq)

Initially the solution of CuSO4 in water will be blue, but when adding HCl the solution will change color to green, since the aqueous solutions of CuCl2 are green. By adding more HCl this color will intensify as the balance shifts to the right, producing more CuCl2 and H2SO4.

On the other hand, adding NH3 to a solution of CuSO4 will produce the following reaction

CuSO4 (aq) + 4NH3 (aq) ⇔ [Cu(NH3)4] SO4 (s)

Thus, by adding NH3 to the CuSO4 solution we will observe the formation of a precipitate corresponding to [Cu(NH3) 4] SO4. When adding more NH3, the formation of more precipitate will be observed as the equilibrium moves to the right, producing a greater quantity of [Cu (NH3) 4] SO4.

6 0

2 years ago