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2 years ago

describe the law of conservation of energy and two examples of energy being transformed from one type to another.

Chemistry

1 answer:

VladimirAG [237]2 years ago

3 0

Answer:

look in the explanation part

Explanation:

In physics and chemistry, the law of conservation of energy states that the total energy of an isolated system remains constant; it is said to be conserved over time. This law means that energy can neither be created nor destroyed; rather, it can only be transformed or transferred from one form to another.

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You have 40 g of a radioisotope. If the half-life of this radioisotope is 4 days, how many grams will remain after 12 days?

anygoal [31]

With reference to radioactive material, half-life is the time required to 50% depletion of initial amount of material.

Given: Initial amount of radioactive material = 40 g
Half life = 4 days.

Therefore, After 4 days, amount of compound left = 40/2 = 20 g
After 8 days, i.e 2 half-life, amount of compound left = 20/2 = 10 g
Finally after 12 days, i.e. 3 half-life, amount of compound left = 10/2 = 5 g

Thus, 5 <span>grams will remain after 12 days</span>

5 0

2 years ago

Complete and balance the following redox reaction in acidic solution As2O3(s) + NO3- (aq) → H3AsO4(aq) + N2O3(aq)

scoray [572]

Answer:

$As_{2}O_{3}(s)+2NO_{3}^{-}(aq)+2H_{2}O(l)+2H^{+}(aq)\rightarrow 2H_{3}AsO_{4}(aq)+N_{2}O_{3}(aq)$

Explanation:

Oxidation: $As_{2}O_{3}(s)\rightarrow H_{3}AsO_{4}(aq)$

Balance As: $As_{2}O_{3}(s)\rightarrow 2H_{3}AsO_{4}(aq)$
Balance H and O in acidic medium: $As_{2}O_{3}(s)+5H_{2}O(l)\rightarrow 2H_{3}AsO_{4}(aq)+4H^{+}(aq)$
Balnce charge: $As_{2}O_{3}(s)+5H_{2}O(l)-4e^{-}\rightarrow 2H_{3}AsO_{4}(aq)+4H^{+}(aq)$ ......(1)

Reduction: $NO_{3}^{-}(aq)\rightarrow N_{2}O_{3}(aq)$

Balance N: $2NO_{3}^{-}(aq)\rightarrow N_{2}O_{3}(aq)$
Balance H and O in acidic medium: $2NO_{3}^{-}(aq)+6H^{+}(aq)\rightarrow N_{2}O_{3}(aq)+3H_{2}O(l)$
Balance charge: $2NO_{3}^{-}(aq)+6H^{+}(aq)+4e^{-}\rightarrow N_{2}O_{3}(aq)+3H_{2}O(l)$ ......(2)

$Equation (1)+Equation (2)$ gives-

$As_{2}O_{3}(s)+2NO_{3}^{-}(aq)+2H_{2}O(l)+2H^{+}(aq)\rightarrow 2H_{3}AsO_{4}(aq)+N_{2}O_{3}(aq)$

3 0

3 years ago

Read 2 more answers

Osmosis is the process responsible for carrying nutrients and water from groundwater supplies to the upper parts of trees. The o

ad-work [718]

Answer:

The molar concentration would have to be 0,81 M.

Explanation:

The osmotic pressure equation is:

$\pi = M R T$

where:

$\pi$ : osmotic pressure [atm]

M: molar concentration [M]

R: gas constant 0,08205 [atm.L/mol.°K]

T: absolute temperature [°K]

To solve the problem, we just clear M from the osmotic pressure equation and then replace our data using the appropiate units. Clearing the variable M we have:

$M = \frac{\pi }{RT}$

We have to use temperature as absolute temperature (in kelvins), T=29+273=302 °K. Now we can replace our values in the equation:

$M = \frac{20,1 atm}{0,08205 \frac{atm.L}{mol.K} * 302 K }$

As we can see, all units will be simplified and we'll have the molar concentration in mol/L.

$M = 0,81 \frac{mol}{L} = 0, 81 M$

7 0

3 years ago

Inessa05 [86]

I don’t understand the question

7 0

3 years ago

Calcium hydride reacts with water to form hydrogen gas according to the unbalanced equation below: CaH2(s) + H2O(l) --> Ca(OH

egoroff_w [7]

Answer:

28 grams CaH₂(s) is required for production of 15L H₂(g) at 25°C and 825Torr.

Explanation:

CaH₂(s) + H₂O(l) => Ca(OH)₂(s) + H₂(g)

Using ideal gas law, PV = nRT

=> moles H₂(g) = PV/RT = [(825/760)Atm](15L)/(0.08206L·Atm·mol⁻¹·K⁻¹)(298K) = 0.6659 mol H₂(g)

From stoichiometry of given equation,

=> 0.6659 mol H₂(g) requires 0.6659 mole CaH₂(s)

Converting moles to grams, multiply by formula weight,

=> 0.6659 mole CaH₂(s) = 0.6659 mole CaH₂(s) x 42g/mole = 27.966 grams CaH₂(s) ≅ 28 grams CaH₂(s) (2 sig. figs.)

3 0

3 years ago