Which equation yields the solutions x=−2 and, №17886136, 02.06.2020 01:50

Mathematics

Which equation yields the solutions x=−2 and x=−10?

Step-by-step answer

P Answered by PhD

2

Options:

A. −x2−2x+35=0

B. 5x2+40x+60=0

C. −4x2+24x−20=0

D. x2+12x+20=0

its not B

sorry it only let me choose once ttm2021

Chemistry

In the reaction shown above, 2.00 x 10^3 g caco3 produce 1.05 x 10^3 g of cao, what is the percent yield?

Step-by-step answer

P Answered by PhD

4

First, we have to get the theoretical yield of CaO:
the balanced equation for the reaction is:
CaCO3(s)→CaO(s) +CO2(g)
covert mass to moles:
moles CaCO3 = mass of CaCO3 / molar mass of CaCO3
= 2x10^3 /100 = 20 moles
the molar ratio between CaCO3 : CaO = 1:1
∴moles of CaO = 1* 20 = 20 moles
∴mass of CaO = moles of CaO * molar mass of CaO
= 20 * 56 = 1120 g
∴the theoritical yield = 1120 g and we have the actual yield =1.05X10^3
∴Percent yield = actual yield / theoritical yield *100
= (1.05x10^3) / 1120 * 100
= 94 %

Chemistry

In the reaction shown above, 2.00 x 10^3 g caco3 produce 1.05 x 10^3 g of cao, what is the percent yield?

Step-by-step answer

P Answered by PhD

4

First, we have to get the theoretical yield of CaO:
the balanced equation for the reaction is:
CaCO3(s)→CaO(s) +CO2(g)
covert mass to moles:
moles CaCO3 = mass of CaCO3 / molar mass of CaCO3
= 2x10^3 /100 = 20 moles
the molar ratio between CaCO3 : CaO = 1:1
∴moles of CaO = 1* 20 = 20 moles
∴mass of CaO = moles of CaO * molar mass of CaO
= 20 * 56 = 1120 g
∴the theoritical yield = 1120 g and we have the actual yield =1.05X10^3
∴Percent yield = actual yield / theoritical yield *100
= (1.05x10^3) / 1120 * 100
= 94 %

Chemistry

A solution is prepared by dissolving 0.26 mol of hydrazoic acid and 0.26 mol of sodium azide in water sufficient to yield 1.00 L of solution. The addition of 0.05 mol of HCl to this buffer solution causes the pH to drop slightly. The pH does not decrease drastically because the HCl reacts with the present in the buffer solution. The Ka of hydrazoic acid is 2.5 x 10^-5.

Step-by-step answer

P Answered by Master

The pH does not decrease drastically because the HCl reacts with the sodium azide (NaN₃) present in the buffer solution.

Explanation:

The buffer solution is formed by 0.26 moles of the weak acid, hydrazoic acid (HN₃), and by 0.26 moles of sodium azide (NaN₃). The equilibrium reaction of this buffer solution is the following:

HN₃(aq) + H₂O(l) ⇄ N₃⁻(aq) + H₃O⁺(aq)

The pH of this solution is:

$pH = pka + log(\frac{[N_{3}^{-}]}{[HN_{3}]}) = -log(2.5 \cdot 10^{-5}) + log(\frac{0.26 mol/1 L}{0.26 mol/1 L}) = 4.60$

When 0.05 moles of HCl is added to the buffer solution, the following reaction takes place:

H₃O⁺(aq) + N₃⁻(aq) ⇄ HN₃(aq) + H₂O(l)

The number of moles of NaN₃ after the reaction with HCl is:

$\eta_{NaN_{3}} = \eta_{i} - \eta_{HCl} = 0.26 moles - 0.05 moles = 0.21 moles$

Now, the number of moles of HN₃ is:

$\eta_{HN_{3}} = \eta_{i} + \eta_{HCl} = 0.26 moles + 0.05 moles = 0.31 moles$

Then, the pH of the buffer solution after the addition of HCl is:

$pH = pka + log(\frac{[N_{3}^{-}]}{[HN_{3}]}) = -log(2.5 \cdot 10^{-5}) + log(\frac{0.21 mol/V_{T}}{0.31 mol/V_{T}}) = 4.43$

The pH of the buffer solution does not decrease drastically, it is 4.60 before the addition of HCl and 4.43 after the addition of HCl.

Therefore, the pH does not decrease drastically because the HCl reacts with the sodium azide (NaN₃) present in the buffer solution.

I hope it helps you!

Geography

Aconfined aquifer has a specific storage of 2.4 x 10-s ft and a thickness of 300 ft. how much water (in acre-ft) would it yield if the water declined 4.8 ft uniformly over a circular area with a radius of 1,500 ft?

Step-by-step answer

P Answered by Specialist

water it yield is 244290.45 ft

Explanation:

given data

specific storage SS = 2.4 × $10^{-5}$ ft

thickness b = 300 ft

water declined ΔH = 4.8 ft

radius r = 1,500 ft

to find out

How much water would it yield

solution

we know that

volume of water is

Vw = S × Area × ΔH ..............1

here S = specific storage × thickness

S = 2.4 × $10^{-5}$ × 300

S = 7.2 × $10^{-3}$

and

area = πr² = π 1,500² = 7068589.471 ft²

so

volume of water is

Vw = 7.2 × $10^{-3}$ × 7068589.471 × 4.8

Vw = 244290.45

so water it yield is 244290.45 ft

Chemistry

A solution is prepared by dissolving 0.26 mol of hydrazoic acid and 0.26 mol of sodium azide in water sufficient to yield 1.00 L of solution. The addition of 0.05 mol of HCl to this buffer solution causes the pH to drop slightly. The pH does not decrease drastically because the HCl reacts with the present in the buffer solution. The Ka of hydrazoic acid is 2.5 x 10^-5.

Step-by-step answer

P Answered by Master

The pH does not decrease drastically because the HCl reacts with the sodium azide (NaN₃) present in the buffer solution.

Explanation:

The buffer solution is formed by 0.26 moles of the weak acid, hydrazoic acid (HN₃), and by 0.26 moles of sodium azide (NaN₃). The equilibrium reaction of this buffer solution is the following:

HN₃(aq) + H₂O(l) ⇄ N₃⁻(aq) + H₃O⁺(aq)

The pH of this solution is:

$pH = pka + log(\frac{[N_{3}^{-}]}{[HN_{3}]}) = -log(2.5 \cdot 10^{-5}) + log(\frac{0.26 mol/1 L}{0.26 mol/1 L}) = 4.60$

When 0.05 moles of HCl is added to the buffer solution, the following reaction takes place:

H₃O⁺(aq) + N₃⁻(aq) ⇄ HN₃(aq) + H₂O(l)

The number of moles of NaN₃ after the reaction with HCl is:

$\eta_{NaN_{3}} = \eta_{i} - \eta_{HCl} = 0.26 moles - 0.05 moles = 0.21 moles$

Now, the number of moles of HN₃ is:

$\eta_{HN_{3}} = \eta_{i} + \eta_{HCl} = 0.26 moles + 0.05 moles = 0.31 moles$

Then, the pH of the buffer solution after the addition of HCl is:

$pH = pka + log(\frac{[N_{3}^{-}]}{[HN_{3}]}) = -log(2.5 \cdot 10^{-5}) + log(\frac{0.21 mol/V_{T}}{0.31 mol/V_{T}}) = 4.43$

The pH of the buffer solution does not decrease drastically, it is 4.60 before the addition of HCl and 4.43 after the addition of HCl.

Therefore, the pH does not decrease drastically because the HCl reacts with the sodium azide (NaN₃) present in the buffer solution.

I hope it helps you!

Geography

Aconfined aquifer has a specific storage of 2.4 x 10-s ft and a thickness of 300 ft. how much water (in acre-ft) would it yield if the water declined 4.8 ft uniformly over a circular area with a radius of 1,500 ft?

Step-by-step answer

P Answered by Specialist

water it yield is 244290.45 ft

Explanation:

given data

specific storage SS = 2.4 × $10^{-5}$ ft

thickness b = 300 ft

water declined ΔH = 4.8 ft

radius r = 1,500 ft

to find out

How much water would it yield

solution

we know that

volume of water is

Vw = S × Area × ΔH ..............1

here S = specific storage × thickness

S = 2.4 × $10^{-5}$ × 300

S = 7.2 × $10^{-3}$

and

area = πr² = π 1,500² = 7068589.471 ft²

so

volume of water is

Vw = 7.2 × $10^{-3}$ × 7068589.471 × 4.8

Vw = 244290.45

so water it yield is 244290.45 ft

Chemistry

1. if you have 4.0 g of a radioactive substance left after 4 half-lives have passed, then how much did you have originally? a. 50 g b. 64 g c. 25 g d. 10 g 2. if a fusion reaction yields 4.50 x 10^9 kj of energy, then how much mass was lost? a. 5.00 x 10^-5 kg b. 1.50 x 10^-6 kg c. 1.50 x 10^-3 kg d. 5.00 x 10^-2 kg 3. for the transmutation shown, which of the following is x? (the first number is on top on the actual answer.) a. 3 1 h b. 6 3 li c. 1 1 h d. 4 2 he 5. which of the following energy yields is most likely to have come from a fission

Step-by-step answer

P Answered by Master

11

Answer 1:
During the radio-active disintegration, when initial concentration of radioactive nuclei is reduced to half, it is referred as half life.
Now, if initial concentration of radio-active element is 64 g. After it crosses 1st half life, it concentration will decrease to 32g. At 2nd half life, concentration will further reduce to 16 g. This concentration will reduce to 8 g, at 3 half life. And, finally at 4th half life, concentration of radioactive element will reduce to 4 g.
Thus ,in present case, correct answer is B 64 g.

Answer 2:
Given that energy produced during nuclear fusion reaction = 4.50 x 10^9 kJ
Now, based on Einstein's Equation, we know that E = mc2
where, E = energy produced
m = mass o f compound
c = speed of light = 3 X 10^8 m/s
Thus, 4.50 x 10^9 x 10^3J = m (3 X 10^8)^2
∴ m = 5 X 10^-5 Kg

Thus, mass lost in the reaction = 5 X 10^-5 Kg

Answer 3:
In nuclear transmutation reaction, a non-radioactive nuclei is collided with a fast moving projectile to convert into radio-active nuclei. Following is the examples of nuclear transmutation reaction
1 1H + 1 0n → 2 1 H + gamma rays

In the above reaction hydrogen, on collision with neutron generates deuterium.
However, other elements listed above will not undergo nuclear transmutation reaction.

Answer 4:
In a nuclear reaction, one nuclei get converted into another nuclei. Nuclear reactions are broadly classified into two types.
a) Nuclear fusion
b) Nuclear fission
In nuclear fusion reaction, two light nuclei under extreme conditions fuse together, while in nuclear fission reaction, a heavy nuclei is bombarded with projectile which breaks the nuclei in smaller nuclei. In either case, mass of product is less than mass of reactant. Thus there is a mass defect. Due to mass defect, tremendous amount of energy is released from the system which is of order of 10^11 Kj/mol. Thus, correct answer is option D.

Chemistry

1. if you have 4.0 g of a radioactive substance left after 4 half-lives have passed, then how much did you have originally? a. 50 g b. 64 g c. 25 g d. 10 g 2. if a fusion reaction yields 4.50 x 10^9 kj of energy, then how much mass was lost? a. 5.00 x 10^-5 kg b. 1.50 x 10^-6 kg c. 1.50 x 10^-3 kg d. 5.00 x 10^-2 kg 3. for the transmutation shown, which of the following is x? (the first number is on top on the actual answer.) a. 3 1 h b. 6 3 li c. 1 1 h d. 4 2 he 5. which of the following energy yields is most likely to have come from a fission

Step-by-step answer

P Answered by Master

11

Answer 1:
During the radio-active disintegration, when initial concentration of radioactive nuclei is reduced to half, it is referred as half life.
Now, if initial concentration of radio-active element is 64 g. After it crosses 1st half life, it concentration will decrease to 32g. At 2nd half life, concentration will further reduce to 16 g. This concentration will reduce to 8 g, at 3 half life. And, finally at 4th half life, concentration of radioactive element will reduce to 4 g.
Thus ,in present case, correct answer is B 64 g.

Answer 2:
Given that energy produced during nuclear fusion reaction = 4.50 x 10^9 kJ
Now, based on Einstein's Equation, we know that E = mc2
where, E = energy produced
m = mass o f compound
c = speed of light = 3 X 10^8 m/s
Thus, 4.50 x 10^9 x 10^3J = m (3 X 10^8)^2
∴ m = 5 X 10^-5 Kg

Thus, mass lost in the reaction = 5 X 10^-5 Kg

Answer 3:
In nuclear transmutation reaction, a non-radioactive nuclei is collided with a fast moving projectile to convert into radio-active nuclei. Following is the examples of nuclear transmutation reaction
1 1H + 1 0n → 2 1 H + gamma rays

In the above reaction hydrogen, on collision with neutron generates deuterium.
However, other elements listed above will not undergo nuclear transmutation reaction.

Answer 4:
In a nuclear reaction, one nuclei get converted into another nuclei. Nuclear reactions are broadly classified into two types.
a) Nuclear fusion
b) Nuclear fission
In nuclear fusion reaction, two light nuclei under extreme conditions fuse together, while in nuclear fission reaction, a heavy nuclei is bombarded with projectile which breaks the nuclei in smaller nuclei. In either case, mass of product is less than mass of reactant. Thus there is a mass defect. Due to mass defect, tremendous amount of energy is released from the system which is of order of 10^11 Kj/mol. Thus, correct answer is option D.

Mathematics

While testing a new pesticide, an agricultural scientist uses two functions to predict the yields of two same-sized potato farms that have different soil conditions. the scientist s predicted yield, in tons, of the first farm can be represented by the following function, where x is the liters of pesticide he uses per acre of the farm. his predicted yield, in tons, of the second farm can be represented by the following function, where x is the liters of pesticide he uses per acre of the farm. if the scientist uses the same amount of pesticide on

Step-by-step answer

P Answered by PhD

5

A. h(x)= -5.86x^2 + 23.37x + 34

Step-by-step explanation:

If the scientist uses the same amount of pesticide on the two farms then the x's of the functions are the same.

Then, the combined yield h(x) of the two farms is just the yield of the first farm plus the yield of the second farm: