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joja [24]
3 years ago
11

A goose is flying south for the winter at a constant speed. Keep in mind that one mile is 1.61 km, and one pound is 454 g.The go

ose has a mass of 20.1 lb and is flying at 9.60 miles/h. What is the kinetic energy of the goose in joules?
Chemistry
1 answer:
ruslelena [56]3 years ago
8 0

Answer:

kinetic energy =  84.0260 joules

Explanation:

given data

one mile = 1.61 km

one pound = 454 g

mass = 20.1 lb = 9.1172 kg

flying velocity = 9.60 miles/h = 9.60 × 1.61 × \frac{5}{18} = 4.2933 m/s  

solution

we get here kinetic energy that is express as

kinetic energy = 0.5 × m × v²   .........................1

put here value and we get

kinetic energy = 0.5 × 9.1172 × 4.2933²

kinetic energy =  84.0260 joules

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You have a saturated solution of BaSO4, a slightly soluble ionic compound. What happens if you add Ba(OH)2, NaNO3, and CuSO4 to
igomit [66]

Answer:

- Addition of Ba(OH)2: favors the formation of a precipitate.

- Undergo a chemical reaction forming soluble species.

- Addition of CuSO4 : favors the formation of a precipitate.

Explanation:

Hello,

In this case, since the dissociation reaction of barium sulfate is:

BaSO_4(s)\rightleftharpoons Ba^{2+}(aq)+SO_4^{2-}(aq)

We must analyze the effect of the common ion:

- By adding barium hydroxide, more barium ions will be added to the equilibrium system so the formation of solid barium sulfate will be favored (reaction shifts leftwards towards reactants).

- By adding sodium nitrate, the following reaction will undergo:

BaSO_4(s)+NaNO_3(aq)\rightarrow Ba(NO_3)_2(aq)+Na_2SO_4(aq)

So the precipitate will turn into other soluble species.

- By adding copper (II) sulfate, more sulfate ions will be added to the equilibrium system so the formation of solid barium sulfate will be favored (reaction shifts leftwards towards reactants).

All of this is supported by the Le Chatelier's principle.

Best regards.

7 0
3 years ago
Consider two gases, A and B, are in a container at room temperature. What effect will the following changes have on the rate of
Oksanka [162]

Answer:

B: increase.

Explanation:

When we are considering two gases A and B in a container at room temperature .

We have to find the change on  rate of reaction when the number of molecules of gases A is doubled

Let [A]=a and [B]=b

A+B\rightarrow product

Rate of reaction

R_1=k[A][B]=kab

We know that concentration is increases with increase in number of moles

When the number of molecules of gases A is doubled then concentration of gases A increases.

Therefore ,[A]=2a

Rate of reaction

R_2=k(2a)(b)=2kab

R_2=2R_1

Hence, the rate of reaction is  2 times the initial rate of reaction.Therefore, the rate of reaction will increase when the number of molecules of gases A is doubled.

Answer: B: increase.

4 0
3 years ago
When atoms gain electrons, they can become larger, because the addition of an electron increases electrostatic repulsion.
astra-53 [7]

Answer:True

Explanation: An anion has a larger radius than a neutral atom because it gains valence electrons. There are added electron/electron repulsions in the valence shell that expand the size of the electron cloud, which results in a larger radius for the anion.

hit the crown for me pls :)

Have a great day

6 0
2 years ago
The following unbalanced equation illustrates the overall reaction by which the body utilizes glucose to produce energy: C6H12O6
s344n2d4d5 [400]

Answer:

the conversion factor is f= 6  mol of glucose/ mol of CO2

Explanation:

First we need to balance the equation:

C6H12O6(s) + O2(g) → CO2(g) + H2O(l) (unbalanced)

C6H12O6(s) + 6O2(g) → 6CO2(g) + 6H2O(l) (balanced)

the conversion factor that allows to calculate the number of moles of CO2 based on moles of glucose is:

f = stoichiometric coefficient of CO2 in balanced reaction / stoichiometric coefficient of glucose in balanced reaction

f = 6 moles of CO2 / 1 mol of glucose = 6  mol of glucose/ mol of CO2

f = 6 mol of CO2/ mol of glucose

for example, for 2 moles of glucose the number of moles of CO2 produced are

n CO2 = f * n gluc = 6 moles of CO2/mol of glucose * 2 moles of glucose= 12 moles of CO2

3 0
3 years ago
How many moles are in 5.6x1025 molecules of calcium oxide?
Vlad1618 [11]

Answer:

<h2>93.02 moles</h2>

Explanation:

To find the number of moles in a substance given it's number of entities we use the formula

n =  \frac{N}{L} \\

where n is the number of moles

N is the number of entities

L is the Avogadro's constant which is

6.02 × 10²³ entities

From the question we have

n =  \frac{5.6 \times  {10}^{25} }{6.02  \times  {10}^{23} }  \\  = 93.023...

We have the final answer as

<h3>93.02 moles</h3>

Hope this helps you

8 0
3 years ago
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