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3 years ago

Of the following, which is the strongest acid?

1 answer:

5 0

Because oxygen is a highly electronegative atom, it pulls electron density towards itself and away from iodine. The more oxygen atoms there are, the more spread out the negative charge is on the conjugate base. And the more spread out the charge is, the more stable the base is. This makes the acid a stronger one. HIo4 will be more acidic according to this

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a man crossed a road 8.25m wide at a speed of 2.01m/s,how long does it take to get man to cross the road

Svetradugi [14.3K]

Answer:

t = 4.1 seconds

Explanation:

It is given that,

Width of road which is to be crossed by a man is 8.25 m, it means it is distance to be covered.

Speed of man is 2.01 m/s

We need to find the time taken by the man to cross the road. It is a concept of speed. Speed of a person is given by total distance covered divided by time taken. So,

$v=\dfrac{d}{t}$

t is time taken

$t=\dfrac{d}{v}\\\\t=\dfrac{8.25}{2.01}\\\\t=4.1\ s$

So, the time taken by the man to cross the road is 4.1 seconds.

5 0

4 years ago

A force F~ = Fx ˆı + Fy ˆ acts on a particle that

Hatshy [7]

Answer:

W = 46 J

Explanation:

We need to find the angle between the two vectors Force vector and displacement vector.

First we will find the angle α of the force vector

$tan\alpha =\frac{1}{8} \\\\\\alpha =7.125 deg\\$

Then we find the angle β of the displacement vector

$tan\beta=\frac{2}{6} \\\\beta = 18.43 deg\\$

With these two angles we can find the angle between the two vectors

∅ = α + β = 25.56 deg

The definition of work is given by the expression

$W=F*d*cos (theta)$

The absolute value of F will be:

$F=\sqrt{8^{2}+1^{2} } \\F= 8.06 N$

The absolute value of d will be:

$d=\sqrt{(6 )^{2}+(2)^{2} } \\d= 6.32m\\$

Now we have:

$W=8.06*6.32*cos(25.56)\\W=46 J$

4 0

3 years ago

Suppose the energy required to freeze 0.250 kg of water were added to the same mass of water at an initial temperature of 1.0 °C

Eduardwww [97]

The answer is a because it is

6 0

3 years ago

What is the gravitational force between two 1 kg objects that are 1 m apart?

lapo4ka [179]

Answer:

$F=6.67\times 10^{-11}\ N$

Explanation:

Given that,

The masses of two objects, m₁ = m₂ = 1 kg

The distance between masses, d = 1 m

We need to find the gravitational force between two masses. The force is given by the relation as follows :

$F=G\dfrac{m_1m_2}{d^2}\\\\F=6.67\times 10^{-11}\times \dfrac{1\times 1}{(1)^2}\\F=6.67\times 10^{-11}\ N$

So, the force between two masses of 1 kg is $6.67\times 10^{-11}\ N$ .

3 0

3 years ago

Uranium was used to generate electricity in a nuclear reactor of a nuclear power station. The nuclear reactor was 29% efficient

Minchanka [31]

Answer:

$C_{lifetime} = 2,618,017,174\,USD$

Explanation:

The energy contained per second in the fuel is:

$\dot E_{fuel} = \frac{800\times 10^{6}\,W}{0.29}$

$\dot E_{fuel} = 2.758\times 10^{9}\,W$

It is know that fission of a gram of uranium liberates $5.8\times 10^{8}\,J$ . Mass flow rate is computed herein:

$\dot m_{U}= \frac{2.758\times 10^{9}\,W}{5.8\times 10^{8}\,\frac{J}{g}}$

$\dot m_{U} = 4.755\,\frac{g}{s}$

The needed quantity of fuel per second is:

$\dot m_{fuel} = \frac{4.755\,\frac{g}{s} }{0.04}$

$\dot m_{fuel} = 118.875\,\frac{g}{s}$

The annual fuel consumption is now obtained:

$\Delta m_{fuel,year} = (0.119\,\frac{kg}{s})\cdot (\frac{3600\,s}{1\,h} )\cdot (\frac{24\,h}{1\,day} )\cdot(\frac{365\,days}{1\,year} )$

$\Delta m_{fuel,year} = 3,752,784\,\frac{kg}{year}$

The cost of fuel for the lifetime of the plant is:

$C_{lifetime} = (3,752,784\,\frac{kg}{year} )\cdot (\frac{0.453\,lb}{1\,kg} )\cdot [(20\,years)\cdot(\frac{12\,USD}{1\,lb} )+(20\,years)\cdot (\frac{65\,USD}{1\,lb} )$

$C_{lifetime} = 2,618,017,174\,USD$

6 0

3 years ago