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STatiana [176]
3 years ago
12

What statement applies to the horizontal rows or periods in the periodic table? A. The properties are all the same. B. The eleme

nts have the same number of electrons. C. The properties change going across each row. D. The elements in each row combine easily.
Physics
1 answer:
Sliva [168]3 years ago
7 0
The right answer for the question that is being asked and shown above is that: "C. <span>. The properties change going across each row. " the </span>statement that applies to the horizontal rows or periods in the periodic table is that t<span>he properties change going across each row. </span>
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A snail crawls 300 cm in 1 hour. Calculate the snail's speed in each of the following units. A) cm/h B) cm/min C) m/h
Crazy boy [7]
A) 300cm/h
B)1 hr=60 min
300/60=5
5cm/min
C)1m=100cm
300/100=3
3m/h
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What type of fat may help lower the risk of heart disease
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Polyunsaturated fatty acids which is Omega-3 fatty acids.
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2.0 mol of monatomic gas A initially has 5000 J of thermal energy. It interacts with 3.0 mol of monatomic gas B, which initially
Naddika [18.5K]

Answer:

 E_particle = 1,129 10⁻²⁰ J / particle

  T= 817.5 K

Explanation:

Energy is a scalar quantity so it is additive, let's look for the total energy of each gas

Gas a

         E_a = 2 5000 = 10000 J

Gas b

         E_b = 3 8000 = 24000 J

When the total system energy is mixed it is

          E_total = E_a + E_b

          E_total = 10000 + 24000 = 34000

The total mass is

           M = m_a + m_b

           M = 2 +3 = 5

The average energy among the entire mass is

           E_averge = E_total / M

            E_averago = 34000/5

            E_average = 6800 J

One mole of matter has Avogadro's number of atoms 6,022 10²³ particles

Therefore, each particle has an energy of

                E_particle = E_averag / 6.022 10²³ = 6800 /6.022 10²³

                E_particle = 1,129 10⁻²⁰ J / particle

For  find the temperature let's use equation

               E = kT

               T = E / k

     

               T = 1,129 10⁻²⁰ / 1,381 10⁻²³

               T = 8.175 102 K

               T= 817.5 K

5 0
3 years ago
12)A black body is heated from 27°C to 127° C. The ratio of their energies of radiations emitted will be
Nat2105 [25]

Answer:

81:256.

Explanation:

Let T denote the absolute temperature of this object.

Calculate the value of T before and after heating:

T(\text{before}) = 27 + 273 = 300\; \rm K.

T(\text{after}) = 127 + 273 = 400\; \rm K.

By the Stefan-Boltzmann Law, the energy that this object emits (over all frequencies) would be proportional to T^4.

Ratio between the absolute temperature of this object before and after heating:

\displaystyle \frac{T(\text{before})}{T(\text{after})} = \frac{3}{4}.

Therefore, by the Stefan-Boltzmann Law, the ratio between the energy that this object emits before and after heating would be:

\displaystyle \left(\frac{T(\text{before})}{T(\text{after})}\right)^{4} = \left(\frac{3}{4}\right)^{4} = \frac{81}{256}.

4 0
3 years ago
Can you plz help me in on 11 12 and 14?
anygoal [31]
I’m pretty sure 14 is mutations
3 0
3 years ago
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