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padilas [110]
2 years ago
6

Which of the following elements could be transmutated into silver by bombardment with a positron?

Chemistry
1 answer:
Elena L [17]2 years ago
6 0

Answer:

I think it's gold but I'm not sure, sorry

Explanation:

good luck tho :)

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During this process of sugar production, carbon dioxide combines with water to
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8 0
3 years ago
How many watts of power (W) do you expend when you exert a force of 10 N that moves a
mihalych1998 [28]

Answer:

Power = 7.5 watt

Explanation:

Given data:

Power expend = ?

Force applied = 10 N

Distance cover = 1.5 m

Time = 2 s

Solution:

Power = work/ time

First of all we will calculate work.

Work = Force × distance

Work = 10 N × 1.5 m

Work = 15 N.m

Now we will calculate the power.

Power = 15 N.m / 2s

N.m/s = 1 watt

Power = 7.5 watt

7 0
3 years ago
Which of the following is not a common way for a lake to form?
VashaNatasha [74]

Answer:

I guess your friendly ya

4 0
3 years ago
Explain why photosynthesis and cellular respiration are the “real circle of life”
BabaBlast [244]

Explanation:Photosynthesis is important because the earth and all of the creatures in the sea and land need the plants to be able to feed themselves with the help of the sun and with cellular respiration, those plants are able to take in the carbon dioxide that we give and make into oxygen with we need. there fore without them we would not be able to exist.

4 0
3 years ago
A sample of N2 gas in a flask is heated from 27 Celcius to 150 Celcius. If the original gas is @ pressure of 1520 torr, what is
Romashka [77]

Answer:

\large \boxed{\text{B.) 2.8 atm}}

Explanation:

The volume and amount are constant, so we can use Gay-Lussac’s Law:

At constant volume, the pressure exerted by a gas is directly proportional to its temperature.

\dfrac{p_{1}}{T_{1}} = \dfrac{p_{2}}{T_{2}}

Data:

p₁ = 1520 Torr; T₁ =   27 °C

p₂ = ?;               T₂ = 150 °C

Calculations:

(a) Convert the temperatures to kelvins

T₁ = (  27 + 273.15) K = 300.15 K

T₂ = (150 + 273.15) K = 423.15 K

(b) Calculate the new pressure

\begin{array}{rcl}\dfrac{1520}{300.15} & = & \dfrac{p_{2}}{423.15}\\\\5.064 & = & \dfrac{p_{2}}{423.15}\\\\5.064\times423.15&=&p_{2}\\p_{2} & = & \text{2143 Torr}\end{array}\\

(c) Convert the pressure to atmospheres

p = \text{2143 Torr} \times \dfrac{\text{1 atm}}{\text{760 Torr}} = \textbf{2.8 atm}\\\\\text{The new pressure reading will be $\large \boxed{\textbf{2.8 atm}}$}

7 0
3 years ago
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