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cluponka [151]
3 years ago
6

What might a physicit choose to study?

Chemistry
1 answer:
AlekseyPX3 years ago
7 0
Thermodynamics, Nuclear Physics, Quantum Physics, Astronomy and Astrophysics
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Consider the balanced equation:
Mamont248 [21]
The answer is 6 moles of water will be produced.
3 0
3 years ago
Read 2 more answers
How many 3-liter balloons could the 12-L helium tank pressurized to 160 atm fill? Keep in mind that an "exhausted" helium tank i
vaieri [72.5K]

Answer:

The 12L helium tank pressurized to 160 atm will fill <em>636 </em>3-liter balloons

Explanation:

It is possible to answer this question using Boyle's law:

P_1V_1=P_2V_2

Where P₁ is the pressure of the tank (160atm), V₁ is the volume of the tank (12L), P₂ is the pressure of the balloons (1atm, atmospheric pressure) And V₂ is the volume this gas will occupy at 1 atm, thus:

160atm×12L = 1atm×V₂

V₂ = 1920L

As the tank will never be empty, the volume of the gas able to fill balloons is the total volume minus 12L, thus the volume of helium able to fill balloons is:

1920L - 12L = 1908L

1908L will fill:

1908L×\frac{1balloon}{3L} = <em>636 balloons</em>

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I hope it helps!

7 0
3 years ago
1. A chemical equation is balanced when *
lubasha [3.4K]

Answer:1 Answer. A chemical equation is balanced when the number of each kind of atom is the same on both sides of the reaction,,

6 0
2 years ago
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Will you be my Magnesium, Silver, Nitrogen but "9.8m/s²" is silent? <br> Please help me here
zmey [24]
I don’t really understand but the symbols of these elements on the periodic table are

MgAgN

and 9.8 m/s^2 is the acceleration of gravity, commonly abbreviated as ‘g’

If you take out the ‘g’s from the periodic table equation(bc the question says 9.8m/s^2 is silent) it spells MAN

So the question could be “Will you be my man?”
4 0
2 years ago
The half-life of Pa-234 is 6.75 hours. If a sample of Pa-234 contains 112.0 g, 1 point
umka2103 [35]

Answer:

28g remain after 13.5 hours

Explanation:

Element decayment follows first order kinetics law:

ln[Pa-234] = -kt + ln [Pa-234]₀ <em>(1)</em>

<em>Where [Pa-234] is concentration after t time, k is rate constant in time, and [Pa-234]₀ is initial concentration</em>

Half-life formula is:

t_{1/2} =  \frac{ln2}{k}

6.75 = ln2 / k

<em>k = 0.1027hours⁻¹</em>

Using rate constant in (1):

ln[Pa-234] = -0.1027hours⁻¹×13.5hours + ln [112.0g]

ln[Pa-234] = 3.332

[Pa-234] = <em>28g after 13.5 hours</em>

<em />

5 0
2 years ago
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