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-Dominant- [34]
3 years ago
6

Imagine the earth's hydrosphere began to dry up and disappear. all of the following would be affected except

Chemistry
2 answers:
RoseWind [281]3 years ago
6 0
Answer most suited would be A
katen-ka-za [31]3 years ago
3 0

A is right. I just took the test on plato.

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A reaction that is NOT thermodynamically favored at low temperatures can become thermodynamically favored at high temperatures
Rom4ik [11]
Explanation:

Its gonna be D
7 0
3 years ago
Which fire symbol signifies ammunition with a mass explosion?
marissa [1.9K]

The answer is Fire symbol 1  

That is Fire symbol 1, signifies ammunition with a mass explosion.

The symbol can be seen in the image attached.

This symbol includes these :

1. Do not use fire fighting unless rescue attempt is planned.

2. If person's safety is in danger, use cover.

This fire symbol 1 signifies ammunition with a mass explosion.


4 0
3 years ago
A helium filled ballon had a volume of 8.50 L on the ground at 20.0 C and a pressure of 750.0 Torr. After the ballon was release
Marrrta [24]

Answer:

V_2=12.1L

Explanation:

Hello!

In this case, according to the given data of volume, pressure and temperature, it is possible to infer this problem can be solved via the combined gas law:

\frac{P_1V_1}{T_1} =\frac{P_2V_2}{T_2}

Thus, regarding the question, we evidence we need V2, but first we make sure the temperatures are in Kelvins:

T_1=20+273=293K\\\\T_2=-40+273=233K

Then, we obtain:

V_2=\frac{P_1V_1T_2}{T_1P_2}\\\\V_2=\frac{0.987atm*8.50L*233K}{293K*0.550atm}\\\\V_2=12.1L

Best regards!

5 0
3 years ago
CH4 and CO₂ are the chemical formulas for:
nadya68 [22]

Answer:

CH4 = Methane

CO2 = Carbon dioxide

<h3>Hope this Helps!!</h3><h3>Hope this Helps!!:)</h3>
6 0
3 years ago
Calcule la densidad del hidrógeno H2 en g/L a 327 mm Hg y 48ºc
tankabanditka [31]
<h3>The density of H₂ = 0.033 g/L</h3><h3>Further explanation</h3>

In general, the gas equation can be written  

\large {\boxed {\bold {PV = nRT}}}

where  

P = pressure, atm , N/m²

V = volume, liter  

n = number of moles  

R = gas constant = 0.082 l.atm / mol K (P= atm, v= liter),or 8,314 J/mol K (P=Pa or N/m², v= m³)

T = temperature, Kelvin  

n = N / No  

n = mole  

No = Avogadro number (6.02.10²³)  

n = m / MW

m = mass  

MW = molecular weight

For density , can be formulated :

\tt \rho=\dfrac{P\times MW}{R\times T}

P = 327 mmHg = 0,430263 atm

R = 0.082 L.atm / mol K

T = 48 ºC = 321.15 K

MW of H₂ =  2.015 g/mol

The density :

\rho=\dfrac{0,430263\times 2.015 }{0.082\times 321.15}\\\\\rho=0.033~g/L

4 0
2 years ago
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