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timofeeve [1]
3 years ago
5

The transfer of energy from objects with higher temperatures to objects with lower temperatures is defined as which of the follo

wing? dispersal heat subduction gradient transfer
Chemistry
1 answer:
timofeeve [1]3 years ago
3 0
<span>All matter contain heat energy, and it can be seen anywhere as long as it contains mass and occupies space. Heat energy is another form energy that transfers particles in a substance through kinetic energy. One way to transfer heat from one object to the other is through the difference in temperature between two objects.</span>
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A gas occupies 25.3 at pressure of 790.5 mmHg. Determine the volume if the pressure is reduced to 8.04 mmHg
taurus [48]

Answer:

2487.51.

Explanation:

As per Boyle's law temperature remaining constant the volume of an ideal gas is inversely proportional to its pressure.

pV= k

therefore, p1V1 = p2V2

here V1 = 25.3, p2 = 8.04mm Hg

pressure p1 = 790.5 mm Hg

this means that

25.3×790.5 = 8.04V2

⇒V2= 2487.51

Hence, the required volume is, 2487.51.

5 0
3 years ago
Brainlist for best answer!!! <br> Why will the whole ecosystem be affected by just one change?
Naddika [18.5K]

Answer:

Critical coordinate drivers incorporate living space alter, climate alter, obtrusive species, over exploitation, and contamination. Most of the coordinate drivers of corruption in biological systems and biodiversity right now stay steady or are developing in escalated in most biological systems.

Explanation:

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3 years ago
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Which gas moves the fastest at STP? (Molar Mass: CO2=44, NO2=46, NH3=17, H2S=34)
Marina86 [1]

C

Sana makatulong answer ko

4 0
3 years ago
Use Q = mcAT
mr Goodwill [35]

Answer:

1).....for the specific heat capacity(c) of water is 4200kg/J°C..

....guven mass(m)=320g(0.32kg)

...change in temperature(ΔT) =35°C

from the formula

Q=mcΔT

Q=0.32Kg x 4200kg/J°C x 35°C

Q=47,040Joules

5 0
3 years ago
What volume (mL) of the partially neutralized stomach acid was neutralized by NaOH during the titration? (portion of 25.00 mL sa
almond37 [142]

The question is incomplete, here is the complete question:

What volume (mL) of the partially neutralized stomach acid having concentration 2 M was neutralized by 0.1 M NaOH during the titration? (portion of 25.00 mL NaOH sample was used; this was the HCl remaining after the antacid tablet did it's job)

<u>Answer:</u> The volume of HCl neutralized is 1.25 mL

<u>Explanation:</u>

To calculate the volume of acid, we use the equation given by neutralization reaction:

n_1M_1V_1=n_2M_2V_2

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of stomach acid which is HCl

n_2,M_2\text{ and }V_2 are the n-factor, molarity and volume of base which is NaOH.

We are given:

n_1=1\\M_1=2M\\V_1=?mL\\n_2=1\\M_2=0.1M\\V_2=25mL

Putting values in above equation, we get:

1\times 2\times V_1=1\times 0.1\times 25\\\\V_1=\frac{1\times 0.1\times 25}{1\times 2}=1.25mL

Hence, the volume of HCl neutralized is 1.25 mL

3 0
3 years ago
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