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4 years ago

A transfer of heat within a liquid or gas that involves warm particles moving in currents is

Chemistry

2 answers:

storchak [24]4 years ago

5 0

The answer is Convection

Convection, which typically occurs within liquids and gases, is a transfer of heat that involves warm particles moving in currents. For example, when a liquid solution is heated, the hottest part of the liquid will rise. As the hot liquid rises, it transfers heat to the top of the solution.

Dominik [7]4 years ago

4 0

I believe the best answer is C. Convection

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Given a 1.50 Liters of solution that contains 62.5 grams of magnesium oxide. What is the Molarity of that solution? Please show

zhenek [66]

Given:
1.50 L
62.5 grams
and the MM of MgO: 40.31 g/mol

Molarity: mol/L

First, find mol.

62.5 g x 1mole ÷ 40.31 g = 1.55 mol

then divide mol and the given liters

1.55mol ÷ 1.50 L= 1.03 M

3 0

3 years ago

An ideal gas sealed in a rigid 4.86-L cylinder, initially at pressure Pi=10.90 atm, is cooled until the pressure in the cylinder

seraphim [82]

Answer:

$\Delta H=-11897J$

Explanation:

Hello,

In this case, it is widely known that for isochoric processes, the change in the enthalpy is computed by:

$\Delta H=\Delta U+V\Delta P$

Whereas the change in the internal energy is computed by:

$\Delta U=nCv\Delta T$

So we compute the initial and final temperatures for one mole of the ideal gas:

$T_1= \frac{P_1V}{nR}=\frac{10.90atm*4.86L}{0.082*n}=\frac{646.02K }{n} \\\\T_2= \frac{P_2V}{nR}=\frac{1.24atm*4.86L}{0.082*n}=\frac{73.49K }{n}$

Next, the change in the internal energy, since the volume-constant specific heat could be assumed as ³/₂R:

$\Delta U=1mol*\frac{3}{2} (8.314\frac{J}{mol*K} )*(73.49K-646.02K )=-7140J$

Then, the volume-pressure product in Joules:

$V\Delta P=4.86L*\frac{1m^3}{1000L} *(1.24atm-10.90atm)*\frac{101325Pa}{1atm} \\\\V\Delta P=-4756.96J$

Finally, the change in the enthalpy for the process:

$\Delta H=-7140J-4757J\\\\\Delta H=-11897J$

Best regards.

7 0

4 years ago

Iridium has two naturally occurring isotopes, Iridium-191 and Iridium-193. If the average atomic mass of orodum s 192.217, what

Lina20 [59]

Answer:

The percent isotopic abundance of Ir-193 is 60.85 %

The percent isotopic abundance of Ir-191 is 39.15 %

Explanation:

we know there are two naturally occurring isotopes of iridium, Ir-191 and Ir-193

First of all we will set the fraction for both isotopes

X for the isotopes having mass 193

1-x for isotopes having mass 191

The average atomic mass is 192.217

we will use the following equation,

193x + 191(1-x) = 192.217

193x + 191 - 191x = 192.217

193x- 191x = 192.217 - 191

2x = 1.217

x= 1.217/2

x= 0.6085

0.6085 × 100 = 60.85 %

60.85% is abundance of Ir-193 because we solve the fraction x.

now we will calculate the abundance of Ir-191.

(1-x)

1-0.6085 =0.3915

0.3915× 100= 39.15 %

6 0

3 years ago

635 mL of a gas is at a pressure of 8.00 atm. What is the volume of the gas at standard pressure (STP)?

Sveta_85 [38]

The new volume at standard temperature and pressure is 5.08 L.

Explanation:

As per the kinetic theory of gases, the volume occupied by gas molecules will be inversely proportional to the pressure of the gas molecules. This is termed as Boyle's law.

So, pressure∝ $\frac{1}{volume}$

Thus, if two pressure and two volumes are given then,

$P_{1} V_{1} = P_{2} V_{2}\\$

Now, we known the values of P₁ = 8 atm, V₁ = 635 mL, P₂ = 1 atm and V₂ we have to determine. We are considering P₂ = 1 atm, because we have to determine V₂ at standard temperature and pressure. And standard pressure is 1 atm.

$8*635*10^{-3} = 1 *V{2} \\\\V_{2} = \frac{8*635*10^{-3} }{1} = 5.08 L$

Thus, the new volume at standard temperature and pressure is 5.08 L.

5 0

3 years ago

Classify the interactions as being direct or indirect competition.

spin [16.1K]

Answer:

.-...

Explanation:

4 0

3 years ago

Read 2 more answers