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3 years ago

Light acts like a

Chemistry

2 answers:

8_murik_8 [283]3 years ago

7 0

Particle
wave are the words you're looking for to fill in the blanks

lukranit [14]3 years ago

5 0

Answer:

Light acts like a <u><em>particle</em></u> when it bounces off surface and acts like a <u><em>wave </em></u>when it bends around objects

Explanation:

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1.42 mol sample of neon gas at a temperature of 13.0 °C is found to occupy a volume of 25.5 liters. The pressure of this gas sam

Dima020 [189]

Answer: 996 mmHg

Explanation:

According to avogadro's law, 1 mole of every substance occupies 22.4 L at NTP, weighs equal to the molecular mass and contains avogadro's number $6.023\times 10^{23}$ of particles.

According to the ideal gas equation:

$PV=nRT$

P = Pressure of the gas = ?

V= Volume of the gas = 25.5 L

T= Temperature of the gas = 13°C = (273+13) K = 286K

R= Gas constant = 0.0821 atmL/K mol

n= moles of gas= 1.42

$P=\frac{nRT}{V}=\frac{1.42\times 0.0821\times 286}{25.5}=1.31atm=996mmHg$ (760mmHg=1atm)

Thus pressure of this gas sample is 996 mm Hg.

3 0

3 years ago

One reaction involved in the conversion of iron ore to the metal is FeO(s) + CO(g) → Fe(s) + CO2(g) Use Hess’s Law to calculate

Ugo [173]

Answer:

$\delta H_{rxn} = -66.0 \ kJ/mole$

Explanation:

Given that:

$3FeO_3_{(s)}+CO_{(g)} \to 2Fe_3O_4_{(s)} +CO_{2(g)} \ \ \delta H = -47.0 \ kJ/mole -- equation (1) \\ \\ \\ Fe_2O_3_{(s)} +3CO_{(g)} \to 2FE_{(s)} + 3CO_{2(g)} \ \ \delta H = -25.0 \ kJ/mole -- equation (2) \\ \\ \\ Fe_3O_4_{(s)} + CO_{(g)} \to 3FeO_{(s)} + CO_{2(g)} \ \delta H = 19.0 \ kJ/mole -- equation (3)$

From equation (3) , multiplying (-1) with equation (3) and interchanging reactant with the product side; we have:

$3FeO_{(s)} + CO_{2(g)} \to Fe_3O_4_{(s)} + CO_{(g)} \ \delta H = -19.0 \ kJ/mole -- equation (4)$

Multiplying (2) with equation (4) ; we have:

$6FeO_{(s)} + 2CO_{2(g)} \to 2Fe_3O_4_{(s)} + 2CO_{(g)} \ \delta H = -38.0 \ kJ/mole -- equation (5)$

From equation (1) ; multiplying (-1) with equation (1); we have:

$2Fe_3O_4_{(s)} +CO_{2(g)} \to 3FeO_3_{(s)}+CO_{(g)} \ \ \delta H = 47.0 \ kJ/mole -- equation (6)$

From equation (2); multiplying (3) with equation (2); we have:

$3 Fe_2O_3_{(s)} +9CO_{(g)} \to 6FE_{(s)} + 9CO_{2(g)} \ \ \delta H = -75.0 \ kJ/mole -- equation (7)$

Now; Adding up equation (5), (6) & (7) ; we get:

$6FeO_{(s)} + 2CO_{2(g)} \to 2Fe_3O_4_{(s)} + 2CO_{(g)} \ \delta H = -38.0 \ kJ/mole -- equation (5)$

$2Fe_3O_4_{(s)} +CO_{2(g)} \to 3FeO_3_{(s)}+CO_{(g)} \ \ \delta H = 47.0 \ kJ/mole -- equation (6)$

$3 Fe_2O_3_{(s)} +9CO_{(g)} \to 6FE_{(s)} + 9CO_{2(g)} \ \ \delta H = -75.0 \ kJ/mole -- equation (7)$

$FeO \ \ \ + \ \ \ CO \ \ \to \ \ \ \ Fe_{(s)} + \ \ CO_{2(g)} \ \ \ \delta H = - 66.0 \ kJ/mole$

<u />

$\delta H_{rxn} = \delta H_1 + \delta H_2 + \delta H_3$ (According to Hess Law)

$\delta H_{rxn} = (-38.0 + 47.0 + (-75.0)) \ kJ/mole$

$\delta H_{rxn} = -66.0 \ kJ/mole$

8 0

3 years ago

Differentiate between nuclear charge and effective nuclear charge

ipn [44]

hope it helps ..............

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3 years ago

True or false atoms with fewer then 4 outer elections lend elections?

Mamont248 [21]

The answer is false ....

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3 years ago

1. The pressure and temperature of a gas are held constant. Which of the following is true for the volume of the gas?

katovenus [111]

C) The volume of the gas is proportional to the number of moles of gas particles.

The Avogadro's law applies to ideal gases with constant pressure and temperature. By that law, the volume of an ideal gas is proportional to the number of moles of particles in that gas.

B) The gas now occupies less volume, and the piston will move downward.

Boyle's Law applies to ideal gases with a constant temperature. The volume of an ideal gas is inversely related to its pressure. A high pressure drives gas particles together, such that they occupy less volume. The gas trapped inside the piston has a smaller volume. As a result, the the piston will move downward.

Alternatively, consider the forces acting on the piston. Both the atmosphere and gravity are dragging the piston down. In order for it to stay in place, the gas below it must exert a pressure to balance the two forces. Now the pressure from outside has increased. The gas inside needs to increase its pressure. It needs a smaller volume to create that extra pressure. As a result, its volume will decrease, and the piston will move downwards.

6 0

3 years ago

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