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3 years ago

Why does sulfur have more than 8 electrons in its valence?

1 answer:

3 0

The octet rules says that an atom can only have a maximum of 8 electrons in its valence shell. This applies from non-metals to metalloids to metals. In this case, sulfur can have more than 8 electrons in its valence when it acquires a larger state specifically in an ionic state. For example SO4 2- where the 2- adds to the number of valence electrons

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In which structure does photosynthesis occur?

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Explanation:

kloroplast

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3 years ago

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A sample of Xe gas is observed to effuse through a pourous barrier in 4.83 minutes. Under the same conditions, the same number o

Solnce55 [7]

Answer:

28.93 g/mol

Explanation:

This is an extension of Graham's Law of Effusion where $\frac{R1}{R2} = \sqrt{\frac{M2}{M1} } = \frac{t2}{t1}$

We're only talking about molar mass and time (t) here so we'll just concentrate on $\sqrt{\frac{M2}{M1} } = \frac{t2}{t1}$ . Notice how the molar mass and time are on the same position, recall effusion is when gas escapes from a container through a small hole. The time it takes it to leave depends on the molar mass. If the gas is heavy, like Xe, it would take a longer time (4.83 minutes). If it was light it would leave in less time, that gives us somewhat an idea what our element could be, we know that it's atleast an element before Xenon.

Let's plug everything in and solve for M2. I chose M2 to be the unknown here because it's easier to have it basically as a whole number already.

$\sqrt{\frac{M2}{131} } = \frac{2.29}{4.83}$

The square root is easier to deal with if you take it out in the first step, so let's remove it by squaring each side by 2, the opposite of square root essentially.

$(\sqrt{\frac{M2}{131} } )^2= (\frac{2.29}{4.83})^2$

${\frac{M2}{131} } = (0.47)^2$

${\frac{M2}{131} } = 0.22$

M2= 0.22 x 131

M2= 28.93 g/mol

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2 years ago

1. Calculate the energy change (q) of the surroundings (water) using the enthalpy equation

Rasek [7]

Answer:

Q1: 728.6 J.

Q2:

a) 668.8 J.

b) 0.3495 J/g°C.

Explanation:

Q1: Calculate the energy change (q) of the surroundings (water) using the enthalpy equation:

The amount of heat absorbed by water = Q = m.c.ΔT.

where, m is the mass of water (m = d x V = (1.0 g/mL)(24.9 mL) = 24.9 g).

c is the specific heat capacity of liquid water = 4.18 J/g°C.

ΔT is the temperature difference = (final T - initial T = 32.2°C - 25.2°C = 7.0°C).

∴ The amount of heat absorbed by water = Q = m.c.ΔT = (24.9 g)(4.18 J/g°C)(7.0°C) = 728.6 J.

Q2: Calculate the energy change (q) of the surroundings (water) using the enthalpy equation

qwater = m × c × ΔT.

We can assume that the specific heat capacity of water is 4.18 J / (g × °C) and the density of water is 1.00 g/mL. calculate the specific heat of the metal. Use the data from your experiment for the unknown metal in your calculation.

a) First part: the energy change (q) of the surroundings (water):

The amount of heat absorbed by water = Q = m.c.ΔT.

where, m is the mass of water (m = d x V = (1.0 g/mL)(25 mL) = 25 g).

c is the specific heat capacity of liquid water = 4.18 J/g°C.

ΔT is the temperature difference = (final T - initial T = 31.6°C - 25.2°C = 6.4°C).

∴ The amount of heat absorbed by water = Q = m.c.ΔT = (25 g)(4.18 J/g°C)(6.4°C) = 668.8 J.

b) second part:

Q water = Q unknown metal.

Q unknown metal = - 668.8 J. (negative sign due to the heat is released from the metal to the surrounding water).

Q unknown metal = - 668.8 J = m.c.ΔT.

m = 27.776 g, c = ??? J/g°C, ΔT = (final T - initial T = 31.6°C - 100.5°C = - 68.9°C).

- 668.8 J = m.c.ΔT = (27.776 g)(c)( - 68.9°C) = - 1914 c.

∴ c = (- 668.8)/(- 1914) = 0.3495 J/g°C.

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3 years ago

Consider the following reversible reaction N2 + 3H2 = 2NH, In a 3 L container, the following amounts are found in equilibrium at

dolphi86 [110]

769km - grams 60000 *

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3 years ago

Lololoslajsiepejhwpa sejwjdalehe

Advocard [28]

Answer:

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Explanation:

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3 years ago