What conditions prevail (are mostly found) in the rain shadow area?

1. Metallic strontium crystallizes in a face-centered cubic lattice, with one Sr atom per lattice point. If the edge length of t

Zarrin [17]

Answer:

$r=215pm$

$N_{Mn}=20$

Explanation:

From the question we are told that:

Edge length of the unit cell $l=608pm$

a)

Generally the equation for The relationship between edge length and radius is mathematically given by

$4r=\sqrt{2a}$

Therefore

$4r=\sqrt{2*608}$

$r=\frac{\sqrt{2*608}}{4}$

$r=215pm$

b)

From the question we are told that:

Density $\rho=7.297$

Edge length of $l=630.0 pm=>630*10^-{10}$

Therefore Volume is given as

$V=l^3$

$V=630*10^-{10}^3$

$V=2.50047*10^{−22}$

Generally the equation for Mass is mathematically given by

$m=Volume*density$

$m=V*\rho$

$m=2.50047*10^{−22}*7.297$

$m=1.83*10^{-21}g$

Therefore Molarity is given as

$n=\frac{M}{Molar M}$

$n=\frac{1.83*10^{-21}g}{55}$

$n=3.32*10^{-23}$

Finally The atoms in a unit cell is

$N_{Mn}=Moles*Avogadro\ constant$

$N_{Mn}=3.32*10^{-23}*6.023*10^{23}$

$N_{Mn}=20$

7 0

3 years ago

For which of the following processes would S° be expected to be most positive? a) O2(g) + 2H2(g)  2H2O(g) b) H2O(l)  H2O(s) c

vaieri [72.5K]

Answer:

Explanation:

a) O₂(g) + 2H₂(g) = 2H₂O

b) H₂O(l) = H₂O(s)

c) NH₃(g) + HCl(g) = NH₄Cl(g)

d) 2NH₄NO₃(s) = 2N₂(g) + O₂(g) + 4H₂O(g)

e) N₂O₄(g) = 2NO₂(g)

ΔS is positive when there is increase in disorderliness. It happens when there is increase in volume .

Increase is volume is maximus in the following reaction.

d) 2NH₄NO₃(s) = 2N₂(g) + O₂(g) + 4H₂O(g)

in this reaction solid NH₄NO₃ is changed to 7 x 22.4 L of gases so there is maximum increase in volume . Hence maximum increase in entropy . Hence ΔS is most positive .

6 0

3 years ago

The decomposition of water produces 50g of Hydrogen gas and 75 g of oxygen gas. How much water

Lostsunrise [7]

84.24 g of water (H₂O)

Explanation:

We have the following chemical reaction:

2 H₂O → 2 H₂ + O₂

Now we calculate the number of moles of products.

number of moles = mass / molar weight

number of moles of H₂ = 50 / 2 = 25 moles

number of moles of O₂ = 75 / 32 = 2.34 moles

We see from the chemical reaction that for every 2 moles of H₂ produced there are 1 mole of O₂ produces for every 25 moles of H₂ produced there are 12.5 moles of O₂ but we only have 2.35 moles of O₂ available. The O₂ will be the limiting quantity from which we devise the following reasoning:

if 2 moles of H₂O produces 1 mole of O₂

then X moles of H₂O produces 2.34 mole of O₂

X = (2 × 2.34) / 1 = 4.68 moles of H₂O

mass = number of moles × molar weight

mass of H₂O = 4.68 × 18 = 84.24 g

Learn more about:

limiting reactant

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4 0

3 years ago

Use the heating curve of 60.0 grams of ice and the list of values to answer the question.

Vesna [10]

Answer:

25kj fdgsgdgdhdhfhfgfhfhfjbk

6 0

3 years ago

True or False: A liquid's volume does not change no matter what the shape of its container.

allsm [11]

Answer:

true

Explanation: becuase if you add the same liquid to a small container and then add it to a big container its going to have the same liquid amount it started with

3 0

3 years ago