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2 years ago

What is the Moleular geometry of Xe0F4

Chemistry

1 answer:

Rudiy272 years ago

3 0

Answer:

Central atom: Xe

Total VSEP: 7

1 x double bond − 1 pair

Revised Total: 6

Geometry: Square pyramidal (based on octahedral)Explanation:

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How does nuclear reactor produce electricty

taurus [48]

With a process called Fission. This generates heat to produce steam.

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3 years ago

Which best explains how the body maintains homeostasis?

MissTica

B. each system works independently to stabilize the body

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2 years ago

A flask with a volume of 3.16 l contains 9.33 grams of an unknown gas at 32.0°c and 1.00 atm. What is the molar mass of the gas?

Inessa05 [86]

Answer:

73.88 g/mol

Explanation:

For this question we have to keep in mind that the unknown substance is a gas, therefore we can use the ideal gas law:

$PV=nRT$

In this case we will have:

P= 1 atm

V= 3.16 L

T = 32 ªC = 305.15 ºK

R= 0.082 $\frac{atm*L}{mol*K}$

n= ?

So, we can solve for "n" (moles):

$1~atm*3.16~L~=~n*0.082~\frac{atm*L}{mol*K}*305.15~K$

$n=\frac{1~atm*3.16~L~}{0.082~\frac{atm*L}{mol*K}*305.15~K}$

$n=0.126~mol$

Now, we have to remember that the molar mass value has "g/mol" units. We already have the grams (9.33 g), so we have to divide by the moles:

$molar~mass=\frac{9.33~grams}{0.126~mol}$

$molar~mass=73.88\frac{grams}{mol}$

7 0

2 years ago

Catalyst

Hoochie [10]

a. volume of NO : 41.785 L

b. mass of H2O : 18 g

c. volume of O2 : 9.52 L

<h3>Further explanation</h3>

Given

Reaction

4 NH₃ (g) + 5 O2 (g) → 4 NO (g) + 6 H2O (l)

Required

a. volume of NO

b. mass of H2O

c. volume of O2

Solution

Assume reactants at STP(0 C, 1 atm)

Products at 1000 C (1273 K)and 1 atm

a. mol ratio NO : O2 from equation : 4 : 5, so mo NO :

$\tt \dfrac{4}{5}\times 0.5=0.4$

volume NO at 1273 K and 1 atm

$\tt V=\dfrac{nRT}{P}=\dfrac{0.4\times 0.08206\times 1273}{1}=41.785~L$

b. 15 L NH3 at STP ( 1mol = 22.4 L)

$\tt \dfrac{15}{22.4}=0.67~mol$

mol ratio NH3 : H2O from equation : 4 : 6, so mol H2O :

$\tt \dfrac{6}{4}\times 0.67=1$

mass H2O(MW = 18 g/mol) :

$\tt mass=mol\times MW=1\times 18=18~g$

c. mol NO at 1273 K and 1 atm :

$\tt n=\dfrac{PV}{RT}=\dfrac{1\times 35.5}{0.08206\times 1273}=0.34$

mol ratio of NO : O2 = 4 : 5, so mol O2 :

$\tt \dfrac{5}{4}\times 0.34=0.425$

Volume O2 at STP :

$\tt 0.425\times 22.4=9.52~L$

5 0

3 years ago

Use the periodic table in the tools bar to answer these questions. How many moles of AgNO3 are present in 1.50 L of a 0.050 M so

IgorC [24]

M= moles de soluto / litros de solucion
moles de soluto = M. litros de solucion

Moles de soluto = 0.050 M x 1.50 L = 0.075 moles de AgNo3

4 0

3 years ago

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