Dibujo de la estructura de lewis con las siguientes iones o moléculas Pbi4 y AgNO3NO

Consider the chemical equation. 2NBr3 + 3NaOH mc008-1.jpg N2 + 3NaBr + 3HOBr If there are 40 mol of NBr3 and 48 mol of NaOH, wha

Vladimir79 [104]

The correct answer o your amazing question is NBr3

6 0

3 years ago

Read 2 more answers

How does adaption allow mice with the mutation to survive better in some areas

Debora [2.8K]

Answer: Mutations can cause instant adaptations, while natural selection is the process by which adaptations occurs over a series of generations. Adaptations are changes or processes of changes by which an organism or species becomes better suited for its environment. A mutation is an alteration of the DNA sequence.

3 0

3 years ago

Copper(II) sulfide is formed when copper and sulfur are heated together. In this reaction,

storchak [24]

Answer:

The mass of copper(II) sulfide formed is:

= 81.24 g

Explanation:

The Balanced chemical equation for this reaction is :

$Cu(s) + S\rightarrow CuS$

given mass= 54 g

Molar mass of Cu = 63.55 g/mol

$Moles = \frac{given\ mass}{Molar\ mass}$

$moles=\frac{54}{63.55}$

Moles of Cu = 0.8497 mol

Given mass = 42 g

Molar mass of S = 32.06 g/mol

$Moles = \frac{given\ mass}{Molar\ mass}$

$moles=\frac{42}{32.06}$

Moles of S = 1.31 mol

Limiting Reagent : The reagent which is present in less amount and consumed in a reaction

First find the limiting reagent :

$Cu + S\rightarrow CuS$

1 mol of Cu require = 1 mol of S

0.8497 mol of Cu should require = 1 x 0.8497 mol

= 0.8497 mol of S

S present in the reaction Medium = 1.31 mol

S Required = 0.8497 mol

S is present in excess and Cu is limiting reagent

All Cu is consumed in the reaction

Amount Cu will decide the amount of CuS formed

$Cu + S\rightarrow CuS$

1 mole of Cu gives = 1 mole of Copper sulfide

0.8497 mol of Cu = 1 x 0.8497 mole of Copper sulfide

= 0.8497

Molar mass of CuS = 95.611 g/mol

$Moles = \frac{given\ mass}{Molar\ mass}$

$0.8497 = \frac{given\ mass}{95.611}$

Mass of CuS = 0.8497 x 95.611

= 81.24 g

3 0

4 years ago

The density of potassium, which has the BCC structure, is 0.855 g/cm3. The atomic weight of potassium is 39.09 g/mol. Calculate

Nimfa-mama [501]

Answer:

lattice parameter = 5.3355x10^-8 cm

atomic radius = 2.3103x10^-8 cm

Explanation:

known data:

p=0.855 g/cm^3

atomic mass = 39.09 g/mol

atoms/cell = 2 atoms

Avogadro number = 6.02x10^23 atom/mol

a) the lattice parameter:

Since potassium has a cubic structure, its volume is equal to:

v = [(atoms/cell)x(atomic mass)/(p)x(Avogadro number)]

substituting values:

v =[(2)x(39.09)/(0.855x6.02x10^23)]=1.5189x10^-22 cm^3

but as the cell volume is

a^3 =v

$a=\sqrt[3]{v}=\sqrt[3]{1.5189x10^{-22} } = 5.3355x10^-8$ cm

for a BCC structure, the atomic radius is equal to

$r=\frac{ax\sqrt{3} }{4}=\frac{5.3355x10^{-8}x\sqrt{3} }{4}=2.3103x10^{-8}cm$

7 0

3 years ago

What form of energy does a compressed spring have? A) heat energy B) kinetic energy C) chemical energy D) potential energy

Akimi4 [234]

Answer:

D) Potential Energy

Explanation:

The compressed spring has elastic potential energy. If it is allowed to de-compress, the elastic potential energy will be converted to kinetic energy.

8 0

3 years ago

Read 2 more answers

Dibujo de la estructura de lewis con las siguientes iones o moléculas Pbi4 y AgNO3NO​

Dibujo de la estructura de lewis con las siguientes iones o moléculas Pbi4 y AgNO3NO