What are the real primary colors

The density of concentrated ammonia, which is 28.0% w/w nh3, is 0.899 g/ml. what volume of this reagent should be diluted to 1.0

vlada-n [284]

Answer: 2.4 ml

Solution :

Molar mass of $NH_3$ = 17 g/mole

Given,: 28% w/w of $NH_3$ solution means 28 g of ammonia in 100 g of solution.

Mass of solution = 100 g

Now we have to calculate the volume of solution.

$Volume=\frac{Mass}{Density}=\frac{100g}{0.899g/ml}=111.2ml$

Molarity : It is defined as the number of moles of solute present in one liter of solution.

$Molarity=\frac{n\times 1000}{V_s}$

where,

n = moles of solute $NH_3=\frac{\text {given mass}}{\text {molar mass}}=\frac{28}{17}=1.65moles$

$V_s$ = volume of solution in liter = 0.11 L

Now put all the given values in the formula of molarity, we get

$Molarity=\frac{1.65moles}{0.11L}=15mole/L$

Using molarity equation:

$M_1V_1=M_2V_2$

$15\times V_1=0.036\times 1.0\times 10^{3}$

$V_1=2.4ml$

6 0

3 years ago

What hybridization is required for central atoms that have a tetrahedral arrangement of electron pairs? A trigo- nal planar arra

Reika [66]

Answer:

sp³;

sp²;

sp;

None;

One;

Two;

They're used to pi bonds.

Explanation:

The central atom in a molecule is generally the one that can make a greater number of bonds. The covalent bonds are made by the sharing of electrons, and, for that, the electron must be alone in the orbital.

To explain this, the hybridization theory was created, which states that, the orbitals are joined to form hybrids ones, and so, by the Hund's law, the electrons are alone in them.

The sigma bonds are done in the hybrids orbitals, and at the pure orbitals, the pi bonds are done. The lone pair of electrons are at a pure orbital. So, to know the hybridization of the central atom, we must know how many sigma bonds it does, and it will be the number of hybrids orbitals (each orbital may have two electrons, thus each bond are done in one orbital).

Double bonds and triple bonds have always only one sigma bond, so the number of sigma bonds is equal to the number of bonds, it's not necessary to know if they are simple, double or triple.

When the arrangement is tetrahedral, the central atom does 4 bonds, so it has 4 sigma bonds, and 4 hybrids orbitals (one of s and three for p), does its hybridization is sp³. Because exists only 3 p orbitals, there are no unhybridized p orbitals in this case.

When the arrangement is trigonal, the central atom does 3 bonds, so it has 3 hybrids orbitals (one of s and two of p), thus the hybridization is sp². So there are one unhybridized p orbitals.

When the arrangement is linear, the central atom does 2 bonds, so it has 2 hybrids orbitals (one of s and one of p), thus the hybridization is sp. So, there are two unhybridized p atoms.

As stated before, the unhybridized p orbitals are used to pi bonds.

5 0

3 years ago

Whoever can answer this I will give all of my points to and brainliest award.

ZanzabumX [31]

i dont see the picture.

7 0

3 years ago

Read 2 more answers

How much positive charge is in 0.7 kg of lithium? with each atom having 3 protons and 3 electrons. The elemental charge is 1.602

hichkok12 [17]

Explanation:

As a neutral lithium atom contains 3 protons and its elemental charge is given as $1.602 \times 10^{-19} C$ . Hence, we will calculate its number of moles as follows.

Moles = $\frac{mass}{\text{molar mass}}$

= $\frac{0.7 \times 1000 g}{7 g/mol}$

= 100 mol

According to mole concept, there are $6.023 \times 10^{23}$ atoms present in 1 mole. So, in 100 mol we will calculate the number of atoms as follows.

No. of atoms = $100 \times 6.023 \times 10^{23}$

= $6.023 \times 10^{25}$ atoms

Since, it is given that charge on 1 atom is as follows.

$3 \times 1.602 \times 10^{-19}C$

= $4.806 \times 10^{-19}C$

Therefore, charge present on $6.023 \times 10^{25}$ atoms will be calculated as follows.

$6.023 \times 10^{25} atoms \times 4.806 \times 10^{-19} C$

$28.95 \times 10^{6}C$

Thus, we can conclude that a positive charge of $28.95 \times 10^{6}C$ is in 0.7 kg of lithium.

3 0

3 years ago

Which of the following would release the most heat? Assume the same mass of in each case. Specific heats of ice, liquid water, a

lesya692 [45]

Answer:

The process which releases most heat is E)

Explanation:

As we know that water freezes at 0ºC and vaporizes at 100ºC, we calculate the heat as follows:

Processes with temperatures < 0ºC : by using specific heat of ice (Sh ice) multiplied by the change in temperature (ΔT= Final Temperature - Initial Temperature)⇒ Sh ice x ΔT

Processes of ice melting (at 0ºC): by using heat of fusion of ice (ΔH fus) multiplied by a conversor factor (1 mol H20= 18 g)⇒ ΔHfus x 1mol/18g

Processes between 0ºC and 100ºC: by using specific heat of liquid water (Sh liq) multiplied by change in temperature ⇒ Sh liq x ΔT

Processes of water evaporation (at 100ºC): by using heat of vaporization (ΔH vap) multiplied by the conversor factor ⇒ ΔH vap x 1mol/18 g

Processes at a temperature >100ºC: by using specific heat of water vapor (Sh vap) multiplied by the change in temperature ⇒ Sh vap x ΔT

A) Water at -25ºC is ice. Ice is heated from -25ºC to 0ºC, then it melts at 0ºC (ice became liquid water) and then liquid water is heated from 0ºC to 70ºC. T

This is the only process in with the heat is absorbed (not releases), so it cannot be the right answer, but we calculate the heat involved to practice:

Heat= (Sh ice x ΔT) + (ΔH fus x 1/18 g) + Sh liq x ΔT

Heat= (2.05 J/g ºC x (0ºC -(-25ºC) ) + (6.01 x 10³ J/mol x 1 mol/18 g) + (4.18 J/g ºC x (70ºC-0ºC)

Heat= 51.25 J + 333,8 J +292.6 J

Heat= 677.65 J (heat is absorbed)

B) Water is cooled from 13ºC to 0ºC, then it is freezed at 0ºC and then the ice is cooled from 0ºC to -2.6 ºC

Heat= (Sh liq x ΔT) + (-ΔH melt x 1/18 g) + (Sh ice x ΔT)

Heat= 4.18 J/ºC x (0ºC- 13ºC) + (-6.01 x 10³ J/mol x 1mol/18 g) + (2.05 J/ºC x (-2.5ºc - 0ºC)

Heat= -54.34 J - 333.8 J + 5.33 J

Heat= -393.47 J (heat is released)

C) Liquid water is cooled from 74ºC to 95ºC

Heat= Sh liq x ΔT

Heat= 4.18 J/ºC x (74ºC - 95ºC)

Heat = -87.78 J (heat is released)

D) Water at 140ºC is in vapor state. Vapor at 140ºC is cooled to 110ºC (still vapor).

Heat = Sh vap x ΔT

Heat= 2.01 J/ºC x (110ºC - 140ºC)

Heat= -60.3 J (heat is released)

E) Vapor at 106ºC is cooled to 100ºC, then it condenses at 100ºC (convertion from gas to liquid), and liquid water is cooled from 100ºC to 88ºC.

Heat= (Sh vap x ΔT) + (-ΔHvap x 1mol/18g) + (Sh liq x ΔT)

Heat= (2.01 J/ºC x (100ºC-106ºC)) - (40.7 x 10³ J/mol x 1mol/18 g) + (4.18 J/ºC x (88ºC -100ºC)

Heat= -2323.32 J (heat is released) <u>THIS IS THE RIGHT ANSWER</u> (the more negative= the more released)

7 0

3 years ago

What are the real primary colors ​

What are the real primary colors