Give the name of the products formed
between Benzene and the following Compounds
Br2,Cl2,HNO3,H2SO4,HCOCl
1-chlore-2,2-dimethylpropane

By using combined gas law , volume will be 1.88 [tex]dm^{3}[/tex].

The combined gas law contains the laws of Charles, Boyle, and Gay-Lussac. When you know the pressure, volume, and temperature of a system and need to modify one of these, you can apply this law to compute the resulting unknown.

Combined gas law is expressed as:

[tex]$\frac{P_{1} V_{1}}{T_{1}}=\frac{P_{2} V_{2}}{T_{2}}$[/tex]

Calculation of volume by using combined gas law

It is given that, [tex]V_{1}[/tex] = 254[tex]dm^{3}[/tex], at STP P = 101.3 kPa and T = 273 K

Now, put the value of given data in above equation.

[tex]V_{2}[/tex] = (254[tex]dm^{3}[/tex]) (101.3 kPa) (26+273) /(150 kPa) (273)

[tex]V_{2}[/tex] = 1.88 [tex]dm^{3}[/tex].

By using combined gas law  volume will be 1.88 [tex]dm^{3}[/tex].

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Answer:

780 g

Explanation:

Step 1: Given data

Moles of sodium phosphate: 4.76 moles

Step 2: Calculate the mass corresponding to 4.76 moles of sodium phosphate

To convert moles to mass, we need a conversion factor. In this case, we will use the molar mass of sodium phosphate, which is 163.94 g/mol.

4.76 mol × 163.94 g/mol = 780 g

Answer:

10.5 L of O₂.

Explanation:

We'll begin by calculating the number of mole in 90 g of C₂H₆. This can be obtained as follow:

Molar mass of C₂H₆ = (12×2) + (6×1)

= 24 + 6

= 30 g/mol

Mass of C₂H₆ = 90 g

Mole of C₂H₆ =?

Mole = mass / molar mass

Mole of C₂H₆ = 90 / 30

Mole of C₂H₆ = 3 moles.

Finally, we shall determine the volume of O₂ required. This is illustrated:

2C₂H₆ + 7O₂ —> 4CO₂ + 6H₂O

From the balanced equation above, we can say that:

2 L of C₂H₆ required 7 L of O₂.

Therefore, 3 L of C₂H₆ will require

= (3 × 7)/ 2 = 10.5 L of O₂.

Thus, 10.5 L of O₂ is required for the reaction.

Answer:

90% of people marry there 7th grade love. since u have read this, u will be told good news tonight. if u don't pass this on nine comments your worst week starts now this isn't fake. apparently if u copy and paste this on ten comments in the next ten minutes you will have the best day of your life tomorrow. you will either get kissed or asked out in the next 53 minutes someone will say i love you.

Answer:

Number of Protons 48

Number of Neutrons 64

Number of Electrons 48

Explanation:

The mass number of an atom is the sum of the number of protons and neutrons in the nucleus of that atom

Answer:

[tex]m_{NH_3}=21.6gNH_3[/tex]

Explanation:

Hello there!

In this case, in agreement to the given chemical reaction, it is possible for us to calculate the mass of NH3 required to remove 57.0 g NO via the stoichiometry based off the 4:6 mole ratio between them:

[tex]m_{NH_3}=57.0g NO*\frac{1molNO}{30.01gNO}*\frac{4molNH_3}{6molNO} *\frac{17.03gNH_3}{1molNH_3} \\\\m_{NH_3}=21.6gNH_3[/tex]

Best regards!

Answer:

29.5 days

Explanation:

orginally 27.3 days but 29.5 days is also correct

Answer:

c

Explanation:

for best accadimes

Answer:

The entropy increases

Explanation:

Just took the quiz

Answer: Hewo, there!

All of the above wold be my answer

Explanation:

Although many factors combine to influence weather, the four main ones are solar radiation, the amount of which changes with Earth's tilt, orbital distance from the sun and latitude, temperature, air pressure and the abundance of water.

Hope this helps you!!

Have a great day

Domains eukarya  would this organism most likely fit.

The domain eukarya comprised of eukaryotes or organisms whose cells contain true nucleus.

It is  the largest of all groups in the classification of life. There are three domains:-

It is the domain of organism called eukaryotes. These are the organism who have a well defined nucleus and membrane bound organelles.

Hence, C) option is correct.

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Answer:

0.0010 M of HC2H302(aq)

Answer:

pretty easy

Explanation:

step 1: Approach her

step 2: Ask her for something (make it look like you're lost even when you aren't)

step 3: engage in a conversation with her

step 4: you talk for hours and she give you her number

step 5: you wake up from your dream

step 6: wait you woke up

step 7: go back to dreaming again.

Answer: Girls are tough. Girls are....attitude, we all know its true.

Explanation

Best thing to start is shush when shes talking.

Answer:

0.322 g

Explanation:

Since our concentration is given in mass per volume percent % (m/v)

% (m/v) = mass of solute in g/volume of solution in mL × 100%

Since % (m/v) = 0.92 and volume of the required solution is 35.0 mL, we find the mass of NaCl from

mass of NaCl = % (m/v)/100 % × volume of solution

= 0.92 % (m/v)/100 %× 35 mL

= 0.0092 × 35

= 0.322 g

The study of a chemical is called chemistry.

The correct answer is 0.322 g

The saline is said that the amount of salt present in the solution. The formula used to solve the question is as follows:-

[tex]\frac{m}{v} = \frac{mass \ of\ solute}{volume\ of\ solution} * 100[/tex]

The data is given as follows:-

Mass of NaCl is as follows:-

 [tex]\frac{0.92}{100} * 35 mL[/tex]

[tex]0.0092 * 35\\= 0.322 g[/tex]

Hence, the correct answer is 0.322g.

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Answer:

Normality factor is defined as the ratio of observed wt. of solute to the theoretical wt. of the solute required to prepare a solution of desired normality. ... The principle states that, “the product of volume and concentration factor of the solution at any dilution is constant provided mass of the solution is constant.”

Explanation:

Answer:

412 g Cl₂

General Formulas and Concepts:

Atomic Structure

Stoichiometry

Explanation:

Step 1: Define

[Given] 3.50 × 10²⁴ molecules Cl₂

[Solve] grams Cl₂

Step 2: Identify Conversions

Avogadro's Number

[PT] Molar Mass of Cl - 35.45 g/mol

Molar Mass of Cl₂ - 2(35.45) = 70.9 g/mol

Step 3: Convert

Step 4: Check

Follow sig fig rules and round. We are given 3 sig figs.

412.072 g Cl₂ ≈ 412 g Cl₂

Answer:

Warm front

Explanation:

A warm front forms when a warm air mass pushes into a cooler air mass, shown in the image to the right (A). Warm fronts often bring stormy weather as the warm air mass at the surface rises above the cool air mass, making clouds and storms. Warm fronts move more slowly than cold fronts because it is more difficult for the warm air to push the cold, dense air across the Earth's surface. Warm fronts often form on the east side of low-pressure systems where warmer air from the south is pushed north.

You will often see high clouds like cirrus, cirrostratus, and middle clouds like altostratus ahead of a warm front. These clouds form in the warm air that is high above the cool air. As the front passes over an area, the clouds become lower, and rain is likely. There can be thunderstorms around the warm front if the air is unstable.

On weather maps, the surface location of a warm front is represented by a solid red line with red, filled-in semicircles along it, like in the map on the right (B). The semicircles indicate the direction that the front is moving. They are on the side of the line where the front is moving. Notice on the map that temperatures at ground level are cooler in front of the front than behind it.

Answer:

3.67

Explanation:

Step 1: Write the balanced equation

CO₂(g) + H₂(g) = CO(g) + H₂O(g)

Step 2: Calculate the value of the pressure equilibrium constant (Kp)

The concentration equilibrium constant (Kc) is 3.67.

We can calculate the value of the pressure equilibrium constant using the following expression.

[tex]Kp = Kc \times (R \times T)^{\Delta n(g)}[/tex]

where,

R: ideal gas constant

T: absolute temperature

Δn(g): moles of gaseous products - moles  of gaseous reactants

Δn(g) = 2 mol - 2 mol = 0

Then,

Kp = Kc × (R × T)⁰ = Kc = 3.67

Answer:

OKAY EVERYONE STOP SAYING PANCREAS THATS WRONG

Explanation:

It is actually thyroid

Answer:

ch3

Explanation:

Answer:

The answer is B

Explanation:

The answer is B because representative particles can only be atoms.

Answer:

2 NH₃(g) + 2 O₂(g) ⇒ N₂O(g) + 3 H₂O(l)   ΔH° = -684 kJ

Explanation:

Let's consider the unbalanced equation that occurs when NH₃(g) reacts with O₂(g) to form N₂O(g) and H₂O(l).

NH₃(g) + O₂(g) ⇒ N₂O(g) + H₂O(l)

We will begin balancing N atoms by multiplying NH₃(g) by 2.

2 NH₃(g) + O₂(g) ⇒ N₂O(g) + H₂O(l)

Then, we will balance H atoms by multiplying H₂O(l) by 3.

2 NH₃(g) + O₂(g) ⇒ N₂O(g) + 3 H₂O(l)

Finally, we will get the balanced equation by multiplying O₂(g) by 2.

2 NH₃(g) + 2 O₂(g) ⇒ N₂O(g) + 3 H₂O(l)

Since 342 kJ of energy are evolved for each mole of NH₃(g), and there are 2 moles of NH₃(g) in the balanced equation, the heat evolved will be2 × 342 kJ = 684 kJ. By convention, when heat is evolved, it takes a negative sign. The balanced thermochemical equation is:

2 NH₃(g) + 2 O₂(g) ⇒ N₂O(g) + 3 H₂O(l)   ΔH° = -684 kJ

Answer:

solution given:

log k=4-[tex] \frac{200}{t} [/tex]

so,

On comparison

∆S°/R = 4

∆ S° = 4R

mousam brother is correct

hope it helps you have a good day

Answer:

The answer is gravity and friction.

Explanation:

Answer:

See Explanation

Explanation:

Metallic bonds involve attraction between electrons and positively charged metal ions. The metals are ionized and electrons form a sea of valence electrons. These loosely bound electrons surround the nuclei of the metals.

The presence of this sea of electrons explains the fact that metals conduct electricity and heat due to the free valence electrons.

Due to the nature of the bonding between metal atoms,metals are malleable and ductile.

Due to the strong electrostatic interaction between metal ions and electrons, the metallic bond is very strong and is very difficult to break thereby accounting for the greater strength of metals as the size of the metallic ion decreases.

Answer:

Volume is reduced to half

Explanation:

Acid base titration are commonly used reactions in a lab, and are ofter used to get pH or different kind of solutions.

The neutralization of an acid base reaction is reached, when the solution (having added an indicator previously) changes its original color. chemically speaking, this occurs when the number of moles of the acid and the base are balanced and equal. In other words the following:

n₁ = n₂   (1)

This expression can also be expressed in function of concentration and volume:

M₁V₁ = M₂V₂   (2)

From here, solving for V₁:

V₁ = M₂V₂  / M₁

Now, this expression is true only when we have the same kind of substance that can lose or gain the same number of hydrogens.

Lets suppose that we have as base NaOH (Monoprotic base) and HCl (monoprotic acid), the titration reaction would be:

NaOH + HCl --------> NaCl + H₂O

As both of the species are monoprotic, the number of moles are the same  when they reach the equilibrium, so, expression  (2) can be used, and calculate volume or concentrations.

However, in this case, a partner made a mistake and use a diprotic acid, in this case, H₂SO₄, In this case, things chance because H₂SO₄ is diprotic, meaning that we need to dissociate two hydrogens in equilibrium, therefore, expression (2) would be something like this.

Acid: 1; Base: 2

H₂SO₄ + 2NaOH ------> Na₂SO₄ + H₂O

nH₂SO₄ = n₁ = 1

nNaOH = n₂ = 2

n₁/n₂ = 1/2

2n₁ = n₂   (3)

Writting this, in function of concentration and volume, it would be:

2M₁V₁ = M₂V₂   (4)

From here, if we solve for the volume of the acid (V₁):

V₁ = M₂V₂ / 2M₁

Therefore, according to this expression, we can see that the volume required of the acid would be half the volume required of the monoprotic acid. For example, if we need 50 mL of Chloridic acid to reach the equivalence point with NaOH, then, with H₂SO₄ it will only need 25 mL. This, of course, assuming that concentrations are the same, and volume of the base used, the same.

Hope this helps

Answer:

i'm pretty sure its the last answer and if not its the 2nd one

Answer:

1. Volume of stock required = 15 mL

Volume of water required = 485 mL

2. Volume of stock required = 5 mL

Volume of water required = 495 mL

3. Volume of stock required = 5.30 mL

Volume of water required = 494.70 mL

4. Volume of stock required = 50 mL

Volume of water required = 450 mL

Explanation:

Using the dilution formula C₁V₁ = C₂V₂

Where C₁ = initial concentration; C₂ = final concentration; V₁ = initial volume; V₂ = final volume

1. C₁ = 5 M NaCl, C₂ = 150 mM NaCl = 0.15 M NaCl

V₁ = ?; V₂ = 500 mL

V₁ = C₂V₂/C₁

V₁ = 0.15 × 500/ 5 = 15 mL

Volume of stock required = 15 mL

Volume of water required = 500 mL - 15 mL

Volume of water required = 485 mL

2. C₁ = 1 M Tris-HCl, C₂ = 50 mM Tris-HCl = 0.05 M Tris-HCl

V₁ = ?; V₂ = 500 mL

V₁ = C₂V₂/C₁

V₁ = 0.05 × 500/ 5 = 5 mL

Volume of stock required = 5 mL

Volume of water required = 500 mL - 5 mL

Volume of water required = 495 mL

3. C₁ = 95% NP-40, C₂ = 1% NP-40

V₁ = ?; V₂ = 500 mL

V₁ = C₂V₂/C₁

V₁ = 1 × 500/ 95 = 5.30 mL

Volume of stock required = 5.30 mL

Volume of water required = 500 mL - 5.30 mL

Volume of water required = 494.70 mL

4. C₁ = 1% SDS, C₂ = 10% SDS

V₁ = ?; V₂ = 500 mL

V₁ = C₂V₂/C₁

V₁ = 1 × 500/ 10 = 50 mL

Volume of stock required = 50 mL

Volume of water required = 500 mL - 50 mL

Volume of water required = 450 mL