(5) Use scientific notation, make the following conversions:
6.20 km to m​

Answer:

When you heat up water, the water molecules start moving around faster and faster. ... So hot water is less dense than cold water. When you put the two together with the hot water on the bottom, the hot water rises to the top, mixing with the cold water along the way and creating purple water.

Explanation:

have a great day :)

Answer:

I THINK it is A

Explanation:

Answer: it is the eardrum caus ei used it on my test and as well got it right

Explanation: because it just is

[tex]hut7utpojghuire\\kjionbenhrmhbnhknokgn\\bihkjihjgnhrk\\\\gnthnrt\\hgnth\\\alpha thtrnhjn4t also this is for fun to me[/tex]

Mass of Barium chloride produced : 2666.56 g

Given

Reaction

6 NaCl + Ba3(PO4)2 → 2 Na3PO4 + 3 BaCl2

1500 g of sodium chloride

Required

Barium chloride produced

Solution

mol NaCl (MW= 58.5 g/mol) :

mol = mass : MW

mol = 1500 g : 58.5 g/mol

mol = 25.64

mol Barium chloride-BaCl2 based on mol NaCl(as a limiting reactant)

From the equation, mol ratio NaCl : BaCl2 = 6 : 3, so mol BaCl2 :

[tex]\tt \dfrac{3}{6}\times 25.64=12.82[/tex]

mass BaCl2(MW=208 g/mol) :

mass = mol x MW

mass = 12.82 x 208

mass = 2666.56 g

Answer: The correct option is B. elements.

Explanation:

A chemical compound is defined as the specie which is formed by the combination of two or more elements. For example: Water, carbon dioxide, etc..

Elements is defined as the chemical specie which consists of only one type of atom. For example: Hydrogen, Oxygen, etc..

Now, when hydrogen element and bromine element combines, it leads to the formation of a heteronuclear chemical compound known as hydrogen bromide having chemical formula HBr.

Hence, the correct option is B. elements.

Hydrogen and bromine are two elements that form hydrogen bromide (HBr).

COMPOUND:

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Answer:

674.26 g of AlI₃

Explanation:

We'll begin by calculating the theoretical yield of aluminum (Al). This can be obtained as follow:

Percentage yield of Al = 67.8%

Actual yield of Al = 30.25 g

Theoretical yield of Al =?

Percentage yield = Actual yield /Theoretical yield × 100/

67.8% = 30.25 / Theoretical yield

67.8 / 100 = 30.25 / Theoretical yield

0.678 = 30.25 / Theoretical yield

Cross multiply

0.678 × Theoretical yield = 30.25

Divide both side by 0.678

Theoretical yield = 30.25 / 0.678

Theoretical yield of Al = 44.62 g

Next, we shall determine the mass of AlI₃ that reacted and the mass of Al produced from the balanced equation. This can be obtained as follow:

AlI₃(s) + 3K(s) → 3KI(s) + Al(s)

Molar mass of AlI₃ = 27 + (3×127)

= 27 + 381 = 408 g/mol

Mass of AlI₃ from the balanced equation = 1 × 408 = 408 g

Molar mass of Al = 27 g/mol

Mass of Al from the balanced equation = 1 × 27 = 27 g

Summary:

From the balanced equation above,

408 g of AlI₃ reacted to produce 27 g of Al.

Finally, we shall determine the mass of

AlI₃ required to produce 44.62 g of Al. This can be obtained as follow:

From the balanced equation above,

408 g of AlI₃ reacted to produce 27 g of Al.

Therefore, Xg of AlI₃ will react to produce 44.62 g of Al i.e

Xg of AlI₃ = (408 × 44.62)/27

Xg of AlI₃ = 674.26 g

Thus, 674.26 g of AlI₃ is needed for the reaction.

Explanation:

For sample A, I put the temperature to 93.2 C and sample B to 6.8 C. Once I dragged them together, I noticed that the numbers were going down the second I put them together! Sample A was getting colder and Sample B was getting warmer. As I let both the samples do their thing, I realized they weren't changing temperatures anymore. Both samples stopped at 50.0 C and have never changed temperatures since then.

Answer:

No; According to the IUPAC , it should be

Answer:

I am pretty sure it is C.

Explanation:

Because the wave frequency's tell us the number of waves there are on that diagram.

Answer:

C. the number of vibrations per second an object is making

Explanation:

Hope this helps. Sorry if it doesn't. T^T

Explanation:

c realsed from the galaxie

Answer:

C

Explanation:

A P E X

Answer:

In atomic physics, the Bohr model or Rutherford–Bohr model, presented by Niels Bohr and Ernest Rutherford in 1913, is a system consisting of a small, dense nucleus surrounded by orbiting electrons—similar to the structure of the Solar System, but with attraction provided by electrostatic forces in place of gravity.

Answer:

BOHR'S MODEL OF ATOM :

In order to overcome the objections raised against Rutherford's model of the atom, Neils Bohr put forward the following postulates about the model of an atom:

These orbits or shells are called energy levels. Energy levels in an atom are shown in Fig.

There orbis for shells are represented by the letters K,L,M,N... or the number n = 1,2,3,4...

ABOUT BOHR :

Neils Bohr (1885-1962) was born in Copenhagen on 7 October 1885. He was appointed professor of physics at Copenhagen University in 1916. He got the Nobel prize for his work on the structure of atom in 1922. Among Professor Bohr's numerous writings. three appearing

s books are:

Answer:Avogadro's number 6.02 x 10^23 of O2 molecules has a mass of 32 gm (atomic weight of O = 16 gm, 2 x 16 = 32).

Explanation:pls brainless

Answer:

density =mass/volume

Explanation:

p=m/v

p means raw

m means mass

v means volume.

Answer:

maintaining an internal body temperature is one of the key components to maintaining homeostasis

It's true since homeostasis is the regulation of body temperature

Answer:

3.746 g/mL

Explanation:

Mass of metal =  15.36 g

Initial Volume = 30.0 mL

Final Volume = 34.1 mL

Volume of metal = Final Volume - Initial Volume = 34.1 mL - 30.0 mL

Volume of metal = 4.1 mL

Density = ?

The relationship between the values is given as;

Density = Mass of metal / Volume of metal

Inserting the values;

Density = 15.36 / 4.1  = 3.746 g/mL

The density of the piece of metal is 3.746 g/ml.

Given the mass of metal is 15.36 g

The initial volume of a cylinder is 30.0 ml

The final volume after putting metal in it is 34.1ml

Step 1-  To find the volume:

The volume of metal = Final Volume – Initial Volume 30.0 mL

Volume of metal = 34.1 – 30.0 = 4.1 ml

Volume of metal = 4.1 mL

Step 2- Find the Density:

Density = Mass of metal is divided by the Volume of metal

[tex]\bold{Density = \dfrac{15.36 }{4.1} =3.746 g/ml}[/tex]  

Thus, the density is 3.746 g/ml.

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Answer:The potential energy of the shared elec- trons is lower in the bond than it would be if the electrons were not shared between two nuclei

Explanation:

The entire essence of chemical combination is the attainment of a lower potential energy.

When two atoms are isolated at a large distance from each other, the potential energy is high.

However as the nuclei of the two atoms approach each other, the potential energy decreases until a minimum is attained. The distance between the nuclei at this point is the bond distance.

Hence, the potential energy of the shared electrons is lower in the bond than it would be if the electrons were not shared between two nuclei.

Answer:

m=621.54g

Explanation:

number of moles= mass of H2O/molar mass of H2O

n=m/mm

make mass the subject of the formula

m=n*mm

find molar mass of water

H2O=1(16)+2(1)

mm= 18.0g/mol. n=34.53mol

hence, m=34.53mol*18.0g/mol

m=621.54g

Answer:

[tex]C_3H_6[/tex]

Explanation:

Hello!

In this case, since the empirical formula is the smallest representation of the molecular formula, it is known that the times in which the empirical formula is into the molecular formula is a whole number and is computed by dividing the molar mass of the molecular formula by that of the empirical formula as shown below:

[tex]\frac{42.08g/mol}{12.01+1.01x2}= \frac{42.08g/mol}{14.03g/mol}=3[/tex]

Thus, the molecular formula times the empirical formula by 3 to obtain:

[tex]C_3H_6[/tex]

Regards!

Answer:

40 kilograms of fuel are required to meet their energy needs for the month of April.

Explanation:

The month of April has 30 days and the full energy needed by the family is determine by the following combustion formula, derived from dimension analysis:

[tex]n\cdot Q = m\cdot L_{c}[/tex] (1)

Where:

[tex]m[/tex] - Mass of the fuel, measured in kilograms.

[tex]L_{c}[/tex] - Calorific value of the fuel, measured in kilojoules per kilogram.

[tex]Q[/tex] - Daily fuel energy, measured in kilojoules.

[tex]n[/tex] - Number of days of the month, dimensionless.

If we know that [tex]Q = 20000\,kJ[/tex], [tex]n = 30[/tex] and [tex]L_{c} = 15000\,\frac{kJ}{kg}[/tex], then the amount of fuel required to meet their energy needs for the month of April is:

[tex]m = \frac{n\cdot Q}{L_{c}}[/tex]

[tex]m = \frac{(30)\cdot (20000\,kJ)}{15000\,\frac{kJ}{kg} }[/tex]

[tex]m = 40\,kg[/tex]

40 kilograms of fuel are required to meet their energy needs for the month of April.

Answer:

It does not matter where the sample of water came from or how it was prepared. Its composition, like that of every other compound, is fixed.

Answer:

NO I DON'T ✌✌✌✌✌✌✌✌✌

Answer:

3 valence electrons

Explanation:

Let us first find the amount increased for everything.

From 1st to 2nd,  1048 - 620 = 428

428 increased from 1st to 2nd.

From 2nd to 3d, 2144 - 1048 = 1096

1096 increased from 2nd to 3rd.

From 3rd to 4th, 8724 - 2144 = 6580

6580 increased from 3rd to 4th.

From 4th  to 5th, 10141 - 8724 = 1417

From 3rd to 4th,  it increased the most. Therefore, we assume that those are the beginning core electrons and not valence electrons. Also, starting from 3rd, they are all core electrons.

For the 1st and 2nd,  we can assume they are valence electrons because the kJ/mol didn't increase that much.

Since the core electrons started from the 3rd, we most likely have 3 valence electrons.

So the final answer is 3.

Hope it helped!

Answer:

Tellurium

Explanation:

Edg 2021

Answer:

the systematic name is called Lithium sulfide.

An inorganic substance composed of the elements lithium (Li) and sulfur (S) is known as lithium sulfide [tex](Li_2S)[/tex].

It is a substance that solidifies in the shape of a crystal. Lithium ions in lithium sulfide have a positive charge, while sulfide ions have a negative charge. The chemical lithium sulfide is important in many industries. It is often employed in the synthesis of various lithium compounds, the manufacture of lithium batteries, and the manufacture of ceramics. Due to their high energy density, lithium batteries are often employed in portable electronics and electric vehicles.

Lithium sulfide is an ionic compound that has a high melting point in terms of physical characteristics. Although it is insoluble in water, many organic solvents can help dissolve it.

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Answer: b!

Explanation:

Answer:

10.80 g of H2

Explanation:

As per the equation, let us calculate the mole ratio. N2+3H2→2NH3. As per the equation one mole of nitrogen

Answer:

The light reflects twice before it finally exits the system. Draw a normal at the point of incidence to the first mirror; measure the angle of incidence (45 degrees); then draw a reflected ray at 45 degrees from the normal. Repeat the process for the second mirror.

Explanation:

Hope that helps!

Answer:

1. nitrous oxide

2. copper sulfide

3. cl2o7

4. MgBr2

Explanation:

hope it helps

Answer:

True? The don't stay exactly still i dont think, but i'd say true.

Explanation:

Answer:

TRUE

Explanation:

In a solid, atoms are packed tightly together and move very slowly. In fact, they do not flow at all: they simply vibrate back and forth.

Key words vibrate, not movement.

Correct me if I'm wrong of course

When Silicone is altered with Phosphorus, 4 of its electrons form covalent bonding and the 5th electron becomes delocalized and thus it increases electrical conduction

Answer:

blue blah blue blah

Explanation:

The question is incomplete, the complete question is;

In Sample Exercise 10.16, we found that one mole of Cl2 confined to 22.41 L at 0 °C deviated slightly from ideal behavior. Calculate the pressure exerted by 1.00 mol Cl2 confined to a smaller volume, 5.00 L, at 25 °C. (a) First use the ideal-gas equation and (b) then use the van der Waals equation for your calculation. (Values for the van der Waals constants are given in Table 10.3.) (c) Why is the difference between the result for an ideal gas and that calculated using the van der Waals equation greater when the gas is confined to 5.00 L compared to 22.4 L?

Answer:

See explanation

Explanation:

Since we have

For 5.00 L using Ideal gas law;

PV= nRT

P = nRT/V

P = 1 * 0.082 * 298 K/5.00

P= 4.887 atm

Using Van der walls equation;

[P + an^2/v^2] *  [V - nb] = nRT

P = (nRT/ [V - nb]) - (an^2/v^2)

P= 1 * 0.082 * 298/5 - (1 * 0.0562) - (6.49(1^2)/5^2)

P = (24.436/4.9438) - 0.2596

P= 4.683 atm

For 22.4 L

P = nRT/V

P = 1 * 0.082 * 298 K/22.4

P= 1.0909 atm

Using Van der Walls equation;

[P + an^2/v^2] *  [V - nb] = nRT

P = (nRT/ [V - nb]) - (an^2/v^2)

P= 1 * 0.082 * 298/22.4 - (1 * 0.0562) - (6.49(1^2)/22.4^2)

P= (24.436/ 22.3438) - 0.0129

P= 1.081 atm

According to Boyle's law, pressure is inversely proportional to volume. Increase in volume leads to decrease in pressure. Therefore the pressure calculated both by the ideal gas equation and Van der Waals equation at 5.00 L is greater than that calculated using both methods at 22.4 L