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nalin [4]
3 years ago
14

If earth is 1.50 x 10^8km from the sun what is the distance in Mm?

Chemistry
1 answer:
Readme [11.4K]3 years ago
8 0

The distance of the earth to the sun in Mm = 1.5 x 10⁵

<h3>Further explanation</h3>

Given

The distance of the earth to the sun : 1.50 x 10⁸ km

Required

The distance in Mm

Solution

In converting units we must pay attention to the conversion factor.

the conversion factor :

1 kilometer(km) = 10⁻³ megameter(Mm)

So the distance conversion :

1.5 x 10⁸ x 10⁻³ = 1.5 x 10⁵ Mm

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The way it can be determined which type of atom we have is to figure out____are contained within the nucleus of the atom A.elect
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protons and neutrons are in the nucleus electrons surround the atom and i have no idea what positrons are i just know they arent in an atom so your answer is B and C

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3 years ago
A forensic scientist reports that a dead woman had high concentrations of a pain-fighting drug in her blood. What type of analys
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C) the scientist made an identification by identifying the amount of drug in her blood and realising it was high
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A blacksmith heated an iron bar to 1445 °C. The blacksmith then tempered the metal by dropping it into 42,800 mL of
Wittaler [7]

Answer:

6626 g

Explanation:

Given that:

Density of water = 1.00 g/ml, volume of water = 42800 ml.

Since density = mass/ volume

mass of water = volume of water * density of water = 42800 ml * 1 g/ml = 42800 g

Initial temperature of water = 22°C and final temperature of water = 45°C.

specific heat capacity for water = 4.184 J/g°C

ΔT water = 45 - 22 = 23°C

For iron:

mass = m,  

specific heat capacity for iron  = 0.444 J/g°C

Initial temperature of iron = 1445°C and final temperature of water = 45°C.

ΔT iron = 45 - 1445 = -1400°C

Quantity of heat (Q) to raised the temperature of a body is given as:

Q = mCΔT

The quantity of heat required to raise the temperature of water is equal to the temperature loss by the iron.

Q water (gain) + Q iron (loss) = 0

Q water = - Q iron

42800 g ×  4.184 J/g°C × 23°C = -m × 0.444 J/g°C × -1400°C

m = 4118729.6/621.6

m = 6626 g

8 0
3 years ago
When you need to produce a variety of diluted solutions of a solute, you can dilute a series of stock solutions. A stock solutio
8090 [49]

Answer:

Volume of stock solution needed = 6.0299 mL

Explanation:

<u> </u>Dilution consists of lowering the amount of solute per unit volume of solution. It is achieved by adding more diluent to the same amount of solute.

This is deduced when thinking that both the dissolution at the beginning and at the end will have the same amount of moles.

<u>Data:</u>

M1 = 6.01 M stock solution concentration

M2 = 0.3624 M diluted solution concentration

V2 =100 mL diluted solution volume

V1 = ? stock solution volume

M1 * V1 = M2 * V2

V1=\frac{M2*V2}{M1} =\frac{0.3624M*100mL}{6.01M} =6.0299 mL

4 0
3 years ago
The following substances dissolve when added to water. Classify the substances according to the strongest solute-solvent interac
Norma-Jean [14]

Answer:

CuCl2-Ion-dipole forces

CuSO4-Ion-dipole forces

NH3-Dipole-dipole forces

CH3OH-Dipole-dipole forces

Explanation:

Water consists of a dipole. The water molecule contains a positive end and a negative end. The positive ion attracts the negative dipole of water while the positive dipole in water interacts with the negative ion of an ionic substance. This explains the dissolution of ionic substances in water.

Copper II chloride and copper sulphate are ionic substances hence they dissolve by the mechanism described above.

Molecules consisting of dipoles dissolves by interaction of the molecule's dipoles with the dipoles in water. For example, methanol interacts with water through hydrogen bonding which is involves molecular dipoles

3 0
3 years ago
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