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o-na [289]
2 years ago
9

I do not quite understand this, anyway someone can help? I'll give brainliest to the correct answer!

Chemistry
2 answers:
Vanyuwa [196]2 years ago
5 0

To solve for number one we must follow these steps:

m = 703.55 g − 345.8 g

m = 357.75 g

Then we solve for the density-

d = 357.75 g / 225 mL

d = 1.59 g/mL

Answer for number 5) 1.59 g/mL

Density= Mass/Volume

therefor- Mass = 65.14g and Volume = 35.4mL

Density = 65.14g/35.4 mL = 1.84 g/mL

Answer for number 6) 1.84 g/mL

Question 7-

g = (0.8765 g/mL)(250.0 mL) = 219.1 g

Answer: 219.1 g

Question 8-

Density = mass/volume

Mass = 1587g

4.5cm x 5.2cm x 6cm = 140.4 cubic-cm

Volume = 140.4 cubic-cm

Mass/volume = 1587/140.4 = 11.3 g/cm^3

Answer: 11.3 g/cm^3

Question 9:

First calculate the volume of iron

49.10mL − 45.50mL = 3.60mL

Density = mass/volume;

density = 28.5g/3.60 mL= 7.92 g/mL

Answer: 7.92 g/mL

Question 10:

Volume = 2500.0g / 10.5g/cm3 = 238.095cm^3

Now we round up the answer to 238cm^3

Answer: 238cm^3

I hope this helped!

Andrews [41]2 years ago
4 0

Answer:

Both flow rapidly down a slope and can be triggered by an earthquake. Landslides contain only rock and soil, while mudflows contain rock, soil, and a high percentage of water.

Explanation:

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If a solution contains 3 moles/liter of sodium chloride (nacl, made of sodium ions and chloride ions), what is the osmolarity of
valkas [14]

<u>Answer:</u> Osmolarity of the sodium chloride solution is 18 Osmol/L.

<u>Explanation:</u>

Osmolarity is defined as the the concentration of the solution which is expressed as the total number of solution particles present in one liter of solvent.

We are given the molarity of the solution which is 3mol/L and to convert it into osmolarity, we will multiply the number of osmoles that are produced by the solute.

Osmole is defined as the particles that contribute to the osmotic pressure of a solution.

The solute given here is sodium chloride (NaCl). Number of osmoles can be determined by the dissociation of this solvent into ions.

The equation given by the dissociation of NaCl:

NaCl\rightarrow Na^++Cl^-

1 mole of sodium chloride produces 2 moles of ions.

So, 3 moles of sodium chloride will produce = (3 × 2) = 6 moles of ions.

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5 0
3 years ago
Determine the temperature change when 150. G block of gold is supplied with 1.00 • 10 ^3 J of heat
IceJOKER [234]

Answer:

∇T = 51.68°C

Explanation:

Mass = 150g

Heat Energy (Q) = 1.0*10³J

Change in temperature ∇T = ?

Q = mc∇T

Q = heat energy

M = mass

C = specific heat capacity of the gold = 0.129j/g°C

∇T = change in temperature

Q = Mc∇T

1.0*10³ = 150 * 0.129 * ∇T

1000 = 19.35∇T

Solve for ∇T

∇T = 1000 / 19.35

∇T = 51.679°C = 51.68°C

The change in temperature of gold was 51.68°C

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3 years ago
In redox half-reactions, a more positive standard reduction potential means I. the oxidized form has a higher affinity for elect
qwelly [4]

Answer:

The 1st and 4th options are correct

I.the oxidized form has a higher affinity for electrons

IV. the greater the tendency for the oxidized form to accept electrons

Explanation:

Half reaction can be described as the oxidation or reduction reaction in a redox reaction.it is In the redox rection there is a change in the oxidation states of Chemical species involved. the oxidized form in the redox has a higher affinity for electrons and the greater the tendency for the oxidized form to accept electrons.

Standard reduction potential which is also referred to as standard cell potential can be described as the potential difference that exist between cathode and anode of the cell. In the standard reduction potential most times the species will be reduced which is usually analysed in a reduction half reaction.

(Standard Hydrogen Electrode) is utilized when determining the Standard reduction or potentials of a chemical specie. this is because of Hydrogen having zero reduction and oxidation potentials, as a result of this a measured potential of any species is compared with that of Hydrogen, the difference helps to know the potential reduction of that particular specie.

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