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3 years ago

Please help!! my grades dying 15 points

Chemistry

1 answer:

AysviL [449]3 years ago

4 0

Answers-in-bold:

There are two common temperature scales. On the Fahrenheit scale, water freezes at 32 degrees. The Celsius scale divides the interval between the freezing and boiling points of water into 100 degrees.

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NEED HELP ASAP

Thepotemich [5.8K]

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1) Nonmetal

2) Calcium (Ca), chemical element, one of the alkaline-earth metals of Group 2 (IIa) of the periodic table.

3) Hafnium

4) 204.3833 u

5) Not sure what you're asking, but oble gas, any of the seven chemical elements that make up Group 18 (VIIIa) of the periodic table. The elements are helium (He), neon (Ne), argon (Ar), krypton (Kr), xenon (Xe), radon (Rn), and oganesson (Og)

6) The metalloids; boron (B), silicon (Si), germanium (Ge), arsenic (As), antimony (Sb), tellurium (Te), polonium (Po) and astatine (At)

7) The Actinide series contains elements with atomic numbers 89 to 103 and is the third group in the periodic table.

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Hope this helps!

8 0

3 years ago

Read 2 more answers

A solution was prepared by dissolving 0.800 g of sulfur S8, in 100.0 g of acetic acid, HC2H3O2. Calculate the freezing point and

sammy [17]

Answer: The freezing point of solution is 16.5°C and the boiling point of solution is 118.2°C

Explanation:

To calculate the molality of solution, we use the equation:

$Molality=\frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ in grams}}$

Where,

$m_{solute}$ = Given mass of solute $(S_8)$ = 0.800 g

$M_{solute}$ = Molar mass of solute $(S-8)$ = 256.52 g/mol

$W_{solvent}$ = Mass of solvent (acetic acid) = 100.0 g

Putting values in above equation, we get:

$\text{Molality of solution}=\frac{0.800\times 1000}{256.52\times 100.0}\\\\\text{Molality of solution}=0.0312m$

Calculation for freezing point of solution:

Depression in freezing point is defined as the difference in the freezing point of water and freezing point of solution.

$\Delta T_f=\text{freezing point of acetic acid}-\text{Freezing point of solution}$

To calculate the depression in freezing point, we use the equation:

$\Delta T_f=iK_fm$

or,

$\text{Freezing point of acetic acid}-\text{Freezing point of solution}=iK_fm$

where,

Freezing point of acetic acid = 16.6°C

i = Vant hoff factor = 1 (for non-electrolyte)

$K_f$ = molal freezing point depression constant = 3.59°C/m

m = molality of solution = 0.0312 m

Putting values in above equation, we get:

$16.6^oC-\text{freezing point of solution}=1\times 3.59^oC/m\times 0.0312m\\\\\text{Freezing point of solution}=16.5^oC$

Hence, the freezing point of solution is 16.5°C

Calculation for boiling point of solution:

Elevation in boiling point is defined as the difference in the boiling point of solution and freezing point of pure solution.

The equation used to calculate elevation in boiling point follows:

$\Delta T_b=\text{Boiling point of solution}-\text{Boiling point of acetic acid}$

To calculate the elevation in boiling point, we use the equation:

$\Delta T_b=iK_bm$

or,

$\text{Boiling point of solution}-\text{Boiling point of acetic acid}=iK_fm$

where,

Boiling point of acetic acid = 118.1°C

i = Vant hoff factor = 1 (for non-electrolyte)

$K_f$ = molal boiling point elevation constant = 3.08°C/m

m = molality of solution = 0.0312 m

Putting values in above equation, we get:

$\text{Boiling point of solution}-118.1^oC=1\times 3.08^oC/m\times 0.0312m\\\\\text{Boiling point of solution}=118.2^oC$

Hence, the boiling point of solution is 118.2°C

5 0

3 years ago

Pls help I’ll mark brainliest pls pls pls

katovenus [111]

The tea was no longer hot or (brewed) so the 5th didn’t dissolve like the others because the tea was hot or warm enough anymore it cooled down. So the sugar won’t dissolve no more.

7 0

2 years ago

Read 2 more answers

potassium, k, has an atomic number of 19 and an atomic mass of 39 how many electrons does a neutral atom of K have?

coldgirl [10]

A neutral atom of potassium has 19 electrons.

8 0

3 years ago

Sulfur dioxide gas (SO2) and oxygen gas (O2) react to form the liquid product of sulfur trioxide (SO3). How much sulfur dioxide

taurus [48]

1) Balanced chemical equation:

2SO2 (g) + O2 (g) -> 2SO3 (l)

2) Molar ratios

2 mol SO2 : 1 mol O2 : 2 mol SO3

3) Convert 6.00 g O2 to moles

number of moles = mass in grams / molar mass

number of moles = 6.00 g / 32 g/mol = 0.1875 mol O2.

4) Use proportions with the molar ratios

=> 2 moles SO2 / 1 mol O2 = x / 0.1875 mol O2

=> x = 0.1875 mol O2 * 2 mol SO2 / 1 mol O2 = 0.375 mol SO2.

5) Convert 0.375 mol SO2 to grams

mass in grams = number of moles * molar mass

molar mass SO2 = 32 g/mol + 2*16 g/mol = 64 g/mol

=> mass SO2 = 0.375 mol * 64 g / mol = 24.0 g

Answer: 24.0 g of SO2 are needed to react completely with 6.00 g O2.

7 0

3 years ago