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liberstina [14]
2 years ago
13

An archeologist finds a 1.62 kg goblet that she believes to be made of pure gold. She determines that the volume of the goblet i

s 205 mL. Calculate the density of the goblet. I was wondering if it is 1.62/205? It is saying it it incorrect though!
Chemistry
1 answer:
katovenus [111]2 years ago
5 0
The  answer is

<span>The density (D) is quotient of mass (m) and volume (V):
</span>D= \frac{m}{V}
The unit is g/cm³

It is given:
m = 1.62 kg = 1620 g
V = 205 mL = 205 cm³
D = ?

Thus:
D= \frac{m}{V} = \frac{1620g}{205cm ^{3} } = 7.90 g/cm ^{3}

The density of the goblet is 7.90 g/cm³.
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James was given a sample of sodium bicarbonate NaHCO3 in a weighing dish to analyze. His teacher asked him to find the percent o
gtnhenbr [62]
This question could be answered easily if the results of the abundance of the other elements are given. You will just have to subtract the sum of all their abundances to 100. Since it's not given, the solution would just be:

Na = 23 g/mol* 1 = 23 g
H = 1 g/mol * 1 = 1 g
C = 12 g/mol * 1 = 12 g
O = 16 g/mol * 3 = 48 g
Total = 84 g

% O = 48/84 * 100 = <em>57.14%</em>

5 0
2 years ago
Read 2 more answers
In a laboratory experiment, the freezing point of an aqueous solution of glucose is found to be -0.325°C, What is the molal conc
Hatshy [7]

0.17 M is the is the molal concentration of this solution

Explanation:

Data given:

freezing point of glucose solution = -0.325 degree celsius

molal concentration of the solution =?

solution is of glucose=?

atomic mass of glucose = 180.01 grams/mole

freezing point of glucose = 146 degrees

freezing point of water = 0 degrees

Kf of glucose = 1.86 °C

ΔT = (freezing point of solvent) - (freezing point of solution)

ΔT = 0.325 degree celsius

molality =?

ΔT = Kfm

rearranging the equation:

m = \frac{0.325}{1.86}

m= 0.17 M

molal concentration of the glucose solution is 0.17 M

3 0
2 years ago
A mixture of N2, O2, and Ar has mole fractions of 0.25, 0.65, and 0.10, respectively. What is the pressure of N2 if the total pr
s344n2d4d5 [400]

Answer:

Partial pressure of nitrogen gas is 0.98 bar.

Explanation:

According to the Dalton's law, the total pressure of the gas is equal to the sum of the partial pressure of the mixture of gases.

P=p_{N_2}+p_{O_2}+p_{Ar}

p_{N_2}=P\times \chi_{N_2}

p_{O_2}=P\times \chi_{O_2}

p_{Ar}=P\times \chi_{Ar}

where,

P = total pressure = 3.9 bar

p_{N_2} = partial pressure of nitrogen gas  

p_{O_2} = partial pressure of oxygen gas  

p_{Ar} = partial pressure of argon gases  

\chi_{N_2} = Mole fraction of nitrogen gas  = 0.25

\chi_{O_2} = Mole fraction of oxygen gas  = 0.65

\chi_{Ar} = Mole fraction of argon gases = 0.10

Partial pressure of nitrogen gas :

p_{N_2}=P\times \chi_{N_2}=3.9 bar\times 0.25 =0.98 bar

Partial pressure of oxygen gas :

p_{N_2}=P\times \chi_{O_2}=3.9 bar\times 0.65=2.54 bar

Partial pressure of argon gas :

p_{N_2}=P\times \chi_{Ar}=3.9 bar\times 0.10=0.39 bar

7 0
2 years ago
Classify each property as associated with a liquid that has strong or weak intermolecular forces Drag the appropriate items to t
Ivan

Answer: intermolecular forces are described as forces that either cause attraction or repulsion between neighbouring particles or molecules

Explanation: liquids with strong intermolecular forces has the following properties:

1.) High boiling point

2.) high surface tension

3.) Low viscosity

4.) Low vapour pressure

While liquids with weak intermolecular forces has the following properties:

1.) Low boiling point

2.)low surface tension

3.) High viscosity

4.) High vapour pressure.

3 0
2 years ago
2Na2O2 + 2CO2 → 2Na2CO3 + O2
ahrayia [7]

the actual yield is the amount of Na₂CO₃ formed after carrying out the experiment

theoretical yield is the amount of Na₂CO₃ that is expected to be formed from the calculations

we need to first find the theoretical yield

2Na₂O₂ + 2CO₂ ---> 2Na₂CO₃ + O₂

molar ratio of Na₂O₂ to Na₂CO₃ is 2:2

number of Na₂O₂ moles reacted is equal to the number of Na₂CO₃ moles formed

number of Na₂O₂ moles reacted is - 7.80 g / 78 g/mol = 0.10 mol

therefore number of Na₂CO₃ moles formed is - 0.10 mol

mass of Na₂CO₃ expected to be formed is - 0.10 mol x 106 g/mol = 10.6 g

therefore theoretical yield is 10.6 g

percent yield = actual yield / theoretical yield  x 100%

81.0  % = actual yield / 10.6 g x 100 %

actual yield = 10.6 x 0.81

actual yield = 8.59 g

therefore actual yield is 8.59 g

7 0
2 years ago
Read 2 more answers
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