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Anna007 [38]
4 years ago
15

A supposedly silver crown is tested to determine its density. it displaces 10.7 ml of water and has a mass of 112 g. part a coul

d the crown be made of silver?
Chemistry
1 answer:
Anni [7]4 years ago
4 0

The volume of water displace by crown is 10.7 mL. This will be equal to the volume of crown. Mass of crown is 112 g.

Density of a substance is defined as mass of substance per unit volume. It is mathematically represented as follows:

d=\frac{m}{V}

Putting the values,

d=\frac{112 g}{10.7 mL}=10.46 g/mL

The theoretical value of density of silver is 10.5 g/cm^{3} also, 1 mL is equal to 1 cm^{3} thus, density will be 10.5 g/mL.

This is approximately equal to the calculated value thus, the crown is made up of silver.


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What is the concentration of a solution with a volume of 1.38 mL that contains 17.36
Elden [556K]

Answer:

C = 107.97 mol/L

Explanation:

Given data:

Volume of solution = 1.38 mL (1.38 mL× 1 L /1000 mL = 0.00138 L)

Mass of ammonium sulfite = 17.36 g

Concentration of solution =?

Solution:

We will calculate the number of moles of ammonium sulfite.

Number of moles = mass/molar mass

Number of moles = 17.36 g / 116.15 g/mol

Number of moles = 0.149 mol

Concentration:

C = n/V

C = concentration

n = number of moles of solute

v = volume in L

C = 0.149 mol / 0.00138 L

C = 107.97 mol/L

5 0
3 years ago
What’s the formula equation for
jeyben [28]
If you have a magnesium for every oxygen, then you have to start with two magnesiums. So the balanced equation is 2 Mg + O2 2 MgO.

And

CaCO3———→CaO + CO2

I hope it helped!

7 0
3 years ago
Read 2 more answers
Calculate the moles and grams of solute in each solution. D. 2.0 L of 0.30M Na2SO4. I already have A, B, and C. Thanks!
e-lub [12.9K]
Find the number of moles
C = n / V
C(Concentration) = 0.30 moles / L
V ( Volume) = 2 L
n = ??
n = C * V
n = 0.30 mol / L * 2 L
n = 0.60 mol


Find the molar mass
2Na = 23 * 2 = 46 grams
1S   = 32 * 1 = 32 grams
O4   = 16 * 4 = 64 grams
Total =            142 grams / mol

Find the mass
n = given mass / molar mass
n = 0.06 mol
molar Mass = 142 grams / mol
given mass = ???

given mass = molar mass * mols
given mass = 142 * 0.6
given mass = 85.2 grams. 

85.2 are in a 2 L solution that has a concentration of 0.6 mol/L


4 0
3 years ago
Read 2 more answers
How many moles of NaCl will react completely with 18.5 L F2 gas at 300.0 K and 1.00 atm?
meriva
Hello!

Data:

P (pressure) = 1 atm
V (volume) = 18.5 L
T (temperature) = 300 K
n (number of mols) = ? (in mol)
R (Gas constant) = 0.082 (atm*L/mol*K)

Apply the data to the Clapeyron equation (ideal gas equation), see:
P*V = n*R*T

1*18.5 = n*0.082*300

18.5 = 24.6n

24.6n = 18.5

n =  \dfrac{18.5}{24.6}

\boxed{\boxed{n \approx 0,752\:mol}}\end{array}}\qquad\checkmark

Note: If the feedback is to be considered, the closest response is 0.751 mol Nacl

_________________
_________________

I hope this helps. =)


3 0
3 years ago
calculate the molarity of MgCl2 in the following solution: 5.34 g of MgCl2 is dissolved and diluted to 214 mL of solution.
anyanavicka [17]
<h3><u>Ⲁⲛ⳽ⲱⲉⲅ</u><u>:</u></h3>

\quad\hookrightarrow\quad \sf {0.262M }

<h3><u>Ⲋⲟⳑⳙⲧⳕⲟⲛ :</u></h3>

Molarity is used to measure the concentration of a solution , so it is also as molar concentration. It is denoted as M or Mol/L

<u>We </u><u>are </u><u>given </u><u>that </u><u>:</u>

  • Weight of \sf MgCl_{2} = 5.34g
  • Volume of solution = 214 ml , or 0.214 L

The molar mass of magnesium chloride ( \sf MgCl_{2} ) is 95.21 g / mol

We can calculate the molarity of the solution by dividing the number of moles of solute by volume of solvent in liter ,i.e:

\quad\longrightarrow\quad \sf  {M = \dfrac{n}{V} } ‎ㅤ‎ㅤ‎ㅤ⸻( 1 )

<em>Where,</em><em> </em>

  • M = molarity
  • n = number of moles
  • V = Volume

We can calculate the number of moles by dividing the actual mass by its molar mass ,i.e:

\quad\longrightarrow\quad \sf { n = \dfrac{w}{m}}‎ㅤ‎ㅤ‎ㅤ‎⸻ ( 2 )

<em>W</em><em>here,</em>

  • n = number of moles
  • m = molar mass
  • w = actual mass

<u>Therefore</u><u>,</u>

\implies\quad \tt {n =\dfrac{w}{m} }

\implies\quad \tt { n =\dfrac{5.35\: g}{95.21\: g /mol}}

\implies\quad{\pmb{ \tt {n = 0.056 mol}} }

<u>P</u><u>utting </u><u>the </u><u>values </u><u>in </u><u>equation </u><u>(</u><u> </u><u>1</u><u> </u><u>)</u><u>:</u>

\implies\quad \tt {M=\dfrac{n}{V} }

\implies\quad \tt { M =\dfrac{0.056\:mol}{0.214\:L}}

\implies\quad\underline{\pmb{ \tt { M = 0.262 \:M }}}

7 0
2 years ago
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