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Degger [83]
2 years ago
10

What is the volume of an object with a mass of 31g and density of 444g/ml​

Chemistry
1 answer:
Tanzania [10]2 years ago
5 0

Answer:

<h2>The answer is 0.07 mL</h2>

Explanation:

The volume of a substance when given the density and mass can be found by using the formula

volume =  \frac{mass}{density} \\

From the question

mass = 31 g

density = 444 g/mL

We have

volume =  \frac{31}{444}   \\  = 0.06981...

We have the final answer as

<h3>0.07 mL</h3>

Hope this helps you

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8 0
3 years ago
If the edge length of the unit cell is 705.2 pm, what is the density of KI in g/cm3.
Kipish [7]

The density  is 3.144 g / cm^3.

<u>Explanation</u>:

If effective number of atom in NaCl type structure, z = 4

a = 705.2 pm ⇒ In centimeter = 705.2 \times 10^-10

Na = 6.023 \times 10^23

density = (molecular weight) (z) / (Na) (a^3)

where molecular weight of KI is 166 g,

           Z represents the atomic number

density = (molecular weight) (z) / (Na) (a^3)              

             = (166 \times 4) / (6.023 \times 10^23) \times (705.2 \times 10^-10)

density = 3.144 g / cm^3.

5 0
2 years ago
A mineral consisted of 29.4% calcium, 23.5% sulfur and 47.1% oxygen. What is the empirical formula?
inessss [21]
Step  one  calculate  the  moles  of  each  element
that  is  moles= %  composition/molar  mass
molar  mass  of    Ca  =  40g/mol,     S=  32  g/mol ,   O=  16 g/mol

moles  of      Ca = 29.4 /40g/mol=0.735 moles,      S= 23.5/32  =0.734 moles,  O= 47.1/16= 2.94   moles

calculate  the mole  ratio  by  dividing   each  mole with  smallest  mole   that   is  0.734
Ca=  0.735/0.734= 1,      S=  0.734/0.734 =1,   O  = 2.94/  0.734= 4
therefore  the  emipical  formula  =  CaSO4
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3 years ago
Name the elements present of carbon monoxide
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2 years ago
A student plans to use a density versus solution concentration standard curve to identify the sodium chloride concentration in a
Art [367]

Answer:

a. 0.50 g, 1.0 g, 1.5 g, 2.0 g, 2.5 g;

g. 0.1 g, 0.5 g, 1.0 g, 1.5 g, 2.0 g

Explanation:

The percent mass is defined as a ratio between the mass of a solute and mass of a solution:

\omega = \frac{m_{solute}}{m_{solution}}\cdot 100\%

Since solution only consists of a solute and solvent, express its mass as:

m_{solution} = m_{solute} + m_{solvent}

Then:

\omega = \frac{m_{solute}}{m_{solution}}\cdot 100\%=\frac{m_{solute}}{m_{solute} + m_{solvent}}

Firstly, solve for how much mass is required to prepare 3.0 %. Let's say, we have x g of the solute:

0.03 = \frac{x}{30.0 + x}\therefore x = 0.03(30.0 + x)

x = 0.90 + 0.03x

0.97x = 0.90\therefore x = 0.93 g

Similarly, solve for 6.0 %, let's say, we have x g of the solute again:

0.06 = \frac{x}{30.0 + x}\therefore x = 0.06(30.0 + x)

x = 1.80 + 0.06x

0.94x = 1.80\therefore x = 1.91 g

Hence, masses should be in a range of 0.93 g to 1.91 g.

7 0
3 years ago
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