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Ymorist [56]
4 years ago
5

An object is either copper or brass. The density of copper is 8.96g/ml. The density of brass is 8.73g/mL. I find that my object

is 10.08
and has a volume of 1.15mL. What is my object made of?
Your answer:
O Copper
O Brass
O Cannot tell from the information given
Chemistry
2 answers:
Brrunno [24]4 years ago
6 0
I like pp hehehehehe hehhe
zvonat [6]4 years ago
6 0
Penis in my butt.......
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PLEASE HELP BEFORE 7 A.M. PACIFIC TIME <br> SEE ATTACHED.
anastassius [24]
For i: 33mL
For ii: 87-88mL
For iii:22.3mL
5 0
3 years ago
Read 2 more answers
What is the noble gas configuration for telleurium
Maru [420]

[Kr] 4d10 5s2 5p4 is the noble gas configuration for telleurium because of the presence of different number of electrons.

<h3>What is telleurium?</h3>

Tellurium is a noble gas element that is non-reactive in nature due to complete outermost shell. It has atomic number of 52 which means that it has 52 number of electrons.

So we can conclude that [Kr] 4d10 5s2 5p4 is the noble gas configuration for telleurium because of the presence of different number of electrons.

Learn more about noble gas here: brainly.com/question/13715159

#SPJ1

4 0
2 years ago
Blue and orange, which color can be best absorbed?
baherus [9]

Blue is the color that can be best absorbed, as orange is the color seen.

6 0
3 years ago
Benzene has a specific gravity of 0.88. if it was spilled in a river what would it do?
Gelneren [198K]
As we know,
                    Density of Benzene  =  876 Kg/m³
And,
                    Density of Water       =  997 Kg/m³
So,
     Specific Gravity is calculated as,

                     Specific Gravity  =  Density of Benzene / Density of Water

                     Specific Gravity    =   876 Kg/m³ / 997 Kg/m³

                     Specific Gravity    =  0.878

Every object having specific gravity less than 1 will float on water and if value is greater than 1 then it will sink.

Benzene being non-polar in nature does not mix with water and due to less density it will float on the surface of water.
4 0
3 years ago
How many milliliters of a 3.4 M NaCl solution would be needed to prepare each solution?
Ksenya-84 [330]

Answer:

a. Approximately 1.3\; \rm mL.

b. Approximately 7.2\; \rm mL.

Explanation:

The unit of concentration "\rm M" is equivalent to "\rm mol \cdot L^{-1}", which means "moles per liter."

However, the volume of both solutions were given in mililiters \rm mL. Convert these volumes to liters:

\displaystyle 45\; \rm mL = 45\; \rm mL \times \frac{1\; \rm L}{1000\; \rm mL} = 0.045\; \rm L.

\displaystyle 330\; \rm mL = 330\; \rm mL \times \frac{1\; \rm L}{1000\; \rm mL} = 0.330\; \rm L.

In a solution of volume V where the concentration of a solute is c, there would be c \cdot V (moles of) formula units of this solute.

Calculate the number of moles of \rm NaCl formula units in each of the two solutions:

Solution in a.:

n = c \cdot V = 0.045\; \rm L \times 0.10\; \rm mol \cdot L^{-1} = 0.0045\; \rm mol.

Solution in b.:

n = c \cdot V = 0.330\; \rm L \times 0.074\; \rm mol \cdot L^{-1} = 0.02442\; \rm mol.

What volume of that 3.4\; \rm M (same as 3.4 \; \rm mol \cdot L^{-1}) \rm NaCl solution would contain that many

For the solution in a.:

\displaystyle V = \frac{n}{c} = \frac{0.0045\; \rm mol}{3.4\; \rm mol \cdot L^{-1}} \approx 0.0013\; \rm L.

Convert the unit of that volume to milliliters:

\displaystyle 0.0013\; \rm L = 0.0013\; \rm L \times \frac{1000\; \rm mL}{1\; \rm L} = 1.3\; \rm mL.

Similarly, for the solution in b.:

\displaystyle V = \frac{n}{c} = \frac{0.02442\; \rm mol}{3.4\; \rm mol \cdot L^{-1}} \approx 0.0072\; \rm L.

Convert the unit of that volume to milliliters:

\displaystyle 0.0072\; \rm L = 0.0072\; \rm L \times \frac{1000\; \rm mL}{1\; \rm L} = 7.2\; \rm mL.

8 0
3 years ago
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