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3 years ago

Can u answer this? Mass: 126 g Volume: 15 cm to the third power Density:?

Chemistry

1 answer:

Andre45 [30]3 years ago

4 0

Answer:

Calculate density

Weight/mass

126

Volume

cm³

Density

8,400

kg/m³

Explanation:

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In Lab 10 you make a stock solution of salicylic acid, and then four dilutions. The stock solution is made by diluting 5.00 ml o

Zigmanuir [339]

Answer:

Stock solution = 1.25 *10^-3 M

Dilution 1 = 5*10^-4 M

Dilution 2= 3.75 * 10^-4 M

Dilution 3 = 2.5 *10^-4 M

Dilution 4 = 1.25 *10^-4 M

Explanation:

Step 1: Data given

The stock solution is made by diluting 5.00 ml of 1.250 x 10-2 M salicylic acid in 50.00 mL of solution.

Step 2: Calculate the concentration of the stock solution:

M1*V1 = M2*V2

⇒ with M1 = the initial concentration = 1.250 *10^-2 M

⇒ with V1 = 5 mL = 5*10^-3 L

⇒ with M2 = TO BE DETERMINED

⇒ with V2 = 50 mL = 50 *10^-3 L

M2 = (M1*V1)/V2

M2 = (1.250 *10^-2 * 5*10^-3 L) / 50 *10^-3

M2 = 0.00125 M = 1.25 *10^-3 M

Step 3: Calculate dilution 1

M2 = (M1*V1)/V2

M2 = (1.25*10^-3 * 10 *10^-3)/(25*10^-3L)

M2 = 0.0005 M = 5*10^-4 M

Step 4: Calculate dilution 2

M2 = (M1*V1)/V2

M2 = (1.25*10^-3 * 10 * 7.5*10^-3)/(25*10^-3)

M2 = 0.000375 M = 3.75 * 10^-4 M

Step 5: Calculate dilution 3

M2 = (M1*V1)/V2

M2 = (1.25*10^-3 * 5*10^-3) /(25*10^-3)

M2 = 0.00025 M = 2.5 *10^-4 M

Step 6: Calculate dilution 4

M2 = (M1*V1)/V2

M2 = (1.25*10^-3 * 2.5*10^-3)/(25*10^-3)

M2 = 0.000125 M = 1.25 *10^-4 M

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3 years ago

The last element in any period always has:

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Answer: The correct option is The properties of a noble gas.

Explanation: There are 7 periods in the periodic table.

The last element of each period are Helium (He), Neon (Ne), Argon (Ar), Krypton (Kr), Xenon (Xe), Radon (Rn) and Ununoctium (Uuo).

The electronic configuration for Helium is $1s^2$ . For He, The outermost electrons are 2.

The electronic configuration for all the other elements is $ns^2ns^6$ ( where, n = 2, 3, 4, 5, 6 and 7 respectively). For all the other gases, the outermost electrons are 8.

All these elements have stable electronic configuration and are not reactive in nature. Hence, they are considered as noble gases.

Therefore, the last element of each period always have the properties of a noble gas.

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