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3 years ago

11

_____ dating places the rocks and rock layers in an area into sequential order. dating gives an exact date on r

ock layers.
Geologic, Absolute

Relative, Geologic

Relative, Absolute

Time, Geologic

1 answer:

tankabanditka [31]3 years ago

4 0

C. relative, absolute

relative dating dates things to the general time period
absolute dating gives the exact date

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When a rock falls from outer space all the way to the groung, its called a?

goblinko [34]

It's called a meteorite.

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3 years ago

Which mineral is softer than fluorite

BigorU [14]

Answer: Calcite is softer than fluorite.

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3 years ago

What is the molarity of NaCL solution that has 58 grams of NaCL dissolved in 4.0 liters of water

Anna007 [38]

14.5 M

Molarity=grams/liters

M=58/4.0

4 0

3 years ago

if i have 4 moles of a gas at a pressure of 5.6 atm and a volume of 12 liters, what is the temperature? (205 K)

Thepotemich [5.8K]

Hello!

If i have three moles of a gas at a pressure of 5.6 atm and a volume of 3 liters, what is the temperature ?

We have the following data:

n (number of moles) = 3 moles

P (pressure) = 5.6 atm

V (volume) = 12 L

T (temperature) = ? (in Kelvin)

R (gas constant) = 0.082 atm.L / mol.K

We apply the data above to the Clapeyron equation (gas equation), let's see:

$P*V = n*R*T$

$5.6*12 = 4*0.082*T$

$67.2 = 0.328\:T$

$0.328\:T = 67.2$

$T = \dfrac{67.2}{0.328}$

$T = 204.8780488...$

$\boxed{\boxed{T \approx 205\:K}}\end{array}}\qquad\checkmark$

Answer:

The temperature is approximately 205 Kelvin

_______________________________

I Hope this helps, greetings ... Dexteright02! =)

3 0

3 years ago

Be sure to answer all parts. For the titration of 10.0 mL of 0.250 M acetic acid with 0.200 M sodium hydroxide, determine the pH

DaniilM [7]

Explanation:

$Molarity=\frac{moles}{Volume(L)}$

Molarity of the acetic acid = 0.250 M

Volume of the acetic acid solution = 10.0 mL = 0.010 L( 1 mL =0.001L)

Moles of acetic acid ;

$n=0.250 M\times 0.010 L=0.0025 mol$

Molarity of the NaOH = 0.200 M

a) Volume of the NaOH solution = 10.0 mL = 0.010 L( 1 mL =0.001L)

Moles of NaOH : $0.200M\times 0.010 L=0.002 mol$

$CH_3COOH+NaOH\rightarrow CH_3COONa+H_2O$

1 mole NaOH neutralizes 1 mole of acetic acid , then 0.002 moles of NaOH will neutralize 0.002 mol of acetic acid.

Moles of acetic acid left un-neutralized = 0.0025 mol - 0.002 = 0.0005 mol

1 mole of acetic acid gives 1 mole of hydrogen ion, then 0.0005 mole of acetic acid will give 0.0005 mole of hydrogen ions.

Moles of hydrogen ion= 0.0005 mol

Volume of the solution = 0.010 L+ 0.010 L = 0.020 L

$[H^+]=\frac{0.0005 mol}{0.020 L}=0.025 M$

The pH of the 10.0 mL of base added to acetic acid solution :

$pH=-\log[H^+]=-\log[0.025 M]=1.60$

b) Volume of the NaOH solution = 12.0 mL = 0.012 L( 1 mL =0.001L)

Moles of NaOH : $0.200M\times 0.012 L=0.0024 mol$

$CH_3COOH+NaOH\rightarrow CH_3COONa+H_2O$

1 mole NaOH neutralizes 1 mole of acetic acid , then 0.0024 moles of NaOH will neutralize 0.0024 mol of acetic acid.

Moles of acetic acid left un-neutralized = 0.0025 mol - 0.0024 = 0.0001 mol

1 mole of acetic acid gives 1 mole of hydrogen ion, then 0.0001 mole of acetic acid will give 0.0001 mole of hydrogen ions.

Moles of hydrogen ion= 0.0001 mol

Volume of the solution = 0.010 L+ 0.012 L = 0.022 L

$[H^+]=\frac{0.0001 mol}{0.022 L}=0.0045 M$

The pH of the 12.0 mL of base added to acetic acid solution :

$pH=-\log[H^+]=-\log[0.0045 M]=2.34$

c) Volume of the NaOH solution = 15.0 mL = 0.015 L( 1 mL =0.001L)

Moles of NaOH : $0.200M\times 0.015 L=0.003 mol$

$CH_3COOH+NaOH\rightarrow CH_3COONa+H_2O$

1 mole NaOH neutralizes 1 mole of acetic acid , then 0.003 moles of NaOH will neutralize 0.003 mol of acetic acid.

All the moles of acetic acid will get neutralized by NaOH and un-neutralized sodium hydroxide will left over.

Moles of NaOH left un-neutralized = 0.003 mol - 0.0025 = 0.0005 mol

1 mole of NaOH gives 1 mole of hydroxide ion, then 0.0005 mole of NaOH acid will give 0.0005 mole of hydroxide ions.

Moles of hydroxide ion= 0.0005 mol

Volume of the solution = 0.010 L+ 0.015 L = 0.025 L

$[OH^-]=\frac{0.0005 mol}{0.025 L}=0.02 M$

The pOH of the 15.0 mL of base added to acetic acid solution :

$pOH=-\log[OH^-]=-\log[0.02 M]=1.70$

The pH of the 15.0 mL of base added to acetic acid solution :

$pH=14-pOH=14-1.70=12.3$

7 0

3 years ago