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3 years ago

Would you expect to find older or newer fossils in rock layers closer to the surface? Why?

Chemistry

1 answer:

Virty [35]3 years ago

3 0

Answer:

Newer

Explanation:

older fossils and rocks are found in lower strata than younger fossils and rock layers because the top layer fossils have not been there as long as the bottom layer fossils to be found deeper in the ground

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Which two substances have no fixed shape and no fixed volume?

zalisa [80]

Answer:

Gases

Explanation:

4 0

3 years ago

A 50.0 mL sample containing Cd2+ and Mn2+ was treated with 56.3 mL of 0.0500 M EDTA. Titration of the excess unreacted EDTA requ

egoroff_w [7]

Answer:

The concentration of Cd2+ is 0.0175 M

The concentration of Mn2+ is 0.0305 M

Explanation:

Step 1: Data given

A 50.0 mL sample contains Cd2+ and Mn2+

volume of 0.05 M EDTA = 56.3 mL

Titration of the excess unreacted EDTA required 13.4 mL of 0.0310 M Ca2+.

Titration of the newly freed EDTA required 28.2 mL of 0.0310 M Ca2+

Step 2: Calculate mole ratio

The reaction of EDTA and any metal ion is 1:1, the number of mol of Cd2+ and Mn2+ in the mixture equals the total number of mol of EDTA minus the number of mol of EDTA consumed in the back titration with Ca2+:

Step 3: Calculate total mol of EDTA

Total EDTA = (56.3 mL EDTA)(0.0500 M EDTA) = 0.002815 mol EDTA

Consumed EDTA = 0.002815 mol – (13.4 mL Ca2+)(0.0310 M Ca2+) = 0.002815 - 0.0004154 = 0.0023996 mol EDTA

Step 4: Calculate total moles of CD2+ and Mn2+

So, the total moles of Cd2+ and Mn2+ must be 0.0023996 mol

Step 5: Calculate remaining moles of Cd2+

The quantity of cadmium must be the same as the quantity of EDTA freed after the reaction with cyanide:

Moles Cd2+ = (28.2 mL Ca2+)(0.0310 M Ca2+) = 0.0008742 mol Cd2+.

Step 6: Calculate remaining moles of Mn2+

The remaining moles must be Mn2+: 0.0023996 - 0.0008742 = 0.0015254 moles Mn2+

Step 7: Calculate initial concentrations

The initial concentrations must have been:

(0.0008742 mol Cd2+)/(50.0 mL) = 0.0175 M Cd2+

(0.0015254 mol Mn2+)/(50.0 mL) = 0.0305 M Mn2+

The concentration of Cd2+ is 0.0175 M

The concentration of Mn2+ is 0.0305 M

4 0

3 years ago

Select The on that most applys Will mark brainliest

sdas [7]

Answer:

A and C

Explanation:

7 0

3 years ago

Gaseous indium dihydride is formed from the elements at elevated temperature:

3241004551 [841]

Answer:

For 1: The value of $Q_p$ for above reaction is 36.83

For 2: The value of $Q_p$ for above reaction is 36.83

For 3: The equilibrium partial pressure of Indium is 0.126 atm

For 4: The equilibrium partial pressure of hydrogen gas is 0.094 atm

For 5: The equilibrium partial pressure of Indium dihydrogen is 0.018 atm

Explanation:

We are given:

Partial pressure of Indium gas = 0.0650 atm

Partial pressure of hydrogen gas = 0.0330 atm

Partial pressure of Indium dihydride = 0.0790 atm

The given chemical equation follows:

$ln(g)+H_2(g)\rightleftharpoons InH_2(g)$

Initial: 0.065 0.033 0.079

At eqllm: 0.065-x 0.033-x 0.079+x

For 1:

The expression of $Q_p$ for above reaction follows:

$Q_p=\frac{p_{InH_2}}{p_{In}\times p_{H_2}}$

Putting values in above equation, we get:

$Q_p=\frac{0.079}{0.065\times 0.033}=36.83$

Hence, the value of $Q_p$ for above reaction is 36.83

For 2:

We are given:

$K_p$ of the reaction = 1.48

There are 3 conditions:

When $K_{p}>Q_p$ ; the reaction is product favored.
When $K_{p}$ ; the reaction is reactant favored.
When $K_{p}=Q_p$ ; the reaction is in equilibrium.

As, $Q_{p}>K_p$ for the given reaction, the reaction is reactant favored.

Hence, the reaction proceed in the backward direction to attain equilibrium

For 3:

The expression of $K_p$ for above reaction follows:

$K_p=\frac{p_{InH_2}}{p_{In}\times p_{H_2}}$

Putting values in above equation, we get:

$1.48=\frac{(0.079+x)}{(0.065-x)\times (0.033-x)}\\\\x=-0.061,0.835$

Neglecting the value of x = 0.835 because the reaction is going backwards. So, by taking this value, the pressure of the reactants will decrease

So, equilibrium partial pressure of Indium = (0.065 - x) = [0.065 - (-0.061)] = 0.126 atm

For 4:

The equilibrium partial pressure of hydrogen gas = (0.033 - x) = [0.033 - (-0.061)] = 0.094 atm

For 5:

The equilibrium partial pressure of Indium dihydrogen = (0.079 + x) = [0.079 + (-0.061)] = 0.018 atm

6 0

3 years ago

Storm surge, high wind, heavy rains which is most dangerous

Ymorist [56]

The answer is storm surge

6 0

3 years ago

Read 2 more answers