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3 years ago

Determine the maximum amount of Fe that was produced during the experiment. Explain how you determined this amount.

Chemistry

1 answer:

dsp733 years ago

5 0

Answer:

5 moles of Fe

Explanation:

The equation of the reaction is;

2 Al(s) + Fe 2O 3(s) --> 2Fe (s) + Al 2O 3 (s)

Now;

1 mole of Fe2O3 require 2 moles of Al

3 moles of Fe2O3 requires 3 × 2/1 = 6 moles of Al

Hence Al is the limiting reactant.

If 2 moles of Al yields 2 moles of Fe

5 moles of Al yields 5 × 2/2 = 5 moles of Fe

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Does alcohol affect enveloped or naked viruses more

slega [8]

Answer:

the protein capsid of naked viruses is less susceptible to environmental . condition

Explanation:

because the envelop is made in part of phospholipids. once the envelop is lysed ,the virus loses its functiovnal receptors and still able to infect susceptible cells.

7 0

3 years ago

Ethanol melts at -114 °C and boils at 78 °C.

Aliun [14]

A) solid
b)liquid
c)liquid
d)gas

5 0

3 years ago

Whoever awser correct I will give you Brainly

vivado [14]

Answer:

Option A.

2Na + 2H2O —> 2NaOH + H2

Explanation:

To know which option is correct, we shall do a head count of the number of atoms present on both side to see which of them is balanced. This is illustrated below below:

For Option A:

2Na + 2H2O —> 2NaOH + H2

Reactant >>>>>>> Product

2 Na >>>>>>>>>>> 2 Na

4 H >>>>>>>>>>>> 4 H

2 O >>>>>>>>>>>> 2 O

Thus, the above equation is balanced.

For Option B:

2Na + 2H2O —> NaOH + H2

Reactant >>>>>>> Product

2 Na >>>>>>>>>>> 1 Na

4 H >>>>>>>>>>>> 3 H

2 O >>>>>>>>>>>> 1 O

Thus, the above equation is not balanced.

For Option C:

2Na + H2O —> 2NaOH + H2

Reactant >>>>>>> Product

2 Na >>>>>>>>>>> 2 Na

2 H >>>>>>>>>>>> 4 H

1 O >>>>>>>>>>>> 2 O

Thus, the above equation is not balanced.

For Option D:

Na + 2H2O —> NaOH + 2H2

Reactant >>>>>>> Product

1 Na >>>>>>>>>>> 1 Na

4 H >>>>>>>>>>>> 5 H

2 O >>>>>>>>>>>> 1 O

Thus, the above equation is not balanced.

From the illustrations made above, only option A is balanced.

7 0

3 years ago

The Ka for formic acid (HCO2H) is 1.8 × 10-4. What is the pH of a 0.35 M aqueous solution of sodium formate (NaHCO2)?

anzhelika [568]

Answer:

9.36

Explanation:

Sodium formate is the conjugate base of formic acid.

Also,

$K_a\times K_b=K_w$

$K_b$ for sodium formate is $K_b=\frac {K_w}{K_a}$

Given that:

$K_a$ of formic acid = $1.8\times 10^{-4}$

And, $K_w=10^{-14}$

So,

$K_b=\frac {10^{-14}}{1.8\times 10^{-4}}$

$K_b=5.5556\times 10^{-11}$

Concentration = 0.35 M

HCOONa ⇒ Na⁺ + HCOO⁻

Consider the ICE take for the formate ion as:

HCOO⁻ + H₂O ⇄ HCOOH + OH⁻

At t=0 0.35 - -

At t =equilibrium (0.35-x) x x

The expression for dissociation constant of sodium formate is:

$K_{b}=\frac {[OH^-][HCOOH]}{[HCOO^-]}$

$5.5556\times 10^{-11}=\frac {x^2}{0.35-x}$

Solving for x, we get:

x = 0.44×10⁻⁵ M

pOH = -log[OH⁻] = -log(0.44×10⁻⁵) = 4.64

pH + pOH = 14

So,

5 0

3 years ago

Hydrogen gas is collected over water at 21 degrees Celsius. At 21 degrees Celsius the vapor pressure of water is 18.7 torr. If t

Alinara [238K]

Answer : The pressure of hydrogen gas is, 739.3 torr

Explanation :

As we are given:

Vapor pressure of water = 18.7 torr

Barometric pressure = 758 torr

Now we have to calculate the pressure of hydrogen gas.

Pressure of hydrogen gas = Barometric pressure - Vapor pressure of water

Pressure of hydrogen gas = 758 torr - 18.7 torr

Pressure of hydrogen gas = 739.3 torr

Therefore, the pressure of hydrogen gas is, 739.3 torr

3 0

3 years ago