PLEASE HELP WILL GIVE BRAINLIEST

If you have 6 moles of reactant A and excess of B and C, how much product E would be formed

pychu [463]

B and C are in excess so amount of E will be determined by A.

Amount of product is determined by limiting reagents - Always.

Hence 6 moles of E will be formed.

Hope this helps!

3 0

3 years ago

What is the mass of a proton

Anni [7]

1.6726219 × 10-27 kilograms
(The -27 is a exponent)

6 0

3 years ago

A container holds a mixture of two gases at The partial pressures of gas A and gas B, respectively, are and If of a third gas is

RSB [31]

Answer:

The total pressure will be the sum of the partial pressure of these three gases.

Explanation:

According to Dalton law of partial pressure,

The total pressure exerted by mixture of gases is equal to the sum of partial pressure of the individual gas.

This expression can be written as,

P(total) = P₁ + P₂ + P₃ + ...... Pₙ

For example:

If the pressure of A is 2 atm and Partial pressure of B is 4 atm the total pressure will be,

P(total) = P₁ + P₂

P(total) = 2 atm + 4atm

P(total) = 6 atm

when third gas is added which exert the partial pressure of 4 atm,

Then total pressure will becomes,

P(total) = P₁ + P₂ + P₃

P(total) = 2 atm + 4atm + 4 atm

P(total) = 10 atm

3 0

3 years ago

in a torricell barometer a pressure of one atmospheric pressure supports q 760 mm coloumn of mercury.if the original tube contai

dybincka [34]

Answer:

The answer to the question is

The height of the mercury fluid column remain the same.

Explanation:

The pressure, P in a column of fluid of height, h is given by

P = (Density p)×(height of fluid column h)×(gravity g)

Therefore, when the diameter is doubled we have

Density of the mercury in the tube with twice the diameter = (Mass of mercury)/(volume of mercury) where the volume of mercury = h×pi×(Diameter×2)^2/4 = h×pi×Diameter^2. Therefore the volume increases by a factor of 4 and therefore the mass increases by a factor of 4 which means that the density remains the same hence

P = p×h1×g = p×h2×g Therefore h1 = h2

The height of the fluid column remain the same

5 0

3 years ago

A natural water with a flow of 3800 m3/d is to be treated with an alum dose of 60 mg/L. Determine the chemical feed rate for the

svet-max [94.6K]

Explanation:

First, we will calculate the feed rate of alum as follows.

$\frac{\text{60 mg alum}}{\text{1 L water}} \times \frac{\text{1000 L water}}{1 m^{3}} \times \frac{3800 m^{3}}{day} \times \frac{\text{1 g alum}}{\text{1000 mg alum}}$

= 228000 g/day

Converting this amount into g/min as follows.

$\frac{228000 g}{1 day} \times \frac{1 day}{1440 min}$

= 158 g/min

Now, the chemical equation will be as follows.

$Al_{2}(SO_{4})_{3}.14H_{2}O \rightarrow 2Al(OH)_{3}(s) + 6H^{+} + 3SO^{2-}_{4} + 8H_{2}O$

$\frac{\text{30 mg alum}}{1 L} \times \frac{\text{1 mmol alum}}{\text{594 mg alum}} \times \frac{\text{3 mmol SO^{2-}_{4}}}{\text{1 mmol alum}}$

= 0.151 mmol $mmol SO^{2-}_{4}/L$

$\frac{0.151 mmol SO^{2-}_{4}}{L} \times \frac{\text{2 meq SO^{2-}_{4}}}{\text{1 mmol SO^{2-}_{4}}} \times \frac{\text{1 meq Alk}}{\text{1 meq SO^{2-}_{4}}} \times \frac{\text{50 mg CaCO_{3}}}{\text{1 meq Alk}}$

= 15.15 mg $CaCO_{3}$ /L

For precipitate:

$Al_{2}(SO_{4})_{3}.14H_{2}O \rightarrow 2Al(OH)_{3}(s) + 6H^{+} + 3SO^{2-}_{4} + 8H_{2}O$

$\frac{\text{30 mg alum}}{1 L} \times \frac{\text{1 mmol alum}}{\text{594 mg alum}} \times \frac{\text{2 mmol Al(OH)_{3}}}{\text{1 mmol alum}} \times \frac{\text{78 mg Al(OH)_{3}}}{\text{1 mmol Al(OH)_{3}}}$

= 7.88 $Al(OH)_{3}/L$

$\frac{7.88 mg Al(OH)_{3}}{1 L} \times \frac{3800 m^{3}}{1 day} \times \frac{1000 L}{1 m^{3}} \times \frac{1 kg}{10^{6} mg}$

= 29.9 $Al(OH)_{3}/day$

3 0

3 years ago