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Sergeeva-Olga [200]

3 years ago

5

An aqueous solution of potassium sulfate is allowed to react with an aqueous solution of calcium nitrate. identify the solid in

the balanced equation.

1 answer:

Maksim231197 [3]3 years ago

6 0

Ans: Calcium sulfate.

K2SO4 (aq) + Ca(NO3)2 (aq) ⇒ 2KNO3 (aq) + CaSO4 (s)

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The Doppler effect related change in wave frequency to what?

Irina18 [472]

Answer:

Relative motion of source and observer.

Explanation:

Doppler Effect : It is the change in the observed frequency of the wave due to<em><u> relative motion of the source and object </u></em>.

8 0

3 years ago

What is the percent of hydrogen by mass in CH4O

WITCHER [35]

Answer:

My guess would be 12.5828

4 0

3 years ago

How many moles of aluminum are needed to make 9 moles of molecular hydrogen? given the reaction: 2 al + 6 hcl → 2 alcl3 + 3h2 6

sergiy2304 [10]

1) Chemical equation

2Al + 6 HCl ---> 2Al Cl3 + 3 H2

2) molar ratios

2 mol Al : 3 moles H2

3) Proportion

2 mol Al / 3mol H2 = x / 9 mol H2

4) Solve for x

x = 9 mol H2 * 2 mol Al / 3 mol H2 = 6 mol Ag

Answer: 6 moles

7 0

3 years ago

Where the oxygen comes from the air (21% O2 and 79% N2). If oxygen is fed from air in excess of the stoichiometric amount requir

guajiro [1.7K]

Answer:

$y_{O2} =4.3$ %

Explanation:

The ethanol combustion reaction is:

$C_{2}H_{5} OH+3O_{2}$ → $2CO_{2}+3H_{2}O$

If we had the amount (x moles) of ethanol, we would calculate the oxygen moles required:

$x*1.10(excess)*\frac{3 O_{2}moles }{etOHmole}$

Dividing the previous equation by x:

$1.10(excess)*\frac{3 O_{2}moles}{etOHmole}=3.30\frac{O_{2}moles}{etOHmole}$

We would need 3.30 oxygen moles per ethanol mole.

Then we apply the composition relation between O2 and N2 in the feed air:

$3.30(O_{2} moles)*\frac{0.79(N_{2} moles)}{0.21(O_{2} moles)}=121.414 (N_{2} moles )$

Then calculate the oxygen moles number leaving the reactor, considering that 0.85 ethanol moles react and the stoichiometry of the reaction:

$3.30(O_{2} moles)-0.85(etOHmoles)*\frac{3(O_{2} moles)}{1(etOHmoles)} =0.75O_{2} moles$

Calculate the number of moles of CO2 and water considering the same:

$0.85(etOHmoles)*\frac{3(H_{2}Omoles)}{1(etOHmoles)}=2.55(H_{2}Omoles)$

$0.85(etOHmoles)*\frac{2(CO_{2}moles)}{1(etOHmoles)}=1.7(CO_{2}moles)$

The total number of moles at the reactor output would be:

$N=1.7(CO2)+12.414(N2)+2.55(H2O)+0.75(O2)\\ N=17.414(Dry-air-moles)$

So, the oxygen mole fraction would be:

$y_{O_{2}}=\frac{0.75}{17.414}=0.0430=4.3$ %

6 0

3 years ago

What is the volume of 0.23kg of pure water

vichka [17]

M = 0,23kg = 230g
d = 1g/cm³

V = 230g / 1g/cm³ = 230cm³ = 0,23L

7 0

3 years ago

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