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Rashid [163]
1 year ago
9

All faculty members are happy to see students help each other. Dumbledore is particularly pleased with Hermione. Though, it shou

ld be mentioned that students should not simply copy off each other. You will not learn anything that way. Snape glares at Ron... Ron slouches in his chair. Snape thinks it’s time for a harder problem. How many milligrams of magnesium reacts with excess HCl to produce 31.2 mL of hydrogen gas at 754 Torr and 25.0◦C. The hydrogen is produced by the following reaction: Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g) Express your answer in milligrams
Chemistry
1 answer:
____ [38]1 year ago
3 0

Answer:

m_{Mg}=30.8mgMg

Explanation:

Hello,

Based on the given chemical reaction, as 31.2 mL of hydrogen are yielded, one computes its moles via the ideal gas equation under the stated conditions as shown below:

n_{H_2}=\frac{PV}{RT}=\frac{754torr*\frac{1atm}{760torr}*0.0312L}{0.082 \frac{atm*L}{mol*K}*298.15K}=1.27x10^{-3}molH_2

Now, since the relationship between hydrogen and magnesium is 1 to 1, one computes its milligrams by following the shown below proportional factor development:

m_{Mg}=1.27x10^{-3}molH_2*\frac{1molMg}{1molH_2}*\frac{24.305gMg}{1molMg}*\frac{1000mgMg}{1gMg}\\m_{Mg}=30.8mgMg

Best regards.

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Zinc is to be electroplated onto both sides of an iron sheet that is 20 cm2 as a galvanized sacrificial anode. It is desired to
Alexandra [31]

Answer:

The time required for the coating is 105 s

Explanation:

Zinc undergoes reduction reaction and absorbs two (2) electron ions.

The expression for the mass change at electrode (m_{ch}) is given as :

\frac{m_{ch}}{M} ZF = It

where;

M = molar mass

Z = ions charge at electrodes

F = Faraday's constant

I = current

A = area

t = time

also; (m_{ch}) = (Ad) \rho ; replacing that into above equation; we have:

\frac{(Ad) \rho}{M} ZF = It  ---- equation (1)

where;

A = area

d = thickness

\rho = density

From the above equation (1); The time required for coating can be calculated as;

[ \frac{20 cm^2 *0.0025 cm*7.13g/cm^3}{65.38g/mol}*2 \frac{moles\ of \ electrons}{mole \ of \ Zn} * 9.65*10^4 \frac{C}{mole \ of \ electrons }  ] = (20 A) t

t = \frac{2100}{20}

= 105 s

8 0
2 years ago
2N2H4(l) + N2O4(l) → 3N2(g) + 4H2O(g) [balanced] How many moles of N2H4 is required to produce 28.3 g of N2? Assume that all rea
JulijaS [17]

Answer: 0.67 moles of N_2H_4

Explanation:

According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number 6.023\times 10^{23} of particles.

To calculate the moles, we use the equation:

\text{Number of moles of nitrogen}=\frac{\text{Given mass}}{\text {Molar mass}}=\frac{28.3}{28.02}=1mole

2N_2H_4(l)+N_2O_4(l)\rightarrow 3N_2(g)+4H_2O(g)

According to stoichiometry:

3 moles of N_2 is produced by 2 moles of N_2H_4

Thus 1 mole of N_2 is produced by= \frac{2}{3}\times 1=0.67moles of N_2H_4

Thus 0.67 moles of N_2H_4 are required to produce 28.3 g of N_2

6 0
2 years ago
Carbonic acid, H2CO3, has two acidic hydrogens. A solution containing an unknown concentration of carbonic acid is titrated with
stepan [7]

Answer:

1) Net ionic equation :

2H^+(aq)+2OH^-(aq)\rightarrow 2H_2O(l)

2) 0.765 M is  the molarity of the carbonic acid solution.

Explanation:

1) In aqueous carbonic acid , carbonate ions and hydrogen ion is present.:

H_2CO_3(aq)\rightarrow 2H^+(aq)+CO_3^{2-}(aq) ..[1]

In aqueous potassium hydroxide , potassium ions and hydroxide ion is present.:

KOH(aq)\rightarrow K^+(aq)+OH^{-}(aq) ..[2]

In aqueous potassium carbonate , potassium ions and carbonate ion is present.:

K_2CO_3(aq)\rightarrow 2K^+(aq)+CO_3^{2-}(aq) ..[3]

H_2CO_3(aq)+2KOH(aq)\rightarrow K_2CO_3(aq)+2H_2O(l)

From one:[1] ,[2] and [3]:

2H^+(aq)+CO_3^{2-}(aq)+2K^+(aq)+2OH^{-}(aq)\rightarrow 2K^+(aq)+CO_3^{2-}(aq)+H_2O(l)

Cancelling common ions on both sides to get net ionic equation :

2H^+(aq)+2OH^-(aq)\rightarrow 2H_2O(l)

2)

To calculate the concentration of acid, we use the equation given by neutralization reaction:

n_1M_1V_1=n_2M_2V_2

where,

n_1,M_1\text{ and }V_1 are the n-factor, molarity and volume of acid which is H_2CO_3

n_2,M_2\text{ and }V_2 are the n-factor, molarity and volume of base which is KOH.

We are given:

n_1=2\\M_1=?\\V_1=50.0 mL\\n_2=1\\M_2=3.840 M\\V_2=20.0 mL

Putting values in above equation, we get:

M_1=\frac{1\times 3.840 M\times 20.0 mL}{2\times 50.2 mL}=0.765 M

0.765 M is  the molarity of the carbonic acid solution.

6 0
2 years ago
Summarize the process a scientist goes through to come up with a<br> satisfactory solution.
maria [59]
Guess and check, test, trial and error, completion.
6 0
1 year ago
The density of o2 gas at 16 degrees Celsius and 1.27atm is?
velikii [3]

Answer:

The density of O₂ gas is 1.71 \frac{g}{L}

Explanation:

Density is a quantity that allows you to measure the amount of mass in a given volume of a substance. So density is defined as the quotient between the mass of a body and the volume it occupies:

density=\frac{mass}{volume}

An ideal gas is characterized by three state variables: absolute pressure (P), volume (V), and absolute temperature (T). The relationship between them constitutes the ideal gas law, an equation that relates the three variables if the amount of substance, number of moles n, remains constant and where R is the molar constant of the gases:

P * V = n * R * T

So, you can get:

\frac{n}{V} =\frac{P}{R*T}

The relationship between number of moles and mass is:

n=\frac{mass}{molar mass}

Replacing:

\frac{\frac{mass}{molar mass} }{V} =\frac{P}{R*T}

\frac{mass}{V*Molar mass} =\frac{P}{R*T}

So:

\frac{mass}{V} =\frac{P*molar mass}{R*T}

Knowing that 1 mol of O has 16 g, the molar mass of O₂ gas is 32 \frac{g}{mol}.

Then:

\frac{mass}{V} =\frac{P*molar mass of O_{2} }{R*T}

In this case you know:

  • P=1.27 atm
  • molar mass of O₂= 32 \frac{g}{mol}.
  • R= 0.0821 \frac{atm*L}{mol*K}
  • T= 16 °C=  289 °K (0°C= 273°K)

Replacing:

density=\frac{mass}{V} =\frac{1.27atm*32\frac{g}{mol}  }{0.0821\frac{atm*L}{mol*K} *289 K}

Solving:

density= 1.71 \frac{g}{L}

<u><em>The density of O₂ gas is 1.71 </em></u>\frac{g}{L}<u><em></em></u>

3 0
1 year ago
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