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3 years ago

Plzzzzzzzzzz Helpppppppp!!!!!!!!!!!!!! 20+PTS and brainliest!!!!!!!!!!!!!!!

Chemistry

1 answer:

jek_recluse [69]3 years ago

3 0

Answer:

Using <em>activity series of metals (search topic)</em>, elements at top of series will replace any metal element below it but elements low on the series will not replace elements above. See reactions in explanation.

Explanation:

Pb(NO₃)₂(aq) + Mg(s) => Mg(NO₃)₂(aq) + Pb(s)

Cu(NO₃)₂(aq) + Mg(s) => Mg(NO₃)₂(aq) + Cu(s)

Fe(NO₃)₂(aq) + Mg(s) => Mg(NO₃)₂(aq) + Fe(s)

Cu(NO₃)₂(aq) + Pb(s) => Pb(NO₃)₂(aq) + Cu(s)

Fe(NO₃)₂(aq) + Pb(s) => No Reaction (Pb will not replace Fe according to the activity series of metals).

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The stopcock connecting a 3.06 L bulb containing methane gas at a pressure of 9.61 atm, and a 6.65 L bulb containing oxygen gas

Contact [7]

Answer : The final pressure in the system is 4.22 atm.

Explanation :

First we have to calculate the moles of methane.

$PV=n_1RT$

where,

P = pressure of gas = 9.61 atm

V = volume of gas = 3.06 L

T = temperature of gas = T

$n_1$ = number of moles of methane gas = ?

R = gas constant

Now put all the given values in the ideal gas equation, we get:

$(9.61atm)\times (3.06L)=n_1\times RT$

$n_1=\frac{29.4}{RT}$

Now we have to calculate the moles of oxygen gas.

$PV=n_2RT$

where,

P = pressure of gas = 1.75 atm

V = volume of gas = 6.65 L

T = temperature of gas = T

$n_2$ = number of moles of oxygen gas = ?

R = gas constant

Now put all the given values in the ideal gas equation, we get:

$(1.75atm)\times (6.65L)=n_2\times RT$

$n_2=\frac{11.6}{RT}$

Now we have to determine the final pressure in the system after mixing the gases.

$P_{total}=(n_1+n_2)\times \frac{RT}{V_{total}}$

where,

$P_{total}$ = final pressure of gas = ?

$V_{total}$ = final volume of gas = (3.06 + 6.65)L = 9.71 L

T = temperature of gas = T

R = gas constant

Now put all the given values in the ideal gas equation, we get:

$P_{total}=(\frac{29.4}{RT}+\frac{11.6}{RT})\times \frac{RT}{9.71L}$

$P_{total}=4.22atm$

Therefore, the final pressure in the system is 4.22 atm.

4 0

3 years ago

4) A 4.00 L balloon is filled with 0.297 moles of helium gas with a pressure

Nadya [2.5K]

Answer: 149

Explanation:

Using the ideal gas equation; PV = nRT

P= Pressure = 0.910 atm, T= Temperature = ?

V= Volume = 4.0L R = Gas constant = 0.08206 L.atm/mol/K

n = number of moles = 0.297

Making 'T' the subject of the formular, we have;

T = P V/ n R = 0.910 x 4 / 0.297 x 0.08206

= 149

6 0

3 years ago

Using the periodic table, find the molecular mass of H2. H2 = g/mole

alekssr [168]

MH₂ = 2×mH = 2×1g = 2 g/mol

5 0

3 years ago

A dependent variable is:

12345 [234]

Answer:

the answer is b

Explanation:

an independent variable is the one that is manipulated

5 0

3 years ago

How many grams of ZrCl4 can be produced if 123 g of ZrSiO4 react with 85.0 g of Cl2?

nydimaria [60]

Answer:

153.3 grams of ZrCl₄ are produced

Explanation:

The equation of the reaction is as follows:

ZrSiO₄ + Cl₂ ----> ZrCl₄ + SiO₂ + O₂

molar mass of ZrSiO₄ = (91 + 32 + 16 * 4) = 187.0 g/mol

molar mass of ZrCl₄ = (91 + 35.5 * 4) = 233.0 g/mol

molar mass of Cl₂ = (35.5 * 2) 71.0 g/mol

From the equation of reaction, 1 mole of ZrSiO₄ reacts with one mole of Cl₂ to produce one mole of ZrCl₄

number of moles of ZrSiO₄ present in 123 g = 123/187 = 0.65 moles

number of moles of Cl₂ present in 85.0 g = 85.0/71.0 = 1.19 moles

therefore, ZrSiO₄ is the limiting reactant

123.0 g of ZrSiO₄ will react with excess Cl₂ to produce 123 * 233/187 grams of ZrCl₄ = 153.3 grams of ZrCl₄

Therefore, 153.3 grams of ZrCl₄ are produced

8 0

3 years ago