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2 years ago

How many moles of CO2 could fit in a 475 mL bag a -22° C and 855 mmHg?

Chemistry

1 answer:

Sergeeva-Olga [200]2 years ago

4 0

Answer:

The amount of moles of CO₂ is 0.026 moles

Explanation:

An ideal gas is a theoretical gas that is considered to be composed of point particles that move randomly and do not interact with each other. Gases in general are ideal when they are at high temperatures and low pressures.

The equation known as the ideal gas equation explains the relationship between the four variables P (Pressure), V (Volume), T (Temperature) and n (Amount of substance). The ideal gas law is expressed mathematically as:

P*V = n*R*T

where P represents the pressure of the gas, V its volume, n the number of moles of gas (which must remain constant), R the constant of the gases and T the temperature of the gas.

In this case:

P= 855 mmHg
V= 475 mL= 0.475 L
n= ?

R= 62.36367 $\frac{mmHg*L}{mol*K}$

T= -22°C= 251 °K

Replacing:

855 mmHg*0.475 L=n*62.36367 $\frac{mmHg*L}{mol*K}$ *251 °K

Solving:

$n=\frac{855 mmHg*0.475 L}{62.36367 \frac{mmHg*L}{mol*K} *251 K}$

n= 0.026 moles

<u><em>The amount of moles of CO₂ is 0.026 moles</em></u>

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Now consider the reaction when 45.0 g NaOH have been added. What amount of NaOH is this, and what amount of FeCl3 can be consume

gayaneshka [121]

The correct answer is that 1.125 mol of NaOH is available, and 60.75 g of FeCl₃ can be consumed.

The mass of NaOH is 45 g

The molar mass of NaOH = 40 g/mol

The moles of NaOH = mass / molar mass

= 45 / 40

= 1.125

Thus, 1.125 mol NaOH is available

3 NaOH + FeCl₃ ⇒ Fe (OH)₃ + 3NaCl

3 mol of NaOH react with 1 mol of FeCl₃

1.125 moles of NaOH will react with x moles of FeCl₃

x = 1.125 / 3

x = 0.375 mol

0.375 mol FeCl₃ can take part in reaction

The molar mass of FeCl₃ is 162 g/mol

The mass of FeCl₃ = moles × mass

= 0.375 × 162

= 60.75 g

Thus, the amount of FeCl₃, which can be consumed is 60.75 g

6 0

2 years ago

You mix 500.0 mL of 0.250 M iron(III) chloride solution with 425.0 mL of 0.350 M barium chloride solution. Assuming the volumes

12345 [234]

Answer:

$M=0.727M$

Explanation:

Hello,

In this case, since iron (III) chloride (FeCl3) and barium chloride (BaCl2) are both chloride-containing compounds, we can compute the moles of chloride from each salt, considering the concentration and volume of the given solutions, and using the mole ratio that is 1:3 and 1:2 for the compound to chlorine:

$n_{Cl^-}=0.50L*0.250\frac{molFeCl_3}{L}*\frac{3molCl^-}{1molFeCl_3}=0.375molCl^- \\\\n_{Cl^-}=0.425L*0.350\frac{molBaCl_2}{L}*\frac{2molCl^-}{1molBaCl_2}=0.2975molCl^-$

So the total mole of chloride ions:

$N_{Cl^-}=0.2975mol+0.375mol=0.6725molCl^-$

And the total volume by adding the volume of each solution in L:

$V=0.500L+0.425L=0.925L$

Finally, the molarity turns out:

$M=\frac{0.6725molCl^-}{0.925L}\\ \\M=0.727M$

Best regards.

5 0

2 years ago

Use the following half-reactions to design a voltaic cell: Sn4+(aq) + 2 e− → Sn2+(aq) Eo = 0.15 V Ag+(aq) + e−→ Ag(s) Eo = 0.80

AVprozaik [17]

Answer:

E° = 0.65 V

Explanation:

Let's consider the following reductions and their respective standard reduction potentials.

Sn⁴⁺(aq) + 2 e⁻ → Sn²⁺(aq) E°red = 0.15 V

Ag⁺(aq) + e⁻ → Ag(s) E°red = 0.80 V

The reaction with the highest reduction potential will occur as a reduction while the other will occur as an oxidation. The corresponding half-reactions are:

Reduction (cathode): Ag⁺(aq) + e⁻ → Ag(s) E°red = 0.80 V

Oxidation (anode): Sn²⁺(aq) → Sn⁴⁺(aq) + 2 e⁻ E°red = 0.15 V

The overall cell potential (E°) is the difference between the standard reduction potential of the cathode and the standard reduction potential of the anode.

E° = E°red, cat - E°red, an = 0.80 V - 0.15 V = 0.65 V

4 0

2 years ago

Calculate the amount, in moles, of PO43- present at equilibrium when excess Sr3(PO4)2 is added to 750. mL 1.2 M Sr(NO3)2(aq). As

Crank

Answer:

1.8 × 10⁻¹⁶ mol

Explanation:

(a) Calculate the solubility of the Sr₃(PO₄)₂

Let s = the solubility of Sr₃(PO₄)₂.

The equation for the equilibrium is

Sr₃(PO₄)₂(s) ⇌ 3Sr²⁺(aq) + 2PO₄³⁻(aq); Ksp = 1.0 × 10⁻³¹

1.2 + 3s 2s

$K_{sp} =\text{[Sr$^{2+}$]$^{3}$[PO$_{4}^{3-}$]$^{2}$} = (1.2 + 3s)^{3}\times (2s)^{2} = 1.0 \times 10^{-31}\\\text{Assume } 3s \ll 1.2\\1.2^{3} \times 4s^{2} = 1.0 \times 10^{-31}\\6.91s^{2} = 1.0 \times 10^{-31}\\s^{2} = \dfrac{1.0 \times 10^{-31}}{6.91} = 1.45 \times 10^{-32}\\\\s = \sqrt{ 1.45 \times 10^{-32}} = 1.20 \times 10^{-16} \text{ mol/L}\\$

(b) Concentration of PO₄³⁻

[PO₄³⁻] = 2s = 2 × 1.20× 10⁻¹⁶ mol·L⁻¹ = 2.41× 10⁻¹⁶ mol·L⁻¹

(c) Moles of PO₄³⁻

Moles = 0.750 L × 2.41 × 10⁻¹⁶ mol·L⁻¹ = 1.8 × 10⁻¹⁶ mol

7 0

2 years ago

What reaction would take place if a hot tungsten filament bulb was surrounded by air

expeople1 [14]

I don't think it wont be a big explosion

4 0

2 years ago

Read 2 more answers