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PtichkaEL [24]
2 years ago
10

Sodium reacts with chlorine gas according to the following reaction: 2Na(s)+Cl2(g)→2NaCl(s) What volume of Cl2 gas, measured at

689 torr and 39 ∘C, is required to form 28 g of NaCl?
Chemistry
1 answer:
asambeis [7]2 years ago
6 0

Answer:6.719Litres of Cl2 gas.

Explanation:According to eqn of rxn

2Na +Cl2=2NaCl

P=689torr=689/760=0.91atm

T=39°C+273=312K

according to stoichiometry of the reaction,1Moles of Cl2 gives 2moles of NaCl

But 28g of NaCl was given,we have to convert this to moles by using the relation, n=mass/MW

MW of NaCl=23+35.5=58.5g/mol

n=28g(mass given of NaCl)/58.5

n=0.479moles of NaCl

Going back to the reaction,

if 1moles of Cl2 produces 2moles of NaCl

x moles of Cl2 will give 0.479moles of NaCl.

x=0.479*1/2

x=0.239moles of Cl2.

To find the volume, we use ideal ggas eqn,PV=nRT

V=nRT/P

V=0.239*0.082*312/0.91

V=6.719Litres

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Answer: 3.75 M

Explanation:

400 mL = 0.4 L

NaOH has a molar mass of around 40 g/mol.

\frac{60 grams}{40 g/mol} = 1.5 moles

Molarity = \frac{1.5 moles}{0.4 Liters} = 3.75 M

8 0
2 years ago
What is a daughter cell
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The cells that result from the reproductive division of one cell during mitosis or meiosis
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A buffer is composed of nh3 and nh4cl. How would this buffer solution control the ph of a solution when a small amount of a stro
Shkiper50 [21]

Answer:

According to Le-chatelier principle, equilibrium will shift towards left to minimize concentration of OH^{-} and keep same equilibrium constant

Explanation:

In this buffer following equilibrium exists -

NH_{3}(aq.)+H_{2}O(l)\rightleftharpoons NH_{4}^{+}(aq.)+OH^{-}(aq.)

So, OH^{-} is involved in the above equilibrium.

When a strong base is added to this buffer, then concentration of OH^{-} increases. Hence, according to Le-chatelier principle, above equilibrium will shift towards left to minimize concentration of OH^{-} and keep same equilibrium constant.

Therefore excess amount of OH^{-} combines with NH_{4}^{+} to produce ammonia and water. So, effect of addition of strong base on pH of buffer gets minimized.

5 0
3 years ago
Strike anywhere matches contain the compound tetraphosphorus trisulfide, which burns to form tetraphosphorus decaoxide and sulfu
erica [24]

Answer:

194.6 mL of SO₂

Explanation:

The reaction that takes place is:

P₄S₃ + 6O₂(g) → P₄O₁₀ + 3SO₂(g)

<u>To solve this problem we need to use PV=nRT</u>, so first let's convert the given units:

  • 23.8 °C → 23.8 + 273.15 = 296.95 K
  • 747 torr → 747/760 = 0.983 atm

We need to calculate V, so in order to do that we calculate n, using the mass of the reactant (P₄S₃):

0.576 g P₄S₃ * \frac{1molP_{4}S_{3}}{220gP_{4}S_{3}} *\frac{3molSO_{2}}{1molP_{4}S_{3}} = 7.85 * 10⁻³ mol SO₂ = n

  • Now we calculate V:

PV=nRT

0.983 atm * V =  7.85 * 10⁻³ mol * 0.082 atm·L·mol⁻¹·K⁻¹ * 296.95 K

V = 0.1946 L

  • Finally we convert L into mL:

0.1946 * 1000 = 194.6 mL

8 0
3 years ago
A 32.5 g iron rod, initially at 22.4 ∘C, is submerged into an unknown mass of water at 63.0 ∘C, in an insulated container. The f
a_sh-v [17]

Answer:

m_{H_2O}=39.0g

Explanation:

Hello,

In this case, is possible to infer that the thermal equilibrium is governed by the following relationship:

\Delta H_{iron}=-\Delta H_{H_2O}\\m_{iron}Cp_{iron}(T_{eq}-T_{iron})=-m_{H_2O}Cp_{H_2O}(T_{eq}-T_{H_2O})

Thus, both iron's and water's heat capacities are: 0.444 and 4.18 J/g°C respectively, so one solves for the mass of water as shown below:

m_{H_2O}=\frac{m_{iron}Cp_{iron}(T_{eq}-T_{iron})}{-Cp_{H_2O}(T_{eq}-T_{H_2O}} \\\\m_{H_2O}=\frac{32.5g*0.444\frac{J}{g^0C}*(59.7-22.4)^0C}{-4.18\frac{J}{g^0C}*(59.7-63.0)^0C} \\\\m_{H_2O}=39.0g

Best regards.

8 0
3 years ago
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