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nikitadnepr [17]
3 years ago
13

Consider the reaction at equilibrium: Mg(OH)2(aq) ←→ MgO(s) + H2O(ℓ) If you add more MgO to the system, how will the system resp

ond? 1. the reaction will shift to the right 2. the reaction will shift to the left 3. nothing will happen 4. not enough information
Chemistry
1 answer:
kobusy [5.1K]3 years ago
4 0

Answer:

2. the reaction will shift to the left

Explanation:

Hello,

By means of the Le Chatelier's principle, when adding a product, the inverse reaction is favored as the products are now considered as reactants, thus, since magnesium oxide is a product for the given chemical reaction, and in addition, it is being added, the equilibrium will shift leftwards, thus, the answer is:

2. the reaction will shift to the left

Best regards.

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3 years ago
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1.00 mole of oxygen at 60 degrees celcius and 5 bar is flowing through a 20cm diameter pipe (inner diameter). What is the veloci
jeka94

Answer:

u = 0.176 m/x

Explanation:

∴ n O2 = 1 mol

∴ mass O2 = 1 mol * 32 g/mol = 32 g

∴ T = 60°C = 333 K

∴ P = 5 bar

⇒ V = RTn/P = (83.14 bar.cm³/mol.K)*(333 K)*(1 mol))/(5 bar)

⇒ V = 5537.124 cm³

∴ d = 20cm

⇒ A = (1/4π)*d² = 314.16 cm²

velocity of the gas (m/x):

  • u = m / ρ*A
  • let time (t) = x sec

∴ ρ = 32 g / 5537.124 cm³ = 5.78 E-3 g/cm³

∴ mass flow rate (m) = 32g / x

⇒ u = (32 g/x) / (( 5.78 E-3 g/cm³)*(314.16cm²))

⇒ u = 17.625 cm/x * ( m/100cm) = 0.176 m/x

4 0
3 years ago
Acid and an alkali combined together will give a salt,if they are mixed in the right amount
algol13
A Neutralisation reaction, the alkali is neutralizing the acid.

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3 years ago
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A 37.2 g sample of copper at 99.8 °C is carefully placed into an insulated container containing 188 g of water at 18.5 °C. Calcu
klasskru [66]

Answer:

T₂ = 19.95°C

Explanation:

From the law of conservation of energy:

Heat\ Lost\ by\ Copper = Heat\ Gained\ by\ Water\\m_cC_c\Delta T_c = m_wC_w\Delta T_w

where,

mc = mass of copper = 37.2 g

Cc = specific heat of copper = 0.385 J/g.°C

mw = mass of water = 188 g

Cw = specific heat of water = 4.184 J/g.°C

ΔTc = Change in temperature of copper = 99.8°C - T₂

ΔTw = Change in temperature of water = T₂ - 18.5°C

T₂ = Final Temperature at Equilibrium = ?

Therefore,

(37.2\ g)(0.385\ J/g.^oC)(99.8\ ^oC-T_2)=(188\ g)(4.184\ J/g.^oC)(T_2-18.5\ ^oC)\\99.8\ ^oC-T_2 = \frac{(188\ g)(4.184\ J/g.^oC)}{(37.2\ g)(0.385\ J/g.^oC)}(T_2-18.5\ ^oC)\\\\99.8\ ^oC-T_2 = (54.92) (T_2-18.5\ ^oC)\\54.92T_2+T_2 = 99.8\ ^oC + 1016.02\ ^oC\\\\T_2 = \frac{1115.82\ ^oC}{55.92}

<u>T₂ = 19.95°C</u>

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3 years ago
Which part of the factory is most similar to the nucleus of a living cell
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The control room is similar to the nucleus, because the nucleus is in charge of cell functions.
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