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3 years ago

Products of MnO + CL2

Chemistry

1 answer:

Alenkinab [10]3 years ago

3 0

If the equation is complete the products would be manganese chloride and oxygen gas would be given off.

MnCl2 + O2

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Balance the chemical equation below: CO2 + H2O + 2678 kJ-> C6H1206 + O2 a. How many moles of H20 are involved in this reactio

11Alexandr11 [23.1K]

Answer:

a) 6 mol H2O

b) this reaction is endothermic

c) when 1 mol of CO2 is used, in the reaction they occur 0.5025 KJ

Explanation:

balanced eq:

6CO2 + 6H2O + 2678 KJ ↔ C6H12O6 + 6O2

6 - C - 6

18 - O - 18

12 - H - 12

a) mol H2O = 6 mol.......from balanced equation.

b) ΔE = 2678 KJ....... this reaction absorbs heat ( ΔE is positive )

c) 1 gramo C6H12O6 ≅ 4 cal

Mw C6H12O6 = 180.156 g/mol

⇒ 1mol CO2 * ( mol C6H12O6 / 6mol CO2 ) =0.166 mol C6H12O6

⇒ 0.166mol C6H12O6 * ( 180.156 g C6H12O6 / mol ) = 30.026g C6H12O6

⇒30.026 gC6H12O6 * ( 4 cal / gC6H12O6 ) * ( Kcal / 1000 cal ) * (4184 J / Kcal ) * ( KJ / 1000 J ) = 0.5025 KJ C6H12O6.

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3 years ago

Sedimentary rocks are changed to metaphoric rocks through seduction and _____________?

Nana76 [90]

Hello,

Here is your answer:

The proper answer to this question is "deposition".

Your answer deposition.

If you need anymore help feel free to ask me!

Hope this helps!

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3 years ago

A mothball, composed of naphthalene (c10h8), has a mass of 1.64 g . part a how many naphthalene molecules does it contain?

OLga [1]

The molar mass of Naphthalene is 128g/mol
Therefore; a mass of 1.64 g of Naphthalene contains'
= 1.64g/128 g
= 0.0128 moles
But, from the Avogadro's law 1 mole of a substance contains 6.022 × 10^23 particles
Therefore 1 mole of Naphthalene contains 6.022×10^23 molecules
Hence; 0.0128 moles × 6.022 ×10^23 molecules
= 7.716 × 10^21 molecules

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3 years ago

What is the part of plant from where <br>new flowers of fruit grows called

Andru [333]

Answer: it is called the ovary

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2 years ago

EXTRA POINTSSS 1. A solution at 25 degrees Celsius is 1.0 × 10–5 M H3O+. What is the concentration of OH– in this solution?

AlekseyPX

Answer:

Concentration of OH⁻:

1.0 × 10⁻⁹ M.

Explanation:

The following equilibrium goes on in aqueous solutions:

$\text{H}_2\text{O}\;(l)\rightleftharpoons \text{H}^{+}\;(aq) + \text{OH}^{-}\;(aq)$ .

The equilibrium constant for this reaction is called the self-ionization constant of water:

$K_w = [\text{H}^{+}]\cdot[\text{OH}^{-}]$ .

Note that water isn't part of this constant.

The value of $K_w$ at 25 °C is $10^{-14}$ . How to memorize this value?

The pH of pure water at 25 °C is 7.
$[\text{H}^{+}] = 10^{-\text{pH}} = 10^{-7}\;\text{mol}\cdot\text{dm}^{-3}$
However, $[\text{OH}^{-}] = [\text{H}^{+}]=10^{-7}\;\text{mol}\cdot\text{dm}^{-3}$ for pure water.
As a result, $K_w = [\text{H}^{+}] \cdot[\text{OH}^{-}] = (10^{-7})^{2} = 10^{-14}$ at 25 °C.

Back to this question. $[\text{H}^{+}]$ is given. 25 °C implies that $K_w = 10^{-14}$ . As a result,

$\displaystyle [\text{OH}^{-}] = \frac{K_w}{[\text{H}^{+}]} = \frac{10^{-14}}{1.0\times 10^{-5}} = 10^{-9} \;\text{mol}\cdot\text{dm}^{-3}$ .

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3 years ago