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klemol [59]
3 years ago
7

HELP PLEASE!!!!!!!!! What if you started with 100 grams of wood and only have 95 grams of ashes produced, is mass still conserve

d? If so where is the missing 5 grams?
Chemistry
1 answer:
Lerok [7]3 years ago
7 0
My guess is smoke. Smoke is matter.
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soldi70 [24.7K]
Northern hemisphere
5 0
4 years ago
Which of the following would introduce bias into an experiment?
Nastasia [14]
I believe it is A or D. since "<span>Scientific bias is the assumption that a </span>theory<span> is true or false without evidence one way or another, or the attempt to dismiss or discourage research efforts to confirm or deny the </span>theory<span> "</span>
5 0
4 years ago
Read 2 more answers
In the Mond process for the purification of nickel, carbon monoxide is reacted with heated nickel to produce Ni(CO)4, which is a
igor_vitrenko [27]

<u>Answer:</u> The equilibrium constant for this reaction is 1.068\times 10^{6}

<u>Explanation:</u>

The equation used to calculate standard Gibbs free change is of a reaction is:

\Delta G^o_{rxn}=\sum [n\times \Delta G^o_{(product)}]-\sum [n\times \Delta G^o_{(reactant)}]

For the given chemical reaction:

Ni(s)+4CO(g)\rightleftharpoons Ni(CO)_4(g)

The equation for the standard Gibbs free change of the above reaction is:

\Delta G^o_{rxn}=[(1\times \Delta G^o_{(Ni(CO)_4(g))})]-[(1\times \Delta G^o_{(Ni(s))})+(4\times \Delta G^o_{(CO(g))})]

We are given:

\Delta G^o_{(Ni(CO)_4(g))}=-587.4kJ/mol\\\Delta G^o_{(Ni(s))}=0kJ/mol\\\Delta G^o_{(CO(g))}=-137.3kJ/mol

Putting values in above equation, we get:

\Delta G^o_{rxn}=[(1\times (-587.4))]-[(1\times (0))+(4\times (-137.3))]\\\\\Delta G^o_{rxn}=-38.2kJ/mol

To calculate the equilibrium constant (at 58°C) for given value of Gibbs free energy, we use the relation:

\Delta G^o=-RT\ln K_{eq}

where,

\Delta G^o = Standard Gibbs free energy = -38.2 kJ/mol = -38200 J/mol  (Conversion factor: 1 kJ = 1000 J )

R = Gas constant = 8.314 J/K mol

T = temperature = 58^oC=[273+58]K=331K

K_{eq} = equilibrium constant at 58°C = ?

Putting values in above equation, we get:

-38200J/mol=-(8.314J/Kmol)\times 331K\times \ln K_{eq}\\\\K_{eq}=e^{13.881}=1.068\times 10^{6}

Hence, the equilibrium constant for this reaction is 1.068\times 10^{6}

4 0
3 years ago
1. Suppose you had 58.44 grams of NaCl and you dissolved it in exactly 2.00 L of solution.
Margarita [4]
.500M or .500 mol/liters
3 0
2 years ago
How many grams of silver chloride are produced from 5.0g of silver nitrate reacting with an excess of
Sav [38]

Answer:

4.21 g of AgCl

3.06 g of BaCl₂ will be needed to complete the reaction

Explanation:

The first step is to determine the reaction.

Reactants: BaCl₂ and AgNO₃

The products will be the silver chloride (AgCl) and the Ba(NO₃)₂

The reaction is: BaCl₂(aq) + 2AgNO₃(aq) → 2AgCl(s) ↓ + Ba(NO₃)₂ (aq)

We determine the silver nitrate moles: 5 g . 1mol / 169.87 g = 0.0294 moles. Now, according to stoichiometry, we know that ratio is 2:2-

2 moles of nitrate can produce 2 moles of chloride, so the 0.0294 moles of silver nitrate, will produce the same amount of chloride.

We convert the moles to mass → 143.32 g / mol . 0.0294 mol = 4.21 g of AgCl.

Now, we consider the BaCl₂.

2 moles of nitrate can react to 1 mol of barium chloride

Then, 0.0294 moles of silver nitrate will react to (0.0294 . 1) /2 = 0.0147 moles. We convert the moles to mass:

0.0147 mol . 208.23 g /1mol = 3.06 g of BaCl₂

6 0
4 years ago
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