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leva [86]
3 years ago
12

Mass of NaHCO3 sample (g): 1.20 g

Chemistry
1 answer:
sasho [114]3 years ago
6 0

Answer: (22.98977 g Na/mol) + (1.007947 g H/mol) + (12.01078 g C/mol) + ((15.99943 g O/mol) x 3) = 84.0067 g NaHCO3/mol

9.

(1.20 g NaHCO3) / (84.0067 g NaHCO3/mol) = 0.0143 mol NaHCO3

10.

Supposing the question is asking about "how many moles" of CO2. And supposing the reaction to be something like:

NaHCO3 + H{+} = Na{+} + H2O + CO2

(0.0143 mol NaHCO3) x (1 mol CO2 / 1 mol NaHCO3) = 0.0143 mol CO2 in theory

11.

n = PV / RT = (1 atm) x (0.250 L) / ((0.0821 L atm/K mol) x (298 K)) = 0.0102 mol CO2

12.

(0.0143 mol - 0.0102 mol) / (0.0143 mol) = 0.287 = 28.7%

Explanation:

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The reaction of Mg Cl2 and KOH can be described as a double substitution type of reaction. This means the cations of the reactants are exchanged in places when the products are formed. In this case, the balanced reaction is expressed
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5 0
3 years ago
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If 13.0 g of MgSO4⋅7H2O is thoroughly heated, what mass of anhydrous magnesium sulfate will remain?
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6.349 g mass of anhydrous magnesium sulfate will remain.

<h3>What are moles?</h3>

A mole is defined as 6.02214076 × 1023 of some chemical unit, be it atoms, molecules, ions, or others. The mole is a convenient unit to use because of the great number of atoms, molecules, or others in any substance.

Molar mass MgSO₄.7 H₂O = 246.52 g/mol

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4 0
1 year ago
Which of the following polyatomic ions will form an ionic compound with two sodium ions? CO32− HCO31− NO21− NO31−
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We have to consider the valencies of the polyatomic ions involved. Recall that it is only a polyatomic ion with a valency of -2 that can form a compound which requires two sodium ions.

When we look closely at the options, we will realize that among all the options, only CO32− has a valency of -2, hence it must be the required answer. In order to be double sure, we put down the ionic reaction equation as follows;

2Na^+(aq) + CO3^2-(aq) ---------> Na2CO3(aq)

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O2 is an element As it contains just one kind of atom, O2 is an element,
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