What is the mass of 0.5 moles of NaCl

Chemistry

2 answers:

jeka57 [31]2 years ago

8 0

Answer:

29.25 g

Explanation:

$mole = \dfrac{mass}{molar \: mass}$

mass = 0.5 × 58.5 [ 23 + 35.5]

= 29.25 gm

Debora [2.8K]2 years ago

6 0

Answer:

29.25g

Explanation:

m=0.5×(23+35.5)

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A fine of 50.0 mL of 0.0900 M CaCl2 reacts with excess sodium carbonate to give 0.366 g of calcium carbonate precipitate. What i

In-s [12.5K]

Answer:

81.26% is the percent yield

Explanation:

Based on the reaction:

CaCl₂ + Na₂CO₃ → 2NaCl + CaCO₃

<em>Where 1 mole of CaCl₂ in excess of sodium carbonate produces 1 mole of calcium carbonate.</em>

<em />

To solve this question we must find the moles of CaCl2 added = Moles CaCO₃ produced (Theoretical yield). The percent yield is:

Actual yield (0.366g) / Theoretical yield * 100

<em>Moles CaCl₂ = Moles CaCO₃:</em>

0.0500L * (0.0900moles / L) = 0.00450 moles of CaCO₃

<em>Theoretical mass -Molar mass CaCO₃ = 100.09g/mol-:</em>

0.00450 moles of CaCO₃ * (100.09g / mol) = 0.450g of CaCO₃

Percent yield = 0.366g / 0.450g * 100

81.26% is the percent yield

3 0

3 years ago

Consider the reaction 2NO(g) 1 O2(g) ¡ 2NO2(g) Suppose that at a particular moment during the reaction nitric oxide (NO) is reac

GalinKa [24]

<h2>a) The rate at which $NO_2$ is formed is 0.066 M/s</h2><h2>b) The rate at which molecular oxygen $O_2$ is reacting is 0.033 M/s</h2>

Explanation:

Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.

$2NO(g)+O_2(g)\rightarrow 2NO_2(g)$

The rate in terms of reactants is given as negative as the concentration of reactants is decreasing with time whereas the rate in terms of products is given as positive as the concentration of products is increasing with time.

Rate in terms of disappearance of $NO$ = $-\frac{1d[NO]}{2dt}$ = 0.066 M/s