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3 years ago

Why are the Flemish called Flemish

Chemistry

1 answer:

Dmitriy789 [7]3 years ago

4 0

Explanation:

The flemish are an ethnic group which belong to the northern part of Belgium called Flanders. They also speak a language called flemish. The major population of Belgium are the flemish. The language they speak is a Low Franconian dialect cluster of the Dutch language.

There was a country called the flanders located in the lowlands of Europe. Any person from this country was considered as flemish previously.

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The quantity of antimony in a sample can be determined by an oxidation–reduction titration with an oxidizing agent.

kotegsom [21]

Answer:

The percentage is k $= 20.8$ %

Explanation:

From the question we are told that

The mass of the stibnite is $m_s = 5.86 \ g$

The volume of KBrO3(aq) is $V = 26.6 mL = 26.6 *10^{-3} \ L$

The concentration of KBrO3(aq) is $C = 0.125 M$

Now the balanced ionic equation for this reaction is

$BrO_3 ^{-}+ 3Sb^{3+} + 6H^{+} \to Br^{1-} + 3Sb^{5+} + 3H_2O$

The number of moles of $BrO_3 ^{-}$ is

$n = C *V$

substituting values

$n = 26.6*10^{-3} * 0.125$

$n = 0.003325 \ mols$

from the reaction we see that 1 mole of $BrO_3 ^{-}$ reacts with 3 moles of $Sb^{3+}$

so 0.003325 moles will react with x moles of $Sb^{3+}$

Therefore

$x = \frac{0.003325 * 3}{1}$

$x = 0.009975 \ mols$

Now the molar mass of $Sb^{3+}$ is a constant with a values of $Z = 121.76 \ g/mol$

Generally the mass of $Sb^{3+}$ is mathematically represented as

$m = x * Z$

substituting values

$m = 1.215 \ g$

The percentage of Sb(antimony) in the overall mass of the stibnite is mathematically evaluated as

k $= \frac{1.215}{5.85 } * 100$

k $= 20.8$ %

4 0

3 years ago

Read 2 more answers

When 100.g Mg3N2 reacts with 75.0 g H2O, what is the maximum theoretical yield of NH3?

Temka [501]

Answer : The correct option is, 23.6 g

Explanation : Given,

Mass of $Mg_3N_2$ = 100.0 g

Mass of $H_2O$ = 75.0 g

Molar mass of $Mg_3N_2$ = 101 g/mol

Molar mass of $H_2O$ = 18 g/mol

First we have to calculate the moles of $Mg_3N_2$ and $H_2O$ .

$\text{Moles of }Mg_3N_2=\frac{\text{Given mass }Mg_3N_2}{\text{Molar mass }Mg_3N_2}$

$\text{Moles of }Mg_3N_2=\frac{100.0g}{101g/mol}=0.990mol$

and,

$\text{Moles of }H_2O=\frac{\text{Given mass }H_2O}{\text{Molar mass }H_2O}$

$\text{Moles of }H_2O=\frac{75.0g}{18g/mol}=4.17mol$

Now we have to calculate the limiting and excess reagent.

The balanced chemical equation is:

$Mg_3N_2(s)+6H_2O(l)\rightarrow 3Mg(OH)_2(aq)+2NH_3(g)$

From the balanced reaction we conclude that

As, 6 moles of $H_2O$ react with 1 mole of $Mg_3N_2$

So, 4.17 moles of $H_2O$ react with $\frac{4.17}{6}=0.695$ moles of $Mg_3N_2$

From this we conclude that, $Mg_3N_2$ is an excess reagent because the given moles are greater than the required moles and $H_2O$ is a limiting reagent and it limits the formation of product.

Now we have to calculate the moles of $NH_3$

From the reaction, we conclude that

As, 6 moles of $H_2O$ react to give 2 moles of $NH_3$

So, 4.17 moles of $H_2O$ react to give $\frac{2}{6}\times 4.17=1.39$ mole of $NH_3$

Now we have to calculate the mass of $NH_3$

$\text{ Mass of }NH_3=\text{ Moles of }NH_3\times \text{ Molar mass of }NH_3$

Molar mass of $NH_3$ = 17 g/mole

$\text{ Mass of }NH_3=(1.39moles)\times (17g/mole)=23.6g$

Therefore, the maximum theoretical yield of $NH_3$ is, 23.6 grams.

4 0

3 years ago

In sexual reproduction, a genetic mutation can ONLY be passed down to the offspring if it occurs in the sex cells.

In-s [12.5K]

The answer to your question is True

8 0

3 years ago

An equilibrium mixture contains 0.750 mol of each of the products (carbon dioxide and hydrogen gas) and 0.200 mol of each of the

sasho [114]

1.302 moles of carbon dioxide would have to be added

<h3>Further explanation</h3>

The equilibrium constant is the value of the product in the equilibrium state of the substance in the right (product) divided by the substance in the left (reactant) with the exponents of each reaction coefficient

The equilibrium constant is based on the concentration (Kc) in a reaction

pA + qB -----> mC + nD

$\large {\boxed {\bold {Kc ~ = ~ \frac {[C] ^ m [D] ^ n} {[A] ^ p [B] ^ q}}}}$