Empirical formula is calculated as follows
calculate the moles of each element, that is % composition/ molar mass
molar masses ( Si= 28.09g/mol , Cl= 35.5 g/mol, I=126.9 g/mol)
moles of silicon = 7.962/28.09g/mol= 0.283 moles
moles of chlorine = 20.10 / 35.5g/mol = 0.566 moles
moles of iodine= 71.94 / 126.9 g/mol= 0.567 moles
divide each mole with smallest mole (0.283)
that is silicon = 0.283/0.283= 1 mole
chlorine = 0.566/0.283= 2 mole
Iodine= o.567/0.283= 2 moles
empirical formula is therefore= SiCl2I2
Answer:175⋅mL of the given sulfuric acid
Explanation:
The melting points of sugar and salt are above 20°C.
<h3>What is melting point?</h3>
The melting points of substances refer to the temperature at which solid substances gain enough energy to become liquids.
The room temperature is approximately 20°C. At this temperature, sugar and salt are solids. This means that the melting temperature of both substances is above 20°C.
More specifically, sugar will melt around 180 °C while salt will melt at a temperature slightly above 800 °C.
More on melting points can be found here: brainly.com/question/25777663
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<span>boron trichloride + water → boric acid + hydrochloric acid</span>
<span>Answer: 0.094%
</span><span>Explanation:
</span>
<span></span><span /><span>
1) Equilibrium chemical equation:
</span><span />
<span>Only the ionization of the formic acid is the important part.
</span><span />
<span>HCOOH(aq) ⇄ HCOO⁻(aq) + H⁺(aq).
</span><span />
<span>2) Mass balance:
</span><span />
<span> HCOOH(aq) HCOO⁻(aq) H⁺(aq).
Start 0.311 0.189
Reaction - x +x +x
Final 0.311 - x 0.189 + x x
3) Acid constant equation:
</span><span />
<span>Ka = [HCOO-] [N+] / [HCOOH] = (0.189 + x) x / (0.311 -x)
</span><span />
<span>= (0.189 + x )x / (0.311 - x) = 0.000177
4) Solve the equation:
You can solve it exactly (it will lead to a quadratic equation so you can use the quadratiic formula). I suggest to use the fact that x is much much smaller than 0.189 and 0.311.
</span><span />
<span>With that approximation the equation to solve becomes:
</span><span>0.1890x / 0.311 = 0.000177, which leads to:</span>
<span /><span>
x = 0.000177 x 0.311 / 0.189 = 2.91 x 10⁻⁴ M
5) With that number, the percent of ionization (alfa) is:
</span><span />
<span>percent of ionization = (moles ionized / initial moles) x 100 =
</span><span>
</span><span>
</span><span>percent ionization = (concentration of ions / initial concentration) x 100 =
</span><span>
</span><span>
</span><span>percent ionization = (0.000291 / 0.311)x 100 = 0.0936% = 0.094%
</span>
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