Which of the following is not part of protozoa?

A reaction between an acid and a base produced lithium chloride (LiCl). What acid and base combination could produce this salt?

dolphi86 [110]

C sulfuric acid and magnesium hydroxide

5 0

3 years ago

Read 2 more answers

When you mixed 20 grams of magnesium and an excess of nitric acid, 1.7 grams of hydrogen was actually produced. What is the perc

Kitty [74]

Based on the balanced chemical reaction presented above, every mole of magnesium (Mg) yields one mole of diatomic hydrogen (H2). When converted to masses, every 24.3 grams of magnesium yields 2 grams of hydrogen.

From the given, there are 20 grams of magnesium available for the reaction. With this amount, the expected yield of hydrogen is 1.646 grams. To calculate the percent yield, divide the actual yield to the hypothetical yield.

*The case is impossible because the actual yield is greater than the theoretical yield.

If we assume that there had been a typographical error and that the actual yield is 0.7 grams instead of 1.7 grams, the percent yield becomes 42.5%. Thus, the answer is letter E.

6 0

3 years ago

Read 2 more answers

1. Hydrogen, Helium, Lithium?

Reika [66]

Answer:

what happened to it, have the gone extinct

6 0

3 years ago

While ethanol (CH3CH2OH) is produced naturally by fermentation, e.g. in beer- and wine-making, industrially it is synthesized by

grin007 [14]

Answer:

$n_{C2H_5OH}^{eq}=14.234mol$

Explanation:

Hello,

In this case, the reaction is:

$C_2H_4+H_2O\rightleftharpoons CH_3CH_2OH$

Thus, the law of mass action turns out:

$Kc=\frac{[CH_3CH_2OH]_{eq}}{[H_2O]_{eq}[CH_2CH_2]_{eq}}$

Thus, since at the beginning there are 29 moles of ethylene and once the equilibrium is reached, there are 16 moles of ethylene, the change $x$ result:

$[CH_2CH_2]_{eq}=29mol-x=16mol\\x=29-16=13mol$

In such a way, the equilibrium constant is then:

$Kc=\frac{\frac{x}{V} }{\frac{16mol}{V}* \frac{3mol}{V}} =\frac{\frac{13mol}{75.0L} }{\frac{16mol}{75.0L}* \frac{3mol}{75.0L}} =20.31$

Thereby, the initial moles for the second equilibrium are modified as shown on the denominator in the modified law of mass action by considering the added 15 moles of ethylene:

$Kc=\frac{\frac{13+x_2}{V} }{\frac{16+15-x_2}{V}* \frac{3-x_2}{V}} =20.31$

Thus, the second change, $x_2$ finally result (solving by solver or quadratic equation):

$x_2=1.234mol$

Finally, such second change equals the moles of ethanol after equilibrium based on the stoichiometry:

$n_{C2H_5OH}^{eq}=x+x_2=13mol+1.234mol\\n_{C2H_5OH}^{eq}=14.234mol$

Best regards.

5 0

3 years ago

Calculate the fraction of lattice sites that are schottky defects for cesium chloride at its melting temperature (645oc). assume

baherus [9]

Answer : 7.87 X $10^{-6}$ .

Explanation : To find the fraction of schottky defects in the given lattice of CsCl,

we use the formula, $\frac{N_{s} }{N}} = exp ( \frac{-Q_{s}}{2KT} )$

on solving with the given values , $Q_{s}= 1.86 eV$ and T as 645 + 273 K and rest are the constants.

$\frac{N_{s} }{N}} = exp ( \frac{-1.86 eV}{2 X (8.62 X 10^{-5}) X (645 +273)} )$

we get the answer as 7.87 X $10^{-6}$ .

6 0

3 years ago

Read 2 more answers