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3 years ago

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Some distilled water is added to an empty beaker. a gram of copper (ii) nitrate is added to the beaker and while the water is be

ing stirred. after a few minutes, what is in the beaker?

1 answer:

Xelga [282]3 years ago

6 0

Copper ions, nitrate ions, and water is in the beaker

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The combustion of 0.25 mol of an unknown organic compound results in the release of 320 kJ of energy. Which of the compounds in

vichka [17]

Answer: C. ethanol

The enthalpy of combustion is the amount of heat produced when one mole of ethanol undergoes complete combustion at 25 ° C and 1 atmosphere pressure, yielding products also at 25 ° C and 1 atm.

<u>The enthalpy of combustion of the unknown compound is</u>

ΔH = - 320 kJ / 0.25 mol = - 1280 kJ / mol

<u>To choose a probable compound according to this combustion enthalpy, we must evaluate the deviation in relation to the values reported in the literature for the three probable compounds</u> (methane, ethylene and ethanol). The deviation (e%) will be calculated according to the following equation,

e% = ( | ΔHx - ΔH | / ΔHx ) x 100%

where ΔHx is the enthalpy of combustion of the probable compound.

The following table shows the combustion enthalpies of the probable compounds and their deviation in relation to the enthalpy of ΔH = - 1280 kJ / mol

Compound Enthalpy of combustion (kJ/mol) Deviation

Methane - 890.7 43.8%

Ehylene -1411.2 9.3%

Ethanol -1368.6 6.5%

According to the previous table, we can say that the most probable compound is ethanol, since it has the smallest deviation in relation to the experimental enthalpy value of combustion.

5 0

3 years ago

Read 2 more answers

Gravitational forces exerted on any two____in the universe

OleMash [197]

Answer:-

Gravitational forces exerted on any two <em><u>bodies</u></em> in the universe

Hope it helps you...

✅

6 0

2 years ago

At 2°C, the vapor pressure of pure water is 23.76 mmHg and that of a certain seawater sample is 23.09 mmHg. Assuming that seawat

wariber [46]

Answer:

0.808 M

Explanation:

Using Raoult's Law

$\frac{P_s}{Pi}= x_i$

where:

$P_s$ = vapor pressure of sea water( solution) = 23.09 mmHg

$P_i$ = vapor pressure of pure water (solute) = 23.76 mmHg

$x_i$ = mole fraction of water

∴

$\frac{23.09}{23.76}= x_i$

$x_i = 0.9718$

$x_i+x_2=1$

$x_2 = 1- x_i$

$x_2 = 1- 0.9718$

$x_2 = 0.0282$

$x_i = \frac{n_i}{n_i+n_2}$ ------ equation (1)

$x_2 = \frac{n_2}{n_i+n_2}$ ------ equation (2)

where; $(n_2) =$ number of moles of sea water

$(n_i) =$ number of moles of pure water

equating above equation 1 and 2; we have :

$\frac{n_2}{n_i}$ $= \frac{0.0282}{0.9178}$

$= 0.02901$

NOW, Molarity = $\frac{moles of sea water}{mass of pure water }*1000$

$= \frac{0.02901}{18}*1000$

$= 0.001616*1000$

$= 1.616 M$

As we assume that the sea water contains only NaCl, if NaCl dissociates to Na⁺ and Cl⁻; we have $\frac{1.616}{2} =0.808 M$

3 0

2 years ago

Read 2 more answers

statement 1,mass is conserved durning the reaction. statement 2 sum of mass and energy is conserved durning the reaction. which

maksim [4K]

Answer:

Explanation:

None of the statement is true for both chemical and nuclear reactions. In chemical reactions, mass is always conserved and the type of atoms are also conserved.

6 0

3 years ago

Calcule a variação da entalpia dessa reação ( 2 NH3 (g) ---> CO(NH2)2 (s) + H2O (L) ) a partir das seguintes equações termoqu

Nitella [24]

ΔH = +438 kJ

We have three equations:

(I) N₂ + 3H₂ → 2NH₃; Δ<em>H</em> = -92 kJ

(II) H₂ +½O₂ → H₂O; Δ<em>H</em> = -286 kJ

(III) CO(NH₂)₂ + ³/₂O₂ → CO₂ + 2H₂O + N₂; Δ<em>H</em> = -632 kJ

From these, we must devise the target equation:

(IV) 2NH₃ + CO₂ → CO(NH₂)₂ + H₂O; Δ<em>H</em> = ?

_________________________________

The target equation has 2NH₃ on the left, so you <em>reverse equation (I)</em>.

When you reverse an equation, you <em>reverse the sign of its ΔH</em>.

(V) 2NH₃ → N₂ + 3H₂; Δ<em>H</em> = +92 kJ

Equation (V) has 1N₂ on the right, and that is not in the target equation.

You need an equation with 1N₂ on the left.

<em>Reverse Equation (III).</em>

(VI) CO₂ + 2H₂O + N₂ → CO(NH₂)₂ + ³/₂O₂; Δ<em>H</em> = +632 kJ

Equation <em>(VI)</em> has ³/₂O₂ on the right, and that is not in the target equation.

You need ³/₂O₂ on the left.

Multiply <em>Equation (II) by three</em>.

When you multiply an equation by three, you <em>multiply its ΔH by thre</em>e.

(VII) 3H₂ +³/₂O₂ → 3H₂O; Δ<em>H</em> = -286 kJ

Now, you add equations (V), (VI), and (VII), <em>cancelling species</em> that appear on opposite sides of the reaction arrows.

When you add equations, you add their Δ<em>H</em> values.

_______________________________________

We get the target equation (IV):

(V) 2NH₃ → <u>N</u>₂ + <u>3H</u>₂; ΔH = + 92 kJ

(VI) CO₂ + <u>2H</u>₂<u>O</u> + <u>N</u>₂ → CO(NH₂)₂ + ³/₂<u>O</u>₂; ΔH = +632 kJ

(VII) <u>3H</u>₂ +³/₂<u>O</u>₂ → <u>3</u>H₂O; ΔH = -286 kJ

(IV) 2NH₃ + CO₂ → CO(NH₂)₂ + H₂O; ΔH = +438 kJ

7 0

2 years ago