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3 years ago

A _ _ has a definite composition that will not vary

Chemistry

1 answer:

emmasim [6.3K]3 years ago

7 0

Answer: Mixtures do not have the same composition, properties, and appearance throughout.

Explanation: It does not vary from sample to sample.

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A certain substance X has a normal freezing point of 5.6 °C and a molal freezing point depression constant Kf-7.78 °C-kg·mol-1.

Harrizon [31]

Answer:

27.60 g urea

Explanation:

The freezing-point depression is expressed by the formula:

ΔT= Kf * m

In this case,

ΔT = 5.6 - (-0.9) = 6.5 °C

Kf = 7.78 °C kg·mol⁻¹

m is the molality of the urea solution in X (mol urea/kg of X)

First we calculate the molality:

6.5 °C = 7.78 °C kg·mol⁻¹ * m

m = 0.84 m

Now we calculate the moles of urea that were dissolved:

550 g X ⇒ 550 / 1000 = 0.550 kg X

0.84 m = mol Urea / 0.550 kg X

mol Urea = 0.46 mol

Finally we calculate the mass of urea, using its molecular weight:

0.46 mol * 60.06 g/mol = 27.60 g urea

7 0

3 years ago

Complete the following single replacement reaction. If they don’t react, just write “NR”

Kipish [7]

Here we have to complete the given single replacement reactions.

The replacement reactions are-

1) Fe (s) + CuCl₂ (aq) → FeCl₂ (aq) + Cu (s)

2) Cu (s) + FeCl₂ (aq) → NA

3) K (s) + NiBr₂ (aq) → NA

4) Ni (s) + KBr (aq) → NiBr₂ (aq) + K (s)

5) Zn (s) + Ca(NO₃)₂ (aq) → Zn(NO₃)₂ (aq) + Ca (s)

6) Ca (s) + Zn(NO₃)₂ (aq) → NA

The replacement reactions can be explained in light of the redox potential.

The standard reduction potential of the half cells involved in these reactions are:

Fe²⁺ + 2e → Fe (E° = -0.441V); Cu²⁺ + 2e → Cu (E° = 0.674V)

Ni²⁺ + 2e → Ni (E° = -0.23V); Zn²⁺ + 2e → Zn (E° = -0.763V)

We know the half cell reactions in which the standard reduction potentials are positive are allowed.

1) The reaction is possible as Cu²⁺/Cu and Fe/Fe²⁺ standard reduction potentials are positive.

2) The reaction is not possible as Cu/Cu²⁺ and Fe²⁺/Fe standard reduction potentials are negative.

3) The reaction is not possible as Ni²⁺/Ni standard reduction potential is negative.

4) The reaction is possible as Ni/Ni²⁺ standard reduction potential is positive.

5) The reaction is possible as Zn/Zn²⁺ standard reduction potential is positive.

6) The reaction is possible as Zn²⁺/Zn standard reduction potential is negative.

4 0

2 years ago

Science 6th grade please help me :)

Gnom [1K]

I’m pretty sure the answer is D :)

6 0

3 years ago

Read 2 more answers

Be sure to answer all parts. Write the equations representing the following processes. In each case, be sure to indicate the phy

barxatty [35]

Answer:

1. $S^{-}(g)+e^{-} \rightarrow S^{2-}(g)$

2. $Ti^{2+}(g) \rightarrow Ti^{3+}(g) \,\, IE = 2652.5\,kJ.mol^{-1}$

3.The electron affinity of $Mg^{2+}$ is zero.

4. $O^{2-}(g) \rightarrow O^{-}(g)+e^{-}$

Explanation:

Electron affinity:

It is defined as the amount of energy change when an electron is added to atom in the gaseous phase.

The electron affinity of $S^{-}$ is as follows.

$S^{-}(g)+e^{-} \rightarrow S^{2-}(g)$

Ionization energy:

Amount of energy required to removal of an electron from an isolated gaseous atom.

The third ionization energy of Titanium is as follows.

$Ti^{2+}(g) \rightarrow Ti^{3+}(g) \,\, IE = 2652.5\,kJ.mol^{-1}$

The electronic configuration of Mg: $1s^{2}2s^{2}2p^{6}3s^{2}$

By the removal of two electrons from a magnesium element we get $Mg^{2+}$ ion.

$Mg^{2+}$ has inert gas configuration i.e, $1s^{2}2s^{2}2p^{6}$

Hence, it does not require more electrons to get stability.

Therefore,the electron affinity of $Mg^{2+}$ is zero.

The ionization energy of $O^{2-}$ is follows.

$O^{2-}(g) \rightarrow O^{-}(g)+e^{-}$

3 0

3 years ago

PLZ HELP! Give me a really (winner) good idea for a kids science fair! My little sister has to do one through my school and I ne

OverLord2011 [107]

Explanation:amiga

extrañar

8 0

3 years ago