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3 years ago

5. Identify each of the following as an element, a mixture, or a compound.

Chemistry

1 answer:

VashaNatasha [74]3 years ago

4 0

Answer:

Aluminum foil ~ element

Air ~ mixture

Water ~ compound

Salt Water ~ mixture

Copper Wire ~ element

Steel ~ element

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The density of liquid mercury is 13.6 g/mL. What is its density in units of lb/in3? (2.5 cm+1 in., 2.205 lbs= 1 kg., 1000 g =1

shusha [124]

Answer:

Density, $\rho=0.49\ lb/in^3$

Explanation:

It is given that the density of liquid mercury is 13.6 g/mL. We need to convert the density into lb/in³.

We know that,

2.205 lbs= 1 kg

1 g = 0.0022 lb

1 mL = 0.0610 in³

$13.6\ \dfrac{g}{mL}=13.6\times \dfrac{0.0022\ lb}{0.0610\ in^3}\\\\=0.49\ lb/in^3$

So, the density of liquid mercury is $0.49\ lb/in^3$ .

3 0

3 years ago

How to do thw part where it says list the elements to period 2 and the metals

marissa [1.9K]

A)2nd period - Li, Be, B,C,N,O,F,Ne
b)Metalls of the 2nd period are Li,Be

7 0

3 years ago

Balance each of the following equations according to the half- reaction method:

lukranit [14]

Answer : The balanced chemical equation in a acidic solution are,

(a) $Sn^{2+}+2Cu^{2+}\rightarrow Sn^{4+}+2Cu^+$

(b) $H_2S+Hg_2^{2+}\rightarrow 2Hg+S+2H^+$

(c) $5CN^-+2ClO_2+H_2O\rightarrow 5CNO^-+2Cl^-+2H^+$

(d) $Fe^{2+}+Ce^{4+}\rightarrow Fe^{3+}+Ce^{3+}$

(e) $2HBrO\rightarrow 3Br^-+2Br+2O_2+H_2O+3H^+$

Explanation :

Redox reaction or Oxidation-reduction reaction : It is defined as the reaction in which the oxidation and reduction reaction takes place simultaneously.

Oxidation reaction : It is defined as the reaction in which a substance looses its electrons. In this, oxidation state of an element increases. Or we can say that in oxidation, the loss of electrons takes place.

Reduction reaction : It is defined as the reaction in which a substance gains electrons. In this, oxidation state of an element decreases. Or we can say that in reduction, the gain of electrons takes place.

(a) The given chemical reaction is,

$Sn^{2+}+Cu^{2+}\rightarrow Sn^{4+}+Cu^+$

The oxidation-reduction half reaction will be :

Oxidation : $Sn^{2+}\rightarrow Sn^{4+}+2e^-$

Reduction : $Cu^{2+}+1e^-\rightarrow Cu^+$

In order to balance the electrons, we multiply the reduction reaction by 2 and then added both equation, we get the balanced redox reaction.

The balanced chemical equation will be,

$Sn^{2+}+2Cu^{2+}\rightarrow Sn^{4+}+2Cu^+$

(b) The given chemical reaction is,

$H_2S+Hg_2^{2+}\rightarrow Hg+S$

The oxidation-reduction half reaction will be :

Oxidation : $H_2S\rightarrow S+2H^++2e^-$

Reduction : $Hg_2^{2+}+2e^-\rightarrow 2Hg$

The electrons in oxidation and reduction reaction are same. Now add both the equation, we get the balanced redox reaction.

The balanced chemical equation in a acidic solution will be,

$H_2S+Hg_2^{2+}\rightarrow 2Hg+S+2H^+$

(c) The given chemical reaction is,

$CN^-+ClO_2\rightarrow CNO^-+Cl^-$

The oxidation-reduction half reaction will be :

Oxidation : $CN^-+H_2O\rightarrow CNO^-+2H^++2e^-$

Reduction : $ClO_2+4H^++5e^-\rightarrow Cl^-+2H_2O$

In order to balance the electrons, we multiply the oxidation reaction by 5 and reduction reaction by 2 and then added both equation, we get the balanced redox reaction.

The balanced chemical equation in a acidic solution will be,

$5CN^-+2ClO_2+H_2O\rightarrow 5CNO^-+2Cl^-+2H^+$

(d) The given chemical reaction is,

$Fe^{2+}+Ce^{4+}\rightarrow Fe^{3+}+Ce^{3+}$

The oxidation-reduction half reaction will be :

Oxidation : $Fe^{2+}\rightarrow Fe^{3+}+1e^-$

Reduction : $Ce^{4+}+1e^-\rightarrow Ce^{3+}$

The electrons in oxidation and reduction reaction are same. Now add both the equation, we get the balanced redox reaction.

The balanced chemical equation will be,

$Fe^{2+}+Ce^{4+}\rightarrow Fe^{3+}+Ce^{3+}$

(e) The given chemical reaction is,

$HBrO\rightarrow Br^-+O_2$

The oxidation-reduction half reaction will be :

Oxidation : $HBrO+H_2O\rightarrow O_2+Br+3H^++3e^-$

Reduction : $HBrO+H^++2e^-\rightarrow Br^-+H_2O$

In order to balance the electrons, we multiply the oxidation reaction by 2 and reduction reaction by 3 and then added both equation, we get the balanced redox reaction.

The balanced chemical equation in a acidic solution will be,

$2HBrO\rightarrow 3Br^-+2Br+2O_2+H_2O+3H^+$

5 0

3 years ago

Calculate the molarity of glucose in a solution that which contains 35 g of glucose in 0.16 kg of phenol

sleet_krkn [62]

C(Molarity) = n of solute/V of solution (mol/L)

glucose(C6H12O6) = 180g/mol
glucose 35g = 35g/(180g/mol) = 0.1944mol

We need density of solution here, and I assume it as density of phenol, 1.07g/mL. (I don't know why the question doesn't contain it)

phenol 0.16kg = 0.16kg/(1.07kg/L) = 0.1495L

0.1944mol/0.1495L = 1.300M(mol/L)

5 0

3 years ago

What determines the degree of polarity in a bond?

Ilia_Sergeevich [38]

Answer:

The relative electronegativity of the two bonded atoms determines the polarity of a bond. If the difference in electronegativities between the two atoms is less than 0.4, the bond is nonpolar covalent. ... If the difference in electronegativities between the two atoms is more than 2.0, the bond is ionic.

5 0

2 years ago