All categories

3 years ago

How many negative ones (-1) would you need to balance out one positive 2 (+2)?

Chemistry

1 answer:

kotegsom [21]3 years ago

7 0

Answer:I do believe it would be two negative ones would balance one positive two

Explanation:

2 + 2(-1)=0

which would be balanced (I think)

You might be interested in

What is the volume of a 200. gram sample of gold if its density is known to be 20.5 g/cm3?

Travka [436]

The volume of a 200 g sample of gold is 9.76 cm³

calculation

volume = mass /density

mass = 200 g

volume = 20.5 g/Cm³

volume is therefore = 200g / 20.5 g/cm³ = 9.76 cm³

3 0

3 years ago

In 1869, mendeleev created a periodic table in which elements were ordered by weight and placed in groups based on their chemica

Tomtit [17]

Answer:

Explanation:

Mendeleev periodic table predicted the properties of undiscovered element like the eka-aluminium.

8 0

2 years ago

A 1.5 M solution of NaOH was made in a laboratory. If the solution made had a volume of 4.5 L, how many grams of NaOH were added

Crank

Answer:

270g

Explanation:

Given parameters:

Concentration of NaOH = 1.5M

Volume = 4.5L

Unknown

Mass of NaOH added = ?

Solution:

To solve the problem, we need to find the number of moles of the NaOH first;

Number of moles = concentration x volume

Number of moles = 1.5 x 4.5 = 6.75mol

Now;

Mass = Number of moles x molar mass

Molar mass of NaOH = 23 + 16 + 1 = 40g/mol

Mass = 6.75 x 40 = 270g

6 0

3 years ago

If a dose of antacid containing 28 g Al(OH)3 reacts to produce 44 g AlCl3, what is the percentage yield of AlCl3

MArishka [77]

Answer:

96%

Explanation:

Step 1: Write the balanced neutralization reaction

Al(OH)₃ + 3 HCl ⇒ AlCl₃ + 3 H₂O

Step 2: Calculate the theoretical yield of AlCl₃

According to the balanced equation, the mass ratio of Al(OH)₃ to AlCl₃ is 81.03:133.34.

28 g Al(OH)₃ × 133.34 g AlCl₃/81.03 g Al(OH)₃ = 46 g AlCl₃

Step 3: Calculate the percent yield of AlCl₃

The real yield of AlCl₃ is 44 g. We can calculate the percent yield using the following expression.

%yield = real yield / theoretical yield × 100%

%yield = 44 g / 46 g × 100% = 96%

3 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