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3 years ago

An object becomes positively charged when it

Chemistry

2 answers:

qwelly [4]3 years ago

7 0

Answer:

When electrons are removed from an object, it becomes positively charged

Hatshy [7]3 years ago

5 0

An electrical charge is created when electrons are transferred to or removed from an object. Because electrons have a negative charge, when they are added to an object, it becomes negatively charged. When electrons are removed from an object, it becomes positively charged.

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What is the relationship between pressure and temperature

hoa [83]

Answer:

The Pressure Temperature Law. This law states that the pressure of a given amount of gas held at constant volume is directly proportional to the Kelvin temperature. With an increase in temperature, the pressure will go up.

Explanation:

7 0

3 years ago

Which combination of an element and an ion will react? View Available Hint(s) Which combination of an element and an ion will re

Aleks04 [339]

Answer: The combination of element ad an ion that will react is $Ni(s)\text{ and }Pt^{2+}(aq.)$

Explanation:

Oxidation reaction is defined as the reaction in which an atom looses its electrons. The oxidation number of the atom gets increased during this reaction.

$X\rightarrow X^{n+}+ne^-$

Reduction reaction is defined as the reaction in which an atom gains electrons. The oxidation number of the atom gets reduced during this reaction.

$X^{n+}+ne^-\rightarrow X$

The substance having highest positive $E^o$ potential will always get reduced and will undergo reduction reaction.

For a reaction to be spontaneous, the standard electrode potential must be positive.

To calculate the $E^o_{cell}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$ ......(1)

For the given options:

Option 1: $Sn(s)\text{ and }Mn^{2+}(aq.)$

Here, tin must undergo oxidation reaction and manganese undergo reduction reaction.

Oxidation half reaction: $Sn(s)\rightarrow Sn^{2+}(aq.)+2e^-;E^o_{Sn^{2+}/Sn}=-0.14V$

Reduction half reaction: $Mn^{2+}(aq.)+2e^-\rightarrow Mn(s);E^o_{Mn^{2+}/Mn}=-1.18V$

Putting values in equation 1, we get:

$E^o_{cell}=-1.18-(-0.14)=-1.04V$

As, the standard potential is coming out to be negative, the given reaction will not take place.

Option 2: $Fe(s)\text{ and }Ca^{2+}(aq.)$

Here, iron must undergo oxidation reaction and calcium undergo reduction reaction.

Oxidation half reaction: $Fe(s)\rightarrow Fe^{2+}(aq.)+2e^-;E^o_{Fe^{2+}/Fe}=-0.44V$

Reduction half reaction: $Ca^{2+}(aq.)+2e^-\rightarrow Ca(s);E^o_{Ca^{2+}/Ca}=-2.87V$

Putting values in equation 1, we get:

$E^o_{cell}=-2.87-(-0.44)=-2.43V$

As, the standard potential is coming out to be negative, the given reaction will not take place.

Option 3: $Ni(s)\text{ and }Pt^{2+}(aq.)$

Here, nickel must undergo oxidation reaction and platinum undergo reduction reaction.

Oxidation half reaction: $Ni(s)\rightarrow Ni^{2+}(aq.)+2e^-;E^o_{Ni^{2+}/Ni}=-0.25V$

Reduction half reaction: $Pt^{2+}(aq.)+2e^-\rightarrow Pt(s);E^o_{Pt^{2+}/Pt}=1.2V$

Putting values in equation 1, we get:

$E^o_{cell}=1.2-(-0.25)=1.45V$

As, the standard potential is coming out to be positive, the given reaction will take place.

Option 4: $H_2(g)\text{ and }Na^{+}(aq.)$

Here, hydrogen must undergo oxidation reaction and sodium undergo reduction reaction.

Oxidation half reaction: $H_2(g)\rightarrow 2H^{+}(aq.)+2e^-;E^o_{2H^{+}/H_2}=0V$

Reduction half reaction: $Na^{+}(aq.)+e^-\rightarrow Na(s);E^o_{Na^{+}/Na}=-0.27V$

Putting values in equation 1, we get:

$E^o_{cell}=-0.27-(-0)=-0.27V$

As, the standard potential is coming out to be negative, the given reaction will not take place.

Hence, the combination of element ad an ion that will react is $Ni(s)\text{ and }Pt^{2+}(aq.)$

8 0

3 years ago

assume that you used a 0.660 g piece of pure aluminum in the synthesis of alum and obtained a 63.5% yield of alum. How many gram

attashe74 [19]

Answer:

1.015g

Explanation:

From the question,

Mass of Al used = 0.660g

The percentage yield = 63.5%

Mass of Alum used = y

To find the mass of Alum used, do the following:

%yield = Mass of Al /Mass and Alum

65% = 0.660/y

65/100 = 0.660/y

Cross multiply

65 x y = 0.660x100

Divide both side by 65

y = (0.660x100) / 65

y = 1.015g

The mass of alum obtained is 1.015g

8 0

4 years ago

How concentrated can lye solution be before its ph rises above calcium hydroxide saturated water?

mel-nik [20]

The solubility of calcium hydroxide (Ca(OH)2) in water at $20^{o}C$ is 0.02335 mol/L. When a Ca(OH)2 solution has a concentration of 0.02335 mol/L at the said temperature, a saturated solution is formed.

Upon dissolution and dissociation of 1 mole Ca(OH)2, 1 mole of calcium ion ( $Ca^{+2}$ ) and 2 moles of hydroxide ions ( $OH^{-1}$ ) are formed. The concentration of the hydroxide ions determine the pH of the solution. This same concentration would be needed in calculating the concentration of lye solution needed.

1. To calculate the pH of the solution, a few approaches are possible. Here, the pOH of the solution is calculated first using the molar concentration of $OH^{-1}$ ions. Because pH + pOH = 14, the pH of the solution can then be calculated by subtracting pOH from 14.

Because 2 moles of $OH^{-1}$ ions are formed per mole of Ca(OH)2 dissolved, the pOH of a saturated Ca(OH)2 solution at $20^{o}C$ is,

*Concentration ( $C_{OH_{1}}$ ) of $OH^{-1}$ ions,
$C_{OH_{1}}$ = 0.02335 mol Ca(OH)2/L * 2 mol $OH^{-1}$ / 1 mol Ca(OH)2
$C_{OH_{1}}$ = 0.0467 mol $OH^{-1}$ /L

*pOH of saturated Ca(OH)2 solution,
 pOH = -log [ $C_{OH_{1}}$ ]
 pOH = -log [0.0467 mol $OH^{-1}$ /L ]
 pOH = 1.33

*pH of saturated Ca(OH)2 solution,
 pH = 14 - 1.33
 pH = 12.67

2. To find the concentration of lye (NaOH) solution that would give the same pH as a saturated Ca(OH)2 solution at the same temperature, the pOH of the saturated Ca(OH)2 solution will be used.

For every mole of NaOH that dissolves and dissociates, 1 mole of $OH^{-1}$ ions is formed. Thus,

*Concentration ( $C_{OH_{2}}$ ) of $OH^{-1}$ ions,
pOH = -log [ $C_{OH_{2}}$ ]
 1.33 = -log [ $C_{OH_{2}}$ ]
 $C_{OH_{2}}$ = 10^-1.33
 $C_{OH_{2}}$ = 0.0467 mol $OH^{-1}$ / L

Because 1 mole of NaOH produces only 1 mole of $OH^{-1}$ ions, the concentration of the $OH^{-1}$ ions should then be equal to the concentration of the lye (NaOH) solution. Thus, 0.0467 mol/L of lye would give the same pH as that of 0.0233 mol/L (saturated) Ca(OH)2 solution at $20^{o}C$ . The pH of both solutions is 12.67.

7 0

3 years ago

What does the number 18.9984 represent in the image?

Yanka [14]

The element symbol in chemistry comprises atomic number and mass number.

Atomic number is written on the top of element and mass number is written at the bottom.

For eg:

If Y is an element that is having atomic number a and mass number x. Then its denotion will be:

ᵃYₓ.

So here ⁹F₁₈.₉₉

Here 18.99 will denote mass number while 9 will denote atomic number.

4 0

3 years ago