If an atom has six protons and four neutrons, what is its charge?

Answers

Answer 1

Answer: +4 that is the answer hoped i helped

Answer 2

Answer: the charge is always the second number.
4 would be the charge.

did this help?

Related Questions

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What is the percent by mass of oxygen in carbon dioxide (CO2)?

Answers

72.7% is the percent by mass of oxygen in carbon dioxide. A percent is obtained by multiplying the result by 100.

One approach to show the concentration for an element within a compound or component in a combination is as a mass percentage. The mass percentage is computed by dividing the total weight of the combination by the mass of each component and multiplying the result by 100%. The mass percent is calculated by dividing the mass that contains the compound and solute by the mass for the element or solute.

Molar mass of oxygen = 32 g/mole

Molar mass of carbon dioxide = 44 g/mole

mass percentage of oxygen =(molar mass of oxygen/molar mass of carbon dioxide)× 100

=(32/44)× 100

=72.7%

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Answer : The percent by mass of oxygen in carbon dioxide is, 72.72%

Solution : Given,

Molar mass of oxygen = 16 g/mole

Molar mass of carbon dioxide = 44 g/mole

As we know that there are 2 atoms of oxygen and 1 atom of carbon present in the carbon dioxide.

So, the molar mass of oxygen, $O_2$ = 2 × 16 = 32 g/mole

Now we have to calculate the percent by mass of oxygen in carbon dioxide.

$\%\text{ by mass of oxygen}=\frac{\text{Molar mass of oxygen}}{\text{Molar mass of }CO_2}* 100$

Now put all the given values in this expression, we get

$\%\text{ by mass of oxygen}=(32g/mole)/(44g/mole)* 100=72.72\%$

Therefore, the percent by mass of oxygen in carbon dioxide is, 72.72%

What is a chemical entity that forms a complex ion around a central atom?

Answers

A coordination complex consists of a central atom or Jon, which usually metallic and is called the coordination centre, and a surrounding array of bound molecules or ions, that are in turn known as ligand or complexing agents

Discuss the shapes of Nh3 and H2O on the basis of VESPER theory?

Answers

Answer:

Ammonia has 4 regions of electron density around the central nitrogen atom (3 bonds and one lone pair). These are arranged in a tetrahedral shape. The resulting molecular shape is trigonal pyramidal with H-N-H angles of 106.7°.

Explanation:

Which pair of elements will form an ionic bond?N and O
K and Br
C and Cl
Fe and Ni

Answers

K and Br can form an ionic bond. Since potassium is a metal (has a small electronegativity) and bromine is a nonmetal (has a large electronegativity), potassium can give an electron to bromine creating potassium bromide.

I hope this helps. Let me know if anything is unclear.

Final answer:

Ionic bonds are formed between a metal and a nonmetal. Among the pairs given, K and Br satisfy these conditions and will form an ionic bond.

Explanation:

The question is asking about ionic bonds, which are formed when one atom donates an electron to another atom. This occurs between a metal and a nonmetal due to their differences in electronegativity.

Looking at the options, N and O are both nonmetals, C and Cl are also both nonmetals, and Fe and Ni are both metals, so they would not form ionic bonds. On the other hand, K (a metal) and Br (a nonmetal) have a difference in electronegativity such that K can donate an electron to Br, forming an ionic bond. Therefore, the pair that would form an ionic bond is K and Br.

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77. A voltaic cell consists of a Zn/Zn2+ half-cell and a Ni/Ni2+ half-cell at 25°C. The initial concentrations of Ni2+ and Zn2+ are 1.50 M and 0.100 M, respectively. a. What is the initial cell potential? b. What is the cell potential when the concentration of Ni2+ has fallen to 0.500 M? c. What are the concentrations of Ni2+ and Zn2+ when the cell potential falls to 0.45 V?

Answers

Answer :

(a) The initial cell potential is, 0.53 V

(b) The cell potential when the concentration of $Ni^(2+)$ has fallen to 0.500 M is, 0.52 V

(c) The concentrations of $Ni^(2+)$ and $Zn^(2+)$ when the cell potential falls to 0.45 V are, 0.01 M and 1.59 M

Explanation :

The values of standard reduction electrode potential of the cell are:

$E^0_([Ni^(2+)/Ni])=-0.23V$

$E^0_([Zn^(2+)/Zn])=-0.76V$

From this we conclude that, the zinc (Zn) undergoes oxidation by loss of electrons and thus act as anode. Nickel (Ni) undergoes reduction by gain of electrons and thus act as cathode.

The half reaction will be:

Reaction at anode (oxidation) : $Zn\rightarrow Zn^(2+)+2e^-$ $E^0_([Zn^(2+)/Zn])=-0.76V$

Reaction at cathode (reduction) : $Ni^(2+)+2e^-\rightarrow Ni$ $E^0_([Ni^(2+)/Ni])=-0.23V$

The balanced cell reaction will be,

$Zn(s)+Ni^(2+)(aq)\rightarrow Zn^(2+)(aq)+Ni(s)$

First we have to calculate the standard electrode potential of the cell.

$E^o=E^o_(cathode)-E^o_(anode)$

$E^o=E^o_([Ni^(2+)/Ni])-E^o_([Zn^(2+)/Zn])$

$E^o=(-0.23V)-(-0.76V)=0.53V$

(a) Now we have to calculate the cell potential.

Using Nernest equation :

$E_(cell)=E^o_(cell)-(0.0592)/(n)\log ([Zn^(2+)])/([Ni^(2+)])$

where,

n = number of electrons in oxidation-reduction reaction = 2

$E_(cell)$ = emf of the cell = ?

Now put all the given values in the above equation, we get:

$E_(cell)=0.53-(0.0592)/(2)\log ((0.100))/((1.50))$

$E_(cell)=0.49V$

(b) Now we have to calculate the cell potential when the concentration of $Ni^(2+)$ has fallen to 0.500 M.

New concentration of $Ni^(2+)$ = 1.50 - x = 0.500

x = 1 M

New concentration of $Zn^(2+)$ = 0.100 + x = 0.100 + 1 = 1.1 M

Using Nernest equation :

$E_(cell)=E^o_(cell)-(0.0592)/(n)\log ([Zn^(2+)])/([Ni^(2+)])$

Now put all the given values in the above equation, we get:

$E_(cell)=0.53-(0.0592)/(2)\log ((1.1))/((0.500))$

$E_(cell)=0.52V$

(c) Now we have to calculate the concentrations of $Ni^(2+)$ and $Zn^(2+)$ when the cell potential falls to 0.45 V.

Using Nernest equation :

$E_(cell)=E^o_(cell)-(0.0592)/(n)\log ([Zn^(2+)+x])/([Ni^(2+)-x])$

Now put all the given values in the above equation, we get:

$0.45=0.53-(0.0592)/(2)\log ((0.100+x))/((1.50-x))$

$x=1.49M$

The concentration of $Ni^(2+)$ = 1.50 - x = 1.50 - 1.49 = 0.01 M

The concentration of $Zn^(2+)$ = 0.100 + x = 0.100 + 1.49 = 1.59 M

SI units are useful because they are all related by multiples of ____.

Answers

Answer:

SI units are useful because they are all related by multiples of 10.

Explanation:

International System of Units (abbreviated SI) is a system of measurement units, which is used almost everywhere in the world.

A unit system is a set of units of consistent, standard and uniform measurement. In other words, a unit system is a set of units (grouped and formally defined) that is used as a standard. Normally, in a system of units, few basic units are defined and from these several derived units are defined.

The International System of Units defines seven basic units, also called fundamental units, which define the corresponding fundamental physical quantities, chosen by convention, and which allow to express any physical quantity in terms or as a combination of them. The other units are obtained by combining the basic units.

The basic units of this system are:

Length-meter (m)
Mass-kilogram (kg)
Time- second (s)
Electric current intensity - ampere (A)
Temperature - Kelvin (K)
Amount of substance - mol (mol)
Light intensity - candela (cd)

The SI is based on the metric system, which is a system of units that are related to each other by multiples or submultiples of 10.

Then, SI units are useful because they are all related by multiples of 10.

SI units are useful as they are all related by multiples of 10.

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