Is an element with 6 elecrons, 6 protons and 6 neutroms an atom, ion or isotope

denpristay [2]

Answer:

b. independent/manipulated variable

Explanation:

Independent/manipulated variable - refers to the variable that is changed by the scientist or an experimenter. Only one variable that is independent is required to ensure a fair test in an excellent experiment. As the independent variable is being changed by an experimenter or scientist, data is being recorded simultaneously as they are collected.

3 0

2 years ago

Calculate the pH of the following simple solutions:

IceJOKER [234]

Answer :

(1) pH = 1.27

(2) pH = 13.35

(3) The given solution is not a buffer.

Explanation :

(1) 53.1 mM HCl

Concentration of HCl = $53.1mM=53.1\times 10^{-3}M$

As HCl is a strong acid. So, it dissociates completely to give hydrogen ion and chloride ion.

So, Concentration of hydrogen ion= $53.1\times 10^{-3}M$

pH : It is defined as the negative logarithm of hydrogen ion concentration.

$pH=-\log [H^+]$

$pH=-\log (53.1\times 10^{-3})$

$pH=1.27$

(2) 0.223 M KOH

Concentration of KOH = 0.223 M

As KOH is a strong base. So, it dissociates completely to give hydroxide ion and potassium ion.

So, Concentration of hydroxide ion= 0.223 M

Now we have to calculate the pOH.

$pOH=-\log [OH^-]$

$pOH=-\log (0.223)$

$pOH=0.65$

Now we have to calculate the pH.

$pH+pOH=14\\\\pH=14-pOH\\\\pH=14-0.65=13.35$

(3) 53.1 mM HCl + 0.223 M KOH

Buffer : It is defined as a solution that maintain the pH of the solution by adding the small amount of acid or a base.

It is not a buffer because HCl is a strong acid and KOH is a strong base. Both dissociates completely.

As we know that the pH of strong acid and strong base solution is always 7.

So, the given solution is not a buffer.

5 0

3 years ago

Consider n2 (g) + 3h2 (g) →→ 2nh3 (g). what is the mass of nitrogen gas required to react with 0.129 g h2?

Flura [38]

The given chemical reaction given above is already balanced such that the number of atoms in the left hand side of the equation is equal to that of the right hand side. Using the dimensional analysis, proper conversion factors and the molar masses,

mass of nitrogen = (0.129 g H₂)(1 mol H₂/2 g H₂)(1 mol N₂/3 mol H₂)(28 g N₂/1 mol N₂)
mass of nitrogen = 0.602 g N₂

Therefore, 0.602 g of nitrogen will be required for he reaction.

6 0

3 years ago

It has been suggested that the surface melting of ice plays a role in enabling speed skaters to achieve peak performance. Carry

vesna_86 [32]

Explanation:

Relation between pressure, latent heat of fusion, and change in volume is as follows.

$\frac{dP}{dT} = \frac{L}{T \times \Delta V}$

Also, $\frac{L}{T} = \Delta S^{fusion}_{m}$

where, $\Delta V^{fusion}_{m}$ is the difference in specific volumes.

Hence, $\frac{dP}{dT} = \frac{\Delta S^{fusion}_{m}}{\Delta V^{fusion}_{m}}$

As, $\Delta S^{fusion}_{m} = \frac{L}{T} = \frac{6010}{273.15}$ = 22.0 J/mol K

And, $\Delta V^{fusion}_{m} = \frac{M}{d_{H_{2}O}} - \frac{M}{d_{ice}}$ ...... (1)

where, $d_{H_{2}O}$ = density of water

$d_{ice}$ = density of ice

M = molar mass of water = $18.02 \times 10^{-3} kg$

Therefore, using formula in equation (1) we will calculate the volume of fusion as follows.

$\Delta V^{fusion}_{m} = \frac{M}{d_{H_{2}O}} - \frac{M}{d_{ice}}$

= $\frac{18.02 \times 10^{-3}}{997} - \frac{18.02 \times 10^{-3}}{920}$

= $-1.51 \times 10^{-6}$

Therefore, calculate the required pressure as follows.

$\frac{dP}{dT} = \frac{22}{-1.51 \times 10^{-6}}$

= $1.45 \times 10^{7} Pa/K$

or, = 145 bar/K

Hence, for change of 1 degree pressure the decrease is 145 bar and for 4.7 degree change dP = $145 \times 4.7 bar$

= 681.5 bar

Thus, we can conclude that pressure should be increased by 681.5 bar to cause 4.7 degree change in melting point.

5 0

3 years ago

A 22.0 mLmL sample of a 1.16 MM potassium sulfate solution is mixed with 14.8 mLmL of a 0.860 MM barium nitrate solution and thi

dezoksy [38]

Answer:

The limiting reactant is Ba(NO3)2

The theoretical yield BaSO4 is 2.97 grams

The percent yield of the reaction is 86.5 %

Explanation:

Step 1: Data given

Volume of a 1.16 M potassium sulfate solution (K2SO4) = 22.0 mL = 0.022 L

Volume of a 0.860 M barium nitrate solution (Ba(NO3)2 = 14.8 mL = 0.0148 L

The solid BaSO4 is collected, dried, and found to have a mass of 2.57 grams

Step 2: The balanced equation

K2SO4(aq) + Ba(NO3)2(aq) → BaSO4(s) + 2KNO3(aq)

Step 3: Calculate moles

Moles = volume * molarity

Moles K2SO4 = 0.022 L * 1.16 M

Moles K2SO4 = 0.02552 moles

Moles Ba(NO3)2 = 0.0148 L * 0.860 M

Moles Ba(NO3)2 = 0.012728 moles

Step 4: Calculate the limiting reactant

For 1 mol K2SO4 we need 1 mol Ba(NO3)2 to produce 1 mol BaSO4 and 2 moles KNO3

Ba(NO3)2 is the limiting reactant. It will completely be consumed. (0.012728 moles) . K2SO4 is in excess. There will remain 0.02552 - 0.012728 = 0.012792 moles

Step 5: Calculate moles BaSO4

‬For 1 mol K2SO4 we need 1 mol Ba(NO3)2 to produce 1 mol BaSO4 and 2 moles KNO3

For 0.012728 moles Ba(NO3)2 we'll have 0.012728 moles BaSO4

Step 6: Calculate mass BaSO4

Mass BasO4 = moles BaSO4 * molar mass BaSO4

Mass BaSO4 = 0.012728 moles * 233.38 g/mol

Mass BaSO4 = 2.97 grams

Step 7: Calculate the percent yield

% yield = (actual yield / theoretical yield ) * 100 %

% yield = ( 2.57 grams / 2.97 grams ) * 100 %

% yield = 86.5 %

The limiting reactant is Ba(NO3)2

The theoretical yield BaSO4 is 2.97 grams

The percent yield of the reaction is 86.5 %

7 0

3 years ago

Is an element with 6 elecrons, 6 protons and 6 neutroms an atom, ion or isotope​

Is an element with 6 elecrons, 6 protons and 6 neutroms an atom, ion or isotope