4. Horizontal rows of the Periodic Table are called:
a, Clusters
Groups
b. Families
d) Periods

Answer: The mass of blue copper sulfate is 3.5 g

Explanation:

Law of conservation of mass states that mass can neither be created nor be destroyed but it can only be transformed from one form to another form.

This also means that total mass on the reactant side must be equal to the total mass on the product side.

The chemical equation for the heating of copper sulfate crystals is:

Let the mass of blue copper sulfate be 'x' grams

We are given:

Mass of copper sulfate powder = 2.1 grams

Mass of water = 1.4 grams

Total mass on reactant side = x

Total mass on product side = (2.1 + 1.4) g

So, by applying law of conservation of mass, we get:

Hence, the mass of blue copper sulfate is 3.5 grams

Answer:

2.35 x 10²⁰ atoms Ga

Explanation:

After converting from mg to g, use the molar mass as the unit converter to convert to moles. Then using Avogadro's number, 6.022 x 10²³ convert from moles to atoms of Ga.

[tex]27.2mgGa*\frac{1g}{1000mg} *\frac{1 mol Ga}{69.72gGa} *\frac{6.022*10^2^3 atoms Ga}{1 molGa} = 2.349 * 10^2^0 atoms Ga[/tex]

Then round to 3 significant figures = 2.35 x 10²⁰ atoms Ga.

The number of atoms in 27.2 mg sample of Ga is 2.35 × 10²⁰ atoms

From the question, we are to calculate the number of atoms in a 27.2 mg sample of Ga.

First, we will determine the number of moles of Ga present

Using the formula,

[tex]Number\ of\ moles = \frac{Mass}{Molar\ mass} [/tex]

Mass = 27.2 mg = 0.0272 g

Molar mass = 69.72 g/mol

Then,

[tex]Number\ of\ moles \ of\ Ga = \frac{0.0272}{69.72} [/tex]

[tex]Number\ of\ moles \ of\ Ga = [/tex] 0.000390132 moles

Now, for the number of atoms present

From the formula

Number of atoms = Number of moles × Avogadro's constant

Then,

Number of Ga atoms = 0.000390132 × 6.022×10²³

Number of Ga atoms = 2.35 × 10²⁰ atoms

Hence, the number of atoms in 27.2 mg sample of Ga is 2.35 × 10²⁰ atoms

Learn more on stoichiometry here: https://brainly.com/question/14464650

Answer:

ARSENIC

Explanation:

It has an atomic number of 33

Answer:

i

[tex]J_{m} = 20 [/tex]

ii

[tex]J_{m} = 22.5 [/tex]

Explanation:

From the question we are told that

  The first temperatures is [tex]T_1 =  25^oC =  25 +273 =298 \ K[/tex]

   The second temperature is  [tex]T_2 =  100^oC =  100 +273 = 373 \ K[/tex]

Generally the equation for  the most highly populated rotational energy level is mathematically represented as

     [tex]J_{m} = [ \frac{RT}{2B}]  ^{\frac{1}{2} } - \frac{1}{2}[/tex]

Here R is the gas constant with value [tex]R =8.314 \ J\cdot K^{-1} \cdot mol^{-1}[/tex]

Also  

      B is given as [tex]B=\ 0.244 \ cm^{-1}[/tex]

   Generally the energy require per mole to move 1 cm is  12 J /mole

So   [tex]0.244 \ cm^{-1}[/tex]  will require x J/mole

           [tex]x =  0.244 *  12[/tex]

=>          [tex]x =  2.928 \ J/mol [/tex]

So at the first temperature

     [tex]J_{m} = [ \frac{8.314 * 298  }{2*  2.928 }]  ^{\frac{1}{2} } - 0.5 [/tex]

=>  [tex]J_{m} = 20 [/tex]

So at the second temperature

           [tex]J_{m} = [ \frac{8.314 * 373  }{2*  2.928 }]  ^{\frac{1}{2} } - 0.5 [/tex]

=>  [tex]J_{m} = 22.5 [/tex]

Answer:

acidic because of electrical issues and the body of electrical equipment

Answer:

a) - 0.2 M

b) - 0.2 M

c)- 0

Explanation:

The chemical formula of copper (II) bromide is CuBr₂. Its molar mass (MM) is calculated as follows:

MM(CuBr₂)= MM(Cu) + (2 x MM(Br) = 63.5 g/mol + (2 x 80 g/mol)= 223.5 g/mol

a). Molarity = moles CuBr₂/1 L solution

moles CuBr₂ = mass/MM = 18.7 g x 1 mol/223.5 g = 0.084 mol

Volume in L = 375 mL x 1 L/1000 mL = 0.375 L

M = 0.084 mol/(0.375 L) = 0.223 M ≅ 0.2 M

b). When is added to water, CuBr₂ dissociates into ions as follows:

CuBr₂ ⇒ Cu²⁺ + 2 Br⁻

We have 1 mol Cu²⁺ (copper (II) cation) per mol of CuBr₂. Thus, the concentration of copper (II) cation is:

0.2 mol CuBr₂ x 1 mol Cu²⁺/mol CuBr₂ = 0.2 M

c). The concentration of acetate anion is 0. There is no acetate anion in the solution (the anion from CuBr₂ is bromide Br⁻).

Answer:

Final volume  = 0.44 L

Explanation:

Given data:

Initial volume of balloon = 1.1 L

Initial pressure = 0.80 atm

Final volume = ?

Final pressure = 2.0 atm

Solution:

The given problem will be solved through the Boly's law,

"The volume of given amount of gas is inversely proportional to its pressure by keeping the temperature and number of moles constant"

Mathematical expression:

P₁V₁ = P₂V₂

P₁ = Initial pressure

V₁ = initial volume

P₂ = final pressure

V₂ = final volume  

Now we will put the values in formula,

P₁V₁ = P₂V₂

0.80 atm × 1.1 L = 2.0 atm × V₂

V₂ = 0.88 atm. L/ 2.0 atm

V₂ = 0.44 L

Answer: Proton - positive charge (+)

Neutron - neutral charge (0)

Electron - negative charge (-)

Explanation:

Answer:

the second answer its science behind it

Answer:

b

Explanation:

Answer:

7.71 × 10⁻⁴ M/s

Explanation:

The initial rate of the reaction can be expressed by using the formula:

[tex]\dfrac{\Delta [O_2]}{\Delta t}[/tex]

where the number of moles of O₂ = [tex]\dfrac{PV}{RT}[/tex]

where;

Pressue P = 1.00 atm

Volume V =5.74mL =  (5.74 /1000) L

Rate R = 0.082 L atm/mol.K

Temperature = 298 K

[tex]= \dfrac{1.00 \ atm \times \dfrac{5.74 }{1000}L}{0.082 \ L \ atm/mol.K \times 298 K}[/tex]

= 2.35 × 10⁻⁴ mol

Δ[O₂] = [tex]\dfrac{moles \ produced - initial \ mole}{\dfrac{5.08 }{1000}L }[/tex]

Δ[O₂] = [tex]\dfrac{2.35 \times 10^{-4} M - 0 M}{\dfrac{5.08 }{1000}}[/tex]

Δ[O₂]  = 0.04626 M

The initial rate = [tex]\dfrac{\Delta [O_2]}{\Delta t}[/tex]

= [tex]\dfrac{0.04626}{60}[/tex]

= 7.71 × 10⁻⁴ M/s

Answer:

4.22mL

Explanation:

V=m/d

v= 18.45g/4.37g/mL

Answer:

QC= [O2]^3/[F2]^10

Explanation:

Answer:

pH = 11

Explanation:

The equilibrium of a weak base as NaCN in water is:

NaCN(aq) + H₂O(l) ⇄ OH⁻(aq) + Na⁺(aq) + HCN(aq)

And kb, the equilibrium constant, is:

Kb = [OH⁻] [HCN] / [NaCN]

Where Kb of NaCN is 2.04x10⁻⁵

In the beginning, the [NaCN] is 0.050mol / L = 0.050M.

Both [OH⁻] and [HCN] are produced from this equilibrium, and its concentration is X, that is:

2.04x10⁻⁵ = [X] [X] / [0.050M]

1.02x10⁻⁶ = X²

X = 1x10⁻³ = [OH⁻]

As pOH = - log [OH⁻]

pOH = 3.00

And pH = 14 - pOH

Answer:..

Explanation:

Answer:

it is Calcium (Ca)

4th period, 2nd group, 2 valence electrons