which two options are examples of chemical changes?
A. A shinny metal bar glows red and expands when heated.

B. white piece of paper turns black and gives off a smell when burned

C. When two clear colorless liquids are mixed together, a white solid forms

D. A white powder mixed with a clear, colorless liquid makes a solution that is also clear, colorless liquid

Answer:..

Explanation:

Answer:

A) enthalpy of reaction

Explanation:

The region C signifies the enthalpy of reaction.

This diagram is the energy profile of an endothermic reaction. In such reaction, heat is absorbed from the surrounding. At the end of the reaction, the heat of product is lesser than that of the reactants.

it is indeed A) enthalpy of reaction

Answer:

Metals and Non-metals

Explanation:

Answer:

Explanation:

We shall use the formula S₁V₁ = S₂V₂

S₁ = .1180 M , V₁ = 25.98 mL

S₂ = ? , V₂ = 52.50 mL

.1180 M  x 25.98 = 52.50 x S₂

S₂ = .0584 M

Molarity of the acid solution = .0584 M .

The concentration of the acid solution is 0.058 M.

The equation of the reaction is;

CH3COOH(aq) + KOH(aq) ----->  CH3COOK(aq) + H2O(l)

The following are known;

Concentration of base CB = 0.1180 M

Volume of base VB =  25.98 mL

Concentration of acid CA = ?

Volume of acid VA = 52.50 mL

Number of moles of acid NA = 1

Number of moles of base NB = 1

Using;

CAVA/CBVB =NA/NB

CAVANB = CBVBNA

CA=  CBVBNA/VANB

CA = 0.1180 M × 25.98 mL × 1/52.50 mL × 1

CA = 0.058 M

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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:

a. 0.183 mol C

b. 0.366 mol H

Explanation:

Assuming total combustion, all of the carbon in the unknown compound turned into carbon dioxide, CO₂.

So first we calculate the CO₂ moles produced, using its molecular weight:

This means in the unknown compound there were 0.183 moles of carbon, C.

Conversely, all of the hydrogen in the unknown compound turned into water, H₂O.

Calculating the H₂O moles:

We multiply the water moles by two, as there are 2 H moles per H₂O mol:

Answer:

4.22mL

Explanation:

V=m/d

v= 18.45g/4.37g/mL

Answer:

What subject is this for?

Explanation:

Answer:

ARSENIC

Explanation:

It has an atomic number of 33

Answer:

C.

Explanation:

You can use the series of elimination for this. First, you look at A. Technology is not an understanding of something new, so we cross that out. Second, you look at B. Technology isn't a series of steps, so we can mark that one off. Third, you look at C. Technology is something created using science for use by society, so we can keep that in mind. Lastly, we check D. Technology doesn't match up to the definition, so we can cross that one out. The answer that would make the most sense would be D.

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:

Ions of opposite charge strongly attract each other; those of like charges repel. As a result ions in an ionic compound are arranged in a particular manner.

Explanation:

Google is smart

Answer:

CHO

Explanation:

Carbon = 41%,  Hydrogen = 4.58%, oxygen = 54.6%

Step 1:

Divide through by their respective relative atomic masses

41/ 12,         4.58/1,         54.6/16

3.41              4.58            3.41

Step 2:

Divide by the lowest ratio:

3.41/3.41,      4.58/3.41,     3.41/3.41

1,                    1,                  1

Hence the empirical formula is CHO

Answer:

The empirical formula of X is C3H4O3.

Explanation:

Answer:

2 C 2 H 2 ( g ) + 5 O 2 ( g ) ⟶ 4 C O 2 ( g ) + 2 H 2 O ( l )- combustion reaction

N H 4 N O 3 ( s ) ⟶ N 2 O ( g ) + 2 H 2 O ( l )- decomposition reaction

C O ( g ) + 2 H 2 ( g ) ⟶ C H 3 O H ( l ) - combination reaction

2 F e ( s ) + 6 H C l ( a q ) ⟶ 2 F e C l 3 ( a q ) + 3 H 2 ( g )- Redox reaction

C a C l 2 ( a q ) + N a 2 C O 3 ( a q ) ⟶ 2 N a C l ( a q ) + C a C O 3 ( s )- double displacement reaction

Explanation:

We can determine the type of reaction by considering the reactants and products.

Combustion is a reaction between a substance and oxygen which produces heat and light. The first reaction is the equation for the combustion of ethyne.

A decomposition reaction is one in which a single reactant breaks down to form products. The second reaction is the decomposition of ammonium nitrate.

A  combination reaction is said to occur when two elements or compounds react to form a single product. The third reaction is the combination of carbon dioxide and methane to form  methanol.

An oxidation-reduction reaction is a reaction in which there is a change in oxidation number of species from left to right of the chemical reaction equation. The fourth reaction is the oxidation of iron (0 to +3 state) and reduction of hydrogen (+1 to 0 state).

A double displacement reaction is a reaction in which ions exchange partners from left to right in the reaction equation. The fifth reaction is a double displacement reaction. Both Na^+ and Ca^2+ exchanged partners from left to right of the reaction equation.

Reactions are the formation of the products from the reactant. The types of reactions are combustion, decomposition, combination, Redox and double displacement.

The reaction is a chemical change in the properties of the reactant that forms the products. It can be of various types based on the formation of the product.

The first reaction is combustion as the reactants react and use oxygen to form heat, carbon dioxide and water. The combustion reaction of ethyne can be shown as,

[tex]\rm 2 C _{2} H _{2} ( g ) + 5 O _{2} ( g ) \rightarrow 4 C O _{2} ( g ) + 2 H _{2} O ( l )[/tex]

The second reaction is decomposition in which a single reactant decomposes to form two or more products. The decomposition of ammonium nitrate can be shown as,

[tex]\rm N H _{4} N O _{3} ( s ) \rightarrow N _{2} O ( g ) + 2 H _{2} O ( l )[/tex]

The third reaction is a combination reaction in which two compound or elements combines to form one product. The combination reaction between carbon monoxide and hydrogen to form methanol can be shown as,

[tex]\rm C O ( g ) + 2 H _{2} ( g ) \rightarrow C H _{3} O H ( l )[/tex]

The fourth reaction is redox and includes the oxidation and the reduction of the species of the reaction. In the reaction, iron undergoes oxidation and hydrogen reduction. The redox reaction can be shown as,

[tex]\rm 2 F e ( s ) + 6 H C l ( a q ) \rightarrow 2 F e C l _{3} ( a q ) + 3 H _{2} ( g )[/tex]

The fifth reaction is a double displacement reaction in which the calcium and sodium interchange their position in the product formation. The reaction can be shown as,

[tex]\rm C a C l _{2} ( a q ) + N a _{2} C O _{3} ( a q ) \rightarrow 2 N a C l ( a q ) + C a C O _{3} ( s )[/tex]

Therefore, the type of reactions is 1. combustion, 2. decomposition, 3. combination, 4. redox and 5. double displacement.

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Answer:

it is Calcium (Ca)

4th period, 2nd group, 2 valence electrons

Answer: Proton - positive charge (+)

Neutron - neutral charge (0)

Electron - negative charge (-)

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:

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

Explanation:

Answer:

the second answer its science behind it

Answer:

b

Explanation:

Answer:

Nervous and Excretory

Explanation:

The nervous system makes you thirsty. The integumentary system makes you sweat. the integumentary system is very similar to the Excretory system.

Nervous and Excretory systems are working together in this scenario.

The nervous system makes you thirsty. The integumentary system makes you sweat. the integumentary system is very similar to the Excretory system.

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Answer:

acidic because of electrical issues and the body of electrical equipment

Answer:

Explanation:

NaNO₃  = Na⁺  + NO₃⁻¹

.497 M                 .497 M

moles of NO₃⁻¹ = .897 x .497 = .4458 moles

Ca( NO₃)₂  = Ca + 2 NO₃⁻¹

.341 M                    2 x .341 M = .682 M

moles of NO₃⁻¹ = .813 x .682 = .5544 moles

Total moles =  .4458 moles  +  .5544 moles

= 1.0002 moles

volume of solution = 897 + 813 = 1710 mL

= 1.710 L

concentration of nitrate ion = 1.0002 / 1.710 M

= .585 M

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

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Answer:

gill lamellae

Explanation:

Gills  in fishes are structures that permit fish to carry out gaseous exchange in water.  Fishes exchange gases  such as oxygen and carbon dioxide using gills.

Fish gills carry out their activity mainly through its major component called the  gill lamellae. These are comb-like filaments which help increase the surface area of the gills in order to facilitate gaseous exchange

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:

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