Writing key words in the margin is part of which note-taking procedure?

Select one:
a. record
b. review
c. reduce
d. recite

Answer:

5;11

Explanation:

Answer:

Explanation:

The new volume can be found by using the formula for Boyle's law which is

[tex]P_1V_1 = P_2V_2[/tex]

where

P1 is the initial pressure

P2 is the final pressure

V1 is the initial volume

V2 is the final volume

Since we are finding the new volume

[tex]V_2 = \frac{P_1V_1}{P_2} \\[/tex]

We have

[tex]V_2 = \frac{300 \times 2}{7} = \frac{600}{7} \\ = 85.714285...[/tex]

We have the final answer as

Hope this helps you

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:

Complete Question

The complete question is shown in the first uploaded image

Answer:

So the math expression is  

             [tex]heat  =  \frac{ 35. 7  KJ *  1900 \ gram }{ 1 \ gram }[/tex]

Explanation:

From the question we are told that

  The heat released for 1 gram of reactant consumed is  [tex]H  =  37.5 \ KJ/g [/tex]

   The mass of reactant considered is  [tex]m =  1.9 \ kg  =  1900 \  g[/tex]

So  if

             [tex]37.5 \ KJ [/tex] is produced for  1 gram

Then

              x kJ is produced for  1900 g  

=>   [tex]x  =  \frac{ 35. 7  KJ *  1900 \ gram }{ 1 \ gram }[/tex]

So the heat released is  

       [tex]heat  =  \frac{ 35. 7  KJ *  1900 \ gram }{ 1 \ gram }[/tex]

Answer:

a. H₂CO₃(aq) + KOH(aq) ⇄ K₂CO₃(aq) + H₂O(l)

b. 0.603 M

Explanation:

Step 1: Write the neutralization reaction

H₂CO₃(aq) + KOH(aq) ⇄ K₂CO₃(aq) + H₂O(l)

Step 2: Calculate the reacting moles of KOH

22.9 mL of 1.430 M KOH react.

0.0229 L × (1.430 mol/L) = 0.0327 mol

Step 3: Calculate the reacting moles of H₂CO₃

The molar ratio of H₂CO₃ to KOH is 1:1. The reacting moles of H₂CO₃ are 1/1 × 0.0327 mol = 0.0327 mol.

Step 4: Calculate the molarity of H₂CO₃

0.0327 moles of H₂CO₃ are in a volume of 54.2 mL. The molarity of H₂CO₃ is:

M = 0.0327 mol/0.0542 L = 0.603 M

Answer:

4

Explanation:

24/6 is 4.

Answer:

Mathematically: D = m/v. If you know what liquid you have, you can look up its density in a table. Once you know that, all you have to do to find the mass of the liquid is to measure its volume. Once you know density and volume, calculate mass using this relationship: mass = density • volume.

Explanation:

Answer:

Ecological islands is just a terminology used, since it isn't surrounded by water like an actual island, instead, it's something isolated from other things, which is the same thing as an island in a way:

Island (geographical) - Land isolated from bigger "main" land masses by water.

Island (ecological) - Land which contains certain isolated features (habitat, plants, etc.).

Explanation:

An ecological island is an area of land, isolated by natural or artificial means from its surrounding land while geographical islands are a body of land surrounded by water.

An ecological island refers to an area of land which is not necessarily an actual island but isolated by natural or artificial means from the surrounding land. A natural micro-habitat exists on sich lands amidst a larger differing ecosystem.

In artificial ecological islands, all non-native species are eradicated, native species are reintroduced and nurtured, and the natural or artificial border is maintained to prevent the reintroduction of non-native species.

A geographical island is a body of land surrounded by water. Continents are also surrounded by water as they are so big but they are not considered islands. Islands may be categorized as either continental or oceanic.

There are so many islands in the lakes, oceans, and rivers around the world. They vary in climate, size, and the types of organisms that inhabit them.

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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:

Exothermic

Explanation:

Nuclear fission means splitting, so there is a lot of energy being released.

Answer:

I think it's A and D

Explanation:

I'm not sure if it's right

Answer:

The answer is B and D

Explanation:

trust fr

Answer:

Kc = 4.86×10⁻⁶

Explanation:

We begin from the equation:

N₂  +  2H₂ ⇄  2NH₃

We start from 3.47×10⁻² moles of N₂(g) and 6.38×10⁻² moles of H₂(g), so when we reach the equilibrium, we get 6.25×10⁻² moles of H₂.

This data means, that in the reaction we made react:

6.38×10⁻² - x = 6.25×10⁻²

x = 1.3×10⁻³ moles of H₂

As stoichiometry is 1:3, we will know that the moles of N₂ that have been reacted were:

1.3×10⁻³ moles / 3 = 4.33×10⁻⁴ moles of N₂

So, in the equilibrium we would have:

3.47×10⁻² moles of N₂ - 4.33×10⁻⁴ moles of N₂ = 0.0343 moles of N₂

How many ammonia, would we have in the equilibrium?

4.33×10⁻⁴ mol . 2 = 8.66×10⁻⁴ moles (from stoichiometry with N₂, 1:2)

(1.3×10⁻³ mol . 2) / 3 = 8.66×10⁻⁴ moles (from stoichiometry with H₂, 2:3)

Let's make the expression for Kc

Kc = [NH₃]³ / [N₂] . [H₂]²

(8.66×10⁻⁴ )³ / (0.0343 . (6.25×10⁻²)² = 4.86×10⁻⁶

Answer:

true. cathode rays are deflected

Calcium (Ca) is in group 2 and period 4 on the periodic table be because Calcium has 2 valence electrons and 4 electron shell. Thus, calcium is a metal like all other group 2 element.

The correct answer to the question is Option C. Calcium, like all group 2 elements, is reactive and a solid at room temperature.

Calcium is a group 2 element majorly because it has 2 valence electrons. It is also in period 4 because it has 4 electron shells.

Being a group 2 element, calcium is a solid at room temperature and also reactive. All elements in the group 2 are metals.

There are other elements in period 4 which are not solid. For example krypton is an element in period 4 and it is a gas and not reactive.

From the above information, we can conclude that the correct answer to the question is:

Option C. Calcium, like all group 2 elements, is reactive and a solid at room temperature.

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Answer:

Mass = 23.232 g

Explanation:

Given data:

Mass of C₈H₁₈ = 7.58 g

Mass of CO₂ produced = ?

Solution:

Chemical equation:

2C₈H₁₈ + 25O₂      →     16CO₂ + 18H₂O

Number of moles of octane:

Number of  moles =  mass/molar mass

Number of moles = 7.58 g/ 114.23 g/mol

Number of moles = 0.066 mol

Now we will compare the moles of CO₂ with octane from balance chemical equation.

                  C₈H₁₈            :           CO₂

                       2              :              16

                 0.066             :           16/2×0.066 = 0.528

Mass of CO₂ produced:

Mass = number of moles × molar mass

Mass = 0.528 mol × 44 g/mol

Mass = 23.232 g

Answer:

ΔH°rxn = -827.5 kJ

Explanation:

Let's consider the following balanced equation.

2 PbS(s) + 3 O₂(g) → 2 PbO(s) + 2 SO₂(g)

We can calculate the standard enthalpy of reaction (ΔH°rxn) from the standard enthalpies of formation (ΔH°f) using the following expression.

ΔH°rxn = [2 mol × ΔH°f(PbO(s)) + 2 mol × ΔH°f(SO₂(g) )] - [2 mol × ΔH°f(PbS(s)) + 3 mol × ΔH°f(O₂(g) )]

ΔH°rxn = [2 mol × ΔH°f(PbO(s)) + 2 mol × ΔH°f(SO₂(g) )] - [2 mol × ΔH°f(PbS(s)) + 3 mol × ΔH°f(O₂(g) )]

ΔH°rxn = [2 mol × (-217.32 kJ/mol) + 2 mol × (-296.83)] - [2 mol × (-100.4) + 3 mol × 0 kJ/mol]

ΔH°rxn = -827.5 kJ

The standard enthalpy of the reaction is -827.5 kJ/mol

The standard enthalpy of reaction [tex]\mathbf{\Delta H^0_{rxn}}[/tex] is the enthalpy change that happens in a system whenever one mole of the matter is converted through a chemical process under normal conditions.

The given reaction can be expressed as:

2PbS(s) + 3O₂(g) → 2PbO(s) + 2SO₂(g)

The standard enthalpy can be represented by the equation:

[tex]\mathbf{\Delta H^0_{rxn} = \sum \Delta _f ^0(products) - \sum \Delta _f^0(reactants)}[/tex]

At standard conditions, the standard enthalpies of formation of the given species are:

∴

[tex]\mathbf{ \Delta H^0_{rxn} = \Big[2 mol \times \Delta H^0_f(PbO(s)) + 2 mol \times \Delta H^0_ f(SO_2(g) )\Big] - \Big[2 mol \times \Delta H^0_f (PbS(s))} + \mathbf{ 3 mol \times \Delta H^0_f(O_2(g) )\Big] }}[/tex]

[tex]\mathbf{\Delta H^0rxn = [2 mol \times (-217.32 kJ/mol) + 2 mol \times (-296.83)] - [2 mol \times (-100.4)} \\ \mathbf{+ 3 mol \times 0 kJ/mol]}}[/tex]

[tex]\mathbf{\Delta H^0rxn = -827.5 \ kJ/mol}}[/tex]

Therefore, we can conclude that the standard enthalpy of the reaction is -827.5 kJ/mol

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Answer:

[tex]a=\frac{f}{d-e}-\frac{b}{c}[/tex]

Explanation:

Hello.

In this case, for the equation:

[tex](a+b/c)(d-e)=f[/tex]

For isolating a, we must first pass (d-e) to divide at the other side as it is initially multiplying:

[tex]a+b/c=\frac{f}{(d-e)}[/tex]

Next, as b/c is adding, it passes to subtract at the other side in order to finally isolate a:

[tex]a=\frac{f}{d-e}-\frac{b}{c}[/tex]

Best regards!

Answer:

Explanation:

Al(s) + 3/2 Br₂(l) = AlBr₃(s)

K = [  AlBr₃] / [ Al] [  Br₂]³/²

K² =  [  AlBr₃]² / [  Al ] ² [ Br₂]³

2 AlBr₃ = 2 Al(s) + 3 Br₂(l) =

K₁ =  [  Al ] ² [ Br₂]³ /  [  AlBr₃]²

K₁ =  ( 1 / K² ) = K⁻²

2 ) 2 Al(s) + 3 Br₂(l) = 2 AlBr₃(s)

K₂ = [ AlBr₃ ]² / [  Al ]² [  Br₂ ]³

K₂ = K²

3 )

AlBr₃(s) =   Al(s) + 3/2 Br₂(l)

K₃  = [ Al ] [ Br₂ ] ³/² / [ AlBr₃ ]

=  ( 1 / K ) = K⁻¹

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:

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:

Red

Explanation:

Acetic acid is an acidic medium. Recall that indicators are organic substances whose color changes in response to change in the pH of the solution. often times, the protonated and deprotonated forms of an indicator have different colors.

However, an equilibrium is set up when an indicator is in acid/ basic medium. Methyl orange is red in acid medium and yellow in  basic medium.

Hence while in acetic acid, the equilibrium lies towards the protonated form of acetic acid, hence the solution appears red.

Answer:

d

Explanation:

collagen is made of proteins

Answer:

it is Calcium (Ca)

4th period, 2nd group, 2 valence electrons

Answer:

36 mol of H2

Explanation:

The balanced equation of the reaction is given as;

3H2 + N2 --> 2NH3

From the reaction;

It takes 3 mol of H2 reacting with 1 mol  of N2 to form 2 mol of NH3

3 mol of H2 = 2 mol of NH3

x mol of H2 = 24 mol of NH3

x =  (24 * 3) / 2 = 36 mol of H2

Answer:

2.75 g

Explanation:

The number of moles of KOH is obtained from the equation;

n=CV

C= concentration

V= volume of solution

n= 0.527 × 25.6/1000

n= 0.0135

Reaction equation;

KOH(aq) + KHC8H4O4(aq) --->(KC8H4O4)K(aq) + H2O(l)

1 mole of KHP reacts with 1 mole of KOH

Hence; 0.0135 moles of KHP reacts with 0.0135 moles of KOH

From;

Number of moles = reacting mass/molar mass

Reacting mass = 0.0135 × 204

= 2.75 g of KHP