The half life of oxygen is 2 minutes. What fraction of a sample of 0.15 will remain after 5 half lives?​

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.

See attached image

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Answer:

The given reaction has been the symbolization for replacement reaction. Thus, option D is correct.

The given general reaction has been:

[tex]\rm AB\;+\;CD\;\rightarrow\;AD\;+\;CB[/tex]

In the given reaction, there has presence of compound AB and CD. The reaction between the two results in the formulation of AD and CB. In the reaction, the B has been more attracted by C and forms bond with C by replacing D.

In the same way, A has been more attracted with D and forms bond with D replacing C.

The type of reaction in which the more reactive element replaces the less reactive element has been termed as replacement reaction. Thus, the given reaction has been the symbolization for replacement reaction. Thus, option D is correct.

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Answer:

Exothermic

Explanation:

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

Answer:

Cellulose is an organic compound with the formula , a polysaccharide consisting of a linear chain of several hundred to many thousands of β linked D-glucose units. Cellulose is an important structural component of the primary cell wall of green plants, many forms of algae and the oomycetes.

Explanation:

Answer:

Hope this helped.

Explanation:

Cellulose is a molecule, consisting of hundreds – and sometimes even thousands – of carbon, hydrogen and oxygen atoms. Cellulose is the main substance in the walls of plant cells, helping plants to remain stiff and upright. Humans cannot digest cellulose, but it is important in the diet as fibre.

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:

True

Explanation:

Because atoms with the same number of valence electrons react in similar ways with other elements.

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:

d

Explanation:

collagen is made of proteins

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:

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

1) Both cells have a cell membrane.

2) A nucleus is present in only Eukaryote cells.

3) I'm pretty sure Prokaryote cells are bacteria.

Hope this helps!! :)

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:

The answer is 16

Explanation:

Answer:

21 is the max but with the bread ypu could make 22 if you had 3 more cheese

Answer:

it is Calcium (Ca)

4th period, 2nd group, 2 valence electrons

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:

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:

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

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:

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:

Explanation:

pH = 10.3

[ H] = 10⁻¹⁰°³

= 5 x 10⁻¹¹ M

concentration of CsOH C = 3 M

pKa of carbonate = 6.35

Ka = 10⁻⁶°³⁵ = 4.46 x 10⁻⁷

Buffer capacity = 2.303 x C x Ka x [ H⁺] / ( Ka + [ H⁺]² )²

= 2.303 x 3 x 4.46 x 10⁻⁷  x 5 x 10⁻¹¹ / ( 4.46 x 10⁻⁷  + 25 x 10⁻²² )²

= 154 x 10⁻¹⁸ / 19.9 x 10⁻¹⁴

= 7.74 x 10⁻⁴ .

Answer:

1.24 L of H₂ at STP .

Explanation:

2Al(s)    +6HCl(aq)    →    2AlCl₃(aq)   +    3H₂(g)

2 moles                                                  3 x 22.4 L

2 x 27 g of Al reacts to give  3 x 22.4 L of H₂ at STP .

1 g of Al will react to give 3 x 22.4 / ( 2 x 27 ) L of H₂ at STP .

= 1.24 L of H₂ at STP .

The volume of hydrogen produced by 1 grams of Al has been 1.24 L.

The balanced chemical reaction has been given as:

[tex]\rm 2\;Al\;+\;6\;HCl\;\rightarrow\;2\;AlCl_3\;+\;3\;H_2[/tex]

From the equation, 2 moles of Aluminum gives 3 moles of Hydrogen

The mass of the compound from moles can be given as:

Mass = moles × molecular mass

Mass of 2 moles Al = 2 × 27 g

Mass of 2  moles Al = 54 g

Mass of 3 moles hydrogen = 3 × 2 g

Mass of 3 moles hydrogen = 6 g

From the equation,

54 g aluminum gives = 6 grams hydrogen

[tex]\rm 1\;gram\;aluminum\;=\;\dfrac{6}{54}\;\times\;1[/tex]

1 gram Aluminum = 0.11 grams hydrogen

The mass of hydrogen produced by 1 gram Al has been 0.11g. The moles equivalent to 0.11g hydrogen has been given as:

Mass = moles × molecular mass

0.11 g = moles × 2 g/mol

Mole of hydrogen = 0.055 mol

The moles of hydrogen produced by 1 gram of Al has been 0.055 mol.

According to the ideal gas equation, any gas at STP has 1 mole equivalent to 22.4 L. So,

1 mol = 22.4 L

0.055 mol = 0.055 × 22.4 L

0.055 mol = 1.244 L.

The volume of hydrogen produced by 1 grams of Al has been 1.24 L.

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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:

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

It is critical to watch or observe what is happening with every step of your experiment