Which of the following is NOT an example of deposition?
sand dune
river delta
cave crystal
valley
This science btw

Answer:

Explanation:

Significant figure implies number of digits that are to be considered. Some rules are required to be considered when writing a given expression to an expected significant figures.

So that:

1) 0.00004050 is 4 significant figures

2) 54.7000 is 6 significant figures

3) 1,000.09 is 6 significant figures

4) 0.039 is 2 significant figures

Answer:

d

Explanation:

collagen is made of proteins

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

Answer:

The sun's light is blocked by the moon.

Explanation:

During the eclipse, the moon rotates right in front of the sun, that's why the eclipse is so rare and only happens every four(?) years/

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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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:

[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!

Dull black surface are good absorbers of radiation.

The process by which energy in the form of electromagnetic radiation is emitted by a heated surface and travels directly to its absorption point at the speed of light.

The rate at which a body absorbs or radiates thermal radiation relies upon the nature of the surface as well.

According to the Kirchhoff's radiation law, the objects, which are good emitters are also good absorbers.

A blackened surface is an excellent absorber as well as an excellent emitter.

A black body refers to a body, which is regarded as a perfect radiator and absorber of electromagnetic radiations, a dull black surface is a good absorber of heat radiation.

Thus, dull black surface is the best surface for absorbing heat radiation.

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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:

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:

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:

true. cathode rays are deflected

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:

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:

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:

Percent error

Explanation:

Percent error is defined as the percentage of the difference between the measured( accurate value) and the known value of a substance.

It is usually calculated by:

Measured value — Known value/ known value * 100%

Percent error is the calculated value that is used to evaluate accuracy for a set of data

Answer:

2016 Km.

Explanation:

The following data were obtained from the question:

Speed (S) = 1.12×10⁴ m/s

Time (t) = 3 mins

Distance (d) =.?

Next, we shall convert 3 mins to seconds. This can be obtained as follow:

1 min = 60 s

Therefore,

3 mins = 3 min × 60 s / 1 min

3 mins = 180 s

Next, we shall determine the distance travelled by the rocket. This is illustrated below:

Speed (S) = 1.12×10⁴ m/s

Time (t) = 180 s

Distance (d) =.?

Speed (S) = distance (d) /time (t)

S = d/t

1.12×10⁴ = d/ 180

Cross multiply

d = 1.12×10⁴ × 180

d = 2016000 m

Finally, we shall convert 2016000 m to Km. This can be obtained as shown below:

1000 m = 1 Km

Therefore,

2016000 m = 2016000 m × 1 Km / 1000 m

2016000 m = 2016 Km

Therefore, the rocket will travel 2016 Km in 3 mins

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:

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:

5;11

Explanation:

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