Which main type of sedimentary rock forms from solutions?

chemical
clastic
organic
shale

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/

Complete Question

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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:

no d

Explanation:

it has a permanent dipole

CH₃F compound contains polar covalent bonds in its structure, hence option D is correct.

This happens when the electronegativity of the relevant atoms differs significantly.

The electronegativities of the carbon and fluorine atoms in CH3F are dissimilar. Because fluorine is more electronegative than carbon, it attracts the shared electrons to it, giving the fluorine atom a partial negative charge and the carbon atom a partial positive charge.

A polar covalent bond is created as a result of this unequal charge distribution. CH₃F compound contains polar covalent bonds in its structure.

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Answer:

try dividing the grams and the volume and see what number matches up with your answer

Answer:

Explanation:

a) O₂(g) + 2H₂(g) = 2H₂O

b) H₂O(l) = H₂O(s)

c) NH₃(g) + HCl(g) =  NH₄Cl(g)

d) 2NH₄NO₃(s) =  2N₂(g) + O₂(g) + 4H₂O(g)

e) N₂O₄(g) =  2NO₂(g)

ΔS is positive when there is increase in disorderliness. It happens when there is increase in volume .

Increase is volume is maximus in the following reaction.

d) 2NH₄NO₃(s) =  2N₂(g) + O₂(g) + 4H₂O(g)

in this reaction solid NH₄NO₃ is changed to 7 x 22.4 L of gases so there is maximum increase in volume . Hence maximum increase in entropy . Hence ΔS is most positive .

Answer:

Reductions in available food and shelter

Explanation:

To solve this we must be knowing each and every concept related to population. Therefore, organisms in over-populated areas do not survive well due to reduction in available food and shelter.

The group of people from whom a quantitative sample is gathered for a research is referred to as a population. Therefore, a population is any collection of people who have anything in common.

A sample is a representative group of a population chosen at random. This is a smaller group that was selected from the population and possesses all of the population's traits. The observations and inferences drawn from the sample data then applied to the entire population.

Therefore, organisms in over-populated areas do not survive well due to reduction in available food and shelter.

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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:

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:

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:

d

Explanation:

collagen is made of proteins

Answer:

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:

The Group 1 elements in the periodic table are known as the alkali metals. They include lithium, sodium and potassium, which all react vigorously with water to produce an alkaline solution.

Explanation:

Answer:

f and a. That is reagents f[HNO3, H2SO4] is used first then followed by reagent a[Br2, FeBr3]

Explanation:

So, in the production of m-bromonitrobenzene[3-Nitro-1-bromobenzene/1-Bromo-3-nitrobenzene] from benzene from the reagents provided by using the fastest reagents that is to say the reagent with the minimum number of steps, we are going to make use of reagents in option 'f' followed by reagents in  option 'a'.

STEP ONE: The first step in the production of m-bromonitrobenzene[3-Nitro-1-bromobenzene/1-Bromo-3-nitrobenzene] from benzene to give the minimum number of steps from the reagents given is nitration by using the reagents in option 'f'' that is  HNO3, H2SO4 to give nitrobenzene. The equation for the chemical reaction is given below as:

C₆H₆ +  HNO₃, H₂SO₄ -------------------------------------------------------> C6H5NO2.

It is the N⁺O₂ that is been used in the reaction mechanism [from the reaction:  HNO₃ + H₂SO₄ < -----------------> N⁺O₂ + HSO₄⁻ + H₂O].

STEP TWO: The next thing to do is to react the nitrobenzene got from the nitration of benzene in step one above with the reagents in option 'a'[Br2, FeBr3]. This step is known as the bromination of nitrobenzene.

C₆H₅NO₂ + Br₂, FeBr3 ------------------------------------------------> C₆H₄BrNO₂.

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:

true. cathode rays are deflected

Answer:

The plant uses carbon dioxide and water to make an entirely new chemical hence chemical change.

Explanation:

Plants create their own 'food', they turn carbon dioxide water and sunlight to create starch and glucose..

Answer:

The properties of carbon dioxide and oxygen are different.

Explanation:

im taking the exam right now and Im pretty sure it's D The properties of carbon dioxide and oxygen are different.

Answer:

1.61 x 10²⁰ molecules

Explanation:

Given parameters:

Mass of the hydrogen gas  = 8.01 x 10⁻⁴g

Unknown

Number of molecules of NH₃ = ?

Solution:

Given reaction:

                     N₂  +  3H₂   →   2NH₃

We have to solve from the known to the unknown

             find the number of moles of the specie;

    number of moles = [tex]\frac{mass}{molar mass}[/tex]

     Molar mass of H₂ = 2(1) = 2g/mol

   Number of moles  = [tex]\frac{0.0008}{2}[/tex]   = 0.0004mol

      0.0004 moles of H₂ will produce [tex]\frac{0.0004 x 2}{3}[/tex]   = 0.00027moles of NH₃

   1 mole of a substance contains 6.02 x 10²³ molecules

     0.00027mole of NH₃ will contain 0.00027 x 6.02 x 10²³

                               = 1.61 x 10²⁰ molecules

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:

There are three basic types of chemical formula, the empirical formula, the molecular formula, and the structural formula.

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:

NH₂O

Explanation:

Given compound:

       N₄H₈O₄

Unknown:

The empirical formula of the compound  = ?

Solution:

The empirical formula of a compound is its simplest formula. It expresses the composition of a the compound in the simplest whole ratio of atoms of the different elements present in the compound.

 For the given compound:

                                  N₄H₈O₄

  Number of moles of N = 4

                                      H = 8

                                      O = 4

 the highest common factor is 4 and we simply divide through by this number;

                                      N = 1

                                      H = 2

                                      O = 1

So, the empirical formula of compound is NH₂O

Answer:

V₂ = 87.24 L

Explanation:

Charle's law states that at constant pressure, the volume of the gas is directly proportional to the temperature. Its mathematical form is given by :

[tex]\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}[/tex]

We have, V₁ = 75, T₁ = 245 K, T₂ = 285, V₂ = ?

Putting all the values, we get :

[tex]V_2=\dfrac{V_1T_2}{T_1}\\\\V_2=\dfrac{75\times 285}{245}\\\\V_2=87.24\ L[/tex]

So, the new volume is 87.24 L.