what following oxide react with both acids and bases to form salts.​

Answers

Answer 1

Answer:

mainly metal oxide use to react with both acid and bases to form salts such as zinc, aluminum etc.


Related Questions

Classify each of the following as either macroscopic, microscopic or particulate:
a. a red blood cell.
b. a sugar molecule.
c. baking powder.

Answers

Answer:

Classify each of the following as either macroscopic, microscopic or particulate:

a. a red blood cell.

b. a sugar molecule.

c. baking powder.

Explanation:

a. A red blood cell is a microscopic particle.

It can be viewed under a microscope.

b. A sugar molecule is also a microscopic substance.

It can be viewed under a microscope.

c. Baking powder is macroscopic substance.

Calculate the moment of inertia of a CH³⁵CL₃ molecule around a rotational axis that contains the C-H bond. The C-Cl bond length is 177pm and the HCCl angle is 107⁰f​

Answers

Answer:

The correct answer is "[tex]4.991\times 10^{-45} \ kg.m^2[/tex]".

Explanation:

According to the question,

[tex]R_{C-Cl} = 177 \ pm[/tex]

or,

         [tex]=1.77\times 10^{-10} \ m[/tex]

[tex]\alpha = 107^{\circ}[/tex]

[tex]m_{Cl}=34.97 \ m.u[/tex]

or,

      [tex]=34.97\times 1.66\times 10^{-27}[/tex]

      [tex]=5.807\times 10^{-26} \ kg[/tex]

The moment of inertia around the rotational axis will be:

⇒  [tex]I=3\times m_{Cl}\times (R_{C-Cl})^2 \ Sin^2 \alpha[/tex]

By putting the values, we get

       [tex]=3\times 5.807\times 10^{-26}\times (1.77\times 10^{-10})^2 \ Sin^2 (107)[/tex]

       [tex]=3\times 5.807\times 10^{-26}\times (1.77\times 10^{-10})^2\times 0.91452[/tex]

       [tex]=4.991\times 10^{-45} \ kg.m^2[/tex]

The doctor has ordered Claforan 1 g in 100 ml D5W to run IV piggyback for 30 minutes twice daily. The pharmacy sends Claforn 2 g in a powdered form, which when reconstituted has a concentration of 180 mg Claforan per ml. How much Claforn will you add to the bag of D5W

Answers

Answer:

0.111 g

Explanation:

1 g = 1000 mg

Doctor ordered the following concentration of Claforan:

C = 1 g/100 mL x 1000 mg/1 g = 10 mg/mL

If we add 2 g iof Claforan, we obtain:

2 g Claforn ---- 180 mg/mL Claforan

To reach a concentration equal to C (10 mg/mL), we need:

10 mg/mL Claforan x 2 g Claforn/(180 mg/mL Claforan) = 0.111 g Claforn

Therefore, we have to add 0,111 g (111 mg) of Claforn to the bag of 100 ml D5W to obtain the ordered concentration of 10 mg/mL Claforan.  

How many molecules in each sample?

64.7 g N2
83 g CCl4
19 g C6H12O6

Answers

Answer:

1.39x10²⁴ molecules N₂.25x10²³ molecules CCl₄6.38x10²² molecules C₆H₁₂O₆

Explanation:

First we convert the given masses into moles, using the compounds' respective molar mass:

64.7 g N₂ ÷ 28 g/mol = 2.31 mol N₂83 g CCl₄ ÷ 153.82 g/mol = 0.540 mol CCl₄19 g C₆H₁₂O₆ ÷ 180 g/mol = 0.106 mol C₆H₁₂O₆

Then we multiply each amount by Avogadro's number, to calculate the number of molecules:

2.31 mol N₂ * 6.023x10²³ molecules/mol = 1.39x10²⁴ molecules0.540 mol CCl₄ * 6.023x10²³ molecules/mol = 3.25x10²³ molecules0.106 mol C₆H₁₂O₆ * 6.023x10²³ molecules/mol = 6.38x10²² molecules

14 protons,14 electrons and 14 neutrons

Answers

The answer is silicon.

the answer is silicon!!

The data shows the number of years that 30 employees worked for an insurance company before retirement. is the population mean for the number of years worked, and % of the employees worked for the company for at least 10 years. (Round off your answers to the nearest integer.)

Answers

Answer:

14

73%

Explanation:

The mean Number of years worked :

. (sum of service years) / employees in the

(8+13+15+3+13+28+4+12+4+26+29+3+10+3+17+13+15+15+23+13+12+1+14+14+17+16+7+27+18+24) /

(417 / 30)

= 13.9 years

= 14 years

The percentage of employees who have worked for atleast 10 years :

Number of employees with service years ≥ 10 years = 22 employees

Total number of employees

Percentage (%) = (22 / 30= * 100% = 0.7333 * 100% = 73.33% = 73%

Match the description with the type of precipitation being described.
1. Its formation requires very strong updrafts
2. Its formation requires falling through a layer of above freezing air
3. Precipitation from cumuliform clouds is typically of this nature
4. Precipitation from stratus clouds is typically of this nature
Options:
a. Hail
b. Drizzle
c. Shower
d. Freezing Rain

Answers

Answer:

1. Its formation requires very strong updrafts = a. Hail

2. Its formation requires falling through a layer of above-freezing air = d. Freezing Rain

3. Precipitation from cumuliform clouds is typically of this nature = c. Shower

4. Precipitation from stratus clouds is typically of this nature = Drizzle

Explanation:

Hail formation requires very strong updrafts, these updrafts are the upward moving air created in a thunderstorm. This period of noticeable thunderstorms creates hails.

Freezing rain requires the presence of warm air, it requires falling through a layer of above-freezing air to the colder air below to produce an ice coating on anything it drops on.

Showers are produced by cumuliform clouds which look like cotton balls. Since cumuliform clouds precipitate too, these clouds can have fluctuating rain in a day in the form of showers.

Drizzle which raises low visibility is considered a type of liquid precipitation since it also falls from a cloud. Drizzle which is obviously smaller in diameter when compared to that of raindrops, however, is common with stratus clouds.

(S)-Pentan-2-ol was treated sequentially with methanesulfonyl chloride (CH3SO2Cl) and then potassium iodide. What is the final product that forms

Answers

Answer:

(S)-Pentan-2-ol was treated sequentially with methanesulfonyl chloride (CH3SO2Cl) and then potassium iodide. What is the final product that forms

Explanation:

Alcohols are poor leaving groups.

To make -OH group a better-leaving group, it should be treated with sulfonyl chlorides.

Then, methane sulfonyl group makes will be substituted on the -OH group and forms sulfonyl esters and makes it a better leaving group.

After that treating with KI proceeds through nucleophilic bimolecular substitution and the final product formed is shown below:.

In a closed system, If a gas is transported to a container with double the volume of the previous container, the gas was held in, what is the gases' new volume?

The volume of the gas is fixed and will not change.
The volume of the gas will be half the original volume.
The volume of the gas will be the original volume squared.
The volume of the gas will be double the original volume.

Answers

Answer:

The volume of the gas is fixed and will not change.

Explanation:

The volume of the gas will not change because there is no change in temperature. Temperature increases the volume of gases enclosed in a container.

Calculate the average atomic mass for X

Answers

Answer:

39.0229 amu

Explanation:

Hello there!

In this case, according to given information, the idea here is to multiply the percent abundance by the mass number of each isotope and then add them all together as shown below:

[tex]=0.0967*38+0.7868*39+0.1134*40+0.0031*41\\\\=3.6746+30.6852+4.536+0.1271\\\\=39.0229amu[/tex]

Regards!

What does the term spontaneous mean in chemical reactions?
A. Producing heat as a product
B. Occurring without added energy
C. Occurring only at high temperatures
D. Occurring in an aqueous solution

Answers

Answer:

B

Explanation:

Spontaneous in chemical reactions means without any external input.

Occurring without added energy. Hence, option B is correct.

What is a spontaneous reaction?

A spontaneous reaction is a reaction that supports the formation of products under the conditions under which the reaction is happening.

Spontaneous Reaction- a reaction that favours the formation of products at the conditions under which the reaction is occurring.

A non-spontaneous reaction can be made spontaneous if it is inside a controlled environment, this is what happens in nuclear power plants that create atomic fusion and fission in chambers that are controlled to control different particles to create nuclear active rays.

Hence, option B is correct.

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What is different between margerine and butter in term of organic chemistry

Answers

Answer:

The most important difference between the two is that butter is derived from dairy and is rich in saturated fats, whereas margarine is made from plant oils. ... If the margarine contains partially hydrogenated oils, it will contain trans fat, even if the label claims that it has 0 g.

Explanation:

(⌒_⌒;)

Perform the following operation and express the answer in scientific notation.
7.296×10² ÷ 9.6×10^-9

Answers

Answer:

7.6×10¹⁰

Explanation:

7.296×10²÷9.6×10⁻⁹

To solve such problem,

We group the whole number ans solved seperately and also group the indices and solve the seperately

Step1 : 7.296/9.6 = 0.76

Step 2: applying the law of indices,

10²÷10⁻⁹ = 10⁽²⁺⁹⁾ = 10¹¹

Therefore,

7.296×10²÷9.6×10⁻⁹ = 0.76×10¹¹ = 7.6×10¹⁰

According to the Arrhenius equation, changing which factors will affect the
rate constant?
A. Temperature and the ideal gas constant
B. The activation energy and the constant A
C. The constant A and the temperature
D. Temperature and activation energy

Answers

Answer:

e−(Ea/RT): the fraction of the molecules present in a gas which have energies equal to or in excess of activation energy at a particular temperature

Answer:

D. Temperature and activation energy is the correct answer

Explanation:

^_^

Please help me ASAP I’ll mark Brainly

Answers

Answer:

1. Vacuole

2. chloroplast

3. Nucleus

4. Plasma membrane - cell membrane

5. Vacuole (same as #1 ?) could be vesicle

Explanation:

Question 16(Multiple Choice Worth 5 points)

(04.01 LC) Which statement is true about the total mass of the reactants during a chemical change?

O It is destroyed during chemical reaction.
O It is less than the total mass of the products. O It is equal to the total mass of the products.
O It is greater than the total mass of the products.​

Answers

Answer:

It is equal to the total mass of the products.

Explanation:

Hope this helps :)

One main difference between the heating of gases on the one hand and solids or liquids on the other is that ___________________. One main difference between the heating of gases on the one hand and solids or liquids on the other is that ___________________. heating of gases depends not only on the temperature difference, but also on the process as well as the amount of gas present. heating of gases depends on temperature difference as well as the amount of gas present. specific heat is not defined for gases. heat cannot be exchanged with gases.

Answers

Answer:

heating of gases depends not only on the temperature difference, but also on the process as well as the amount of gas present.

Explanation:

The work done when a gas is heated does not only depends on the initial and final states of the gas but also on the process used to achieve the change of state of the gas.

Several processes can be applied in changing the state of a gas such as; adiabatic process, isobaric process, isochoric process and isothermal process.

Hence, the heating of a gas, depends not only on the temperature difference, as well as the amount of gas present according to the ideal gas laws but also on the process used to achieve the change of state.

1 or 2 topics or two lessons should be explained in an illustrated childrens book minimum of 10 pages must have 3 or more sentences

Answers

Answer:

Yes because same topic are long

Water has a density of 1.00 g/mL. If you put an object that has a density of 0.79 g/mL into water, it will sink to the
bottom.
ANSWER:

True

False

Answers

Answer:

False

Explanation:

An object with a density greater than 1.00g/mL (greater than the density of water) will sink. An object with a density less than the density of water, will float.

If the water has a density of 1.00 g/mL. If you put an object that has a density of 0.79 g/mL into water, it will sink to the bottom, this statement is false.

What is density?

The density of an actual content is its mass per unit volume. The most common symbol for density is d, but the Latin letter D can also be used.

Three of an object's most fundamental properties are mass, volume, and density. Mass describes how heavy something is, volume describes its size, and density is defined as mass divided by volume.

The density of something is a measure of how heavy it is in relation to its size. When an artifact is more dense than water, it plunges; when an object is less dense than water, it floats.

Density is a property of a substance that is independent of the amount of substance.

As in the given scenario, water is having density 1 g/mL and object in having density less then it so it will float on water.

Thus, the given statement is false as the material will not sink, rather it will float on water.

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I have an unknown volume of gas held at a temperature of 115 K in a container with a pressure of 60atm. If by increasing the temperature to 225 K and decreasing the pressure to 30. atm causes the volume of the gas to be 29 liters, how many liters of gas did I start with?
SHOW YOUR WORK

Answers

Explanation:

here is the answer to your question.

Methyl orange can change color by transitioning from one chromophore to another. When added to a clear solution and the solution turns red, it is determined to be a(n) __________ in its __________ stable form.

Answers

Answer:

acidic titration in its stable form

Explanation:

Methyl orange can change its color in titration solution. The yellow color is towards alkaline solution and red color is towards acidic solution. The Ph value of solution will change during this chemical process.

The mass of a single tantalum atom is 3.01×10-22 grams. How many tantalum atoms would there be in 37.1 milligrams of tantalum?

Answers

Answer: There are [tex]1.23 \times 10^{22}[/tex] atoms present in 37.1 mg of tantalum.

Explanation:

Given: Mass of single tantalum atom = [tex]3.01 \times 10^{-22} g[/tex]

Mass of tantalum atoms = 37.1 mg (1 mg = 0.001 g) = 0.0371 g

Therefore, number of tantalum atoms present in 0.0371 grams is calculated as follows.

[tex]No. of atoms = \frac{0.0371 g}{3.01 \times 10^{-22}}\\= 1.23 \times 10^{20}[/tex]

Thus, we can conclude that there are [tex]1.23 \times 10^{22}[/tex] atoms present in 37.1 mg of tantalum.

There are [tex]1.23\times 10^{20}[/tex] atoms of tantalum in 37.1 mg of tantalum.

Explanation:

Given:

Mass of single atom of tantalum =[tex]3.01\times 10^{-22} g[/tex]

To find:

The number of atoms of tantalum in 37.1 milligrams.

Solution:

Mass of tantalum = 37.1 mg

[tex]1 mg = 0.001 g\\37.1 mg=37.1\times 0.001 g=0.0371 g[/tex]

The number of atoms in 0.0371 grams of tantalum = N

Mass of a single atom of tantalum = [tex]3.01\times 10^{-22} g[/tex]

Then a mass of N atoms of tantalum will be:

[tex]0.0371 g=N\times 3.01\times 10^{-22} g\\N=\frac{0.0371 g}{ 3.01\times 10^{-22} g}\\=1.23\times 10^{20}[/tex]

There are [tex]1.23\times 10^{20}[/tex] atoms of tantalum in 37.1 mg of tantalum.

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a sample of copper was heated at 275.1 C and placed into 272 g of water at 21.0 C. The temperature of the water rose at 29.7 C. How many grams of copper were in the sample

Answers

Answer:

104.8 g

Explanation:

From the question given above, the following data were obtained:

Initial temperature of copper (T꜀) = 275.1 °C

Mass of water (Mᵥᵥ) = 272 g

Initial temperature of water (Tᵥᵥ) = 21 °C

Equilibrium temperature (Tₑ) = 29.7 °C

Mass of copper (M꜀) =?

NOTE:

Specific heat capacity of copper (C꜀) = 0.385 J/gºC

Specific heat capacity of water (Cᵥᵥ) = 4.184 J/gºC

Finally, we shall determine the mass of the copper in the sample. This can be obtained as follow:

Heat loss by copper = Heat gained by water

M꜀C꜀(T꜀ – Tₑ) = MᵥᵥCᵥᵥ(Tₑ – Tᵥᵥ)

M꜀ × 0.385 (275.1 – 29.7) = 272 × 4.184(29.7 – 21)

M꜀ × 0.385 × 245.4 = 1138.048 × 8.7

M꜀ × 94.479 = 9901.0176

Divide both side by 94.479

M꜀ = 9901.0176 / 94.479

M꜀ = 104.8 g

Thus, the mass of the copper in the sample is 104.8 g

A
(c) 2 C(s) + MnO2(s)
Mn(s) + 2 CO(g)
O combination reaction
O decomposition reaction
O combustion reaction
O single-displacement reaction

Answers

Answer: The reaction, [tex]2C(s) + MnO_{2}(s) \rightarrow Mn(s) + 2CO(g)[/tex] is a single-displacement reaction.

Explanation:

A chemical reaction in which one element of a compound is replaced by another element participating in the reaction.

For example, [tex]2C(s) + MnO_{2}(s) \rightarrow Mn(s) + 2CO(g)[/tex]

Here, the element manganese is replaced by carbon atom. As only one element gets replaced so, it is a single-displacement reaction.

Thus, we can conclude that [tex]2C(s) + MnO_{2}(s) \rightarrow Mn(s) + 2CO(g)[/tex] is a single-displacement reaction.

Na2CO3 reacts with dil.HCl to produce NaCl, H2O and CO2. If 21.2 g of pure Na2CO3 are added in a solution containing 21.9g HCl , a. Find the limiting reagent. (2) b. Calculate the number of moles of excess reagent left over.(2) c. Calculate the number of molecules of H2O formed.(1) d. Calculate volume of CO2 gas produced at 270C and 760mm Hg pressure.(2) e. Write significance of limiting reagent​

Answers

Answer:

See explanation

Explanation:

Equation of the reaction;

Na2CO3(aq) + 2HCl(aq) -------> 2NaCl(aq) + H2O(l) + CO2(g)

Number of moles of Na2CO3 = 21.2g/106g/mol = 0.2 moles Na2CO3

Number of moles of HCl = 21.9g/36.5g/mol = 0.6 moles of HCl

1 mole of Na2CO3 reacts with 2 moles of HCl

0.2 moles of Na2CO3 reacts with 0.2 × 2/1 = 0.4 moles of HCl

Hence Na2CO3 is the limiting reactant

Since there is 0.6 moles of HCl present, the number of moles of excess reagent=

0.6 moles - 0.4 moles = 0.2 moles of HCl

1 mole of Na2CO3 forms 1 mole of water

0.2 moles of Na2CO3 forms 0.2 moles of water

Number of molecules of water formed = 0.2 moles × 6.02 × 10^23 = 1.2 × 10^23 molecules of water

1 mole of Na2CO3 yields 1 mole of CO2

0.2 moles of Na2CO3 yields 0.2 moles of CO2

1 mole of CO2 occupies 22.4 L

0.2 moles of CO2 occupies 0.2 × 22.4 = 4.48 L at STP

Hence;

V1=4.48 L

T1 = 273 K

P1= 760 mmHg

T2 = 27°C + 273 = 300 K

P2 = 760 mmHg

V2 =

P1V1/T1 = P2V2/T2

P1V1T2 = P2V2T1

V2 = P1V1T2/P2T1

V2 = 760 × 4.48 × 300/760 × 273

V2= 4.9 L

The limiting reactant is the reactant that determines the amount of product formed in a reaction. When the limiting reactant is exhausted, the reaction stops.

When electrons in a molecule are not found between a pair of atoms but move throughout the molecule, this is called Group of answer choices

Answers

Answer:

delocalised electrons

Explanation:

they are called delocalised electrons because that can move freely in the molecule

A buffer is prepared containing 0.75 M NH3 and 0.20 M NH4 . Calculate the pH of the buffer using the Kb for NH3. g

Answers

Answer:

pH=8.676

Explanation:

Given:

0.75 M [tex]NH_{3}[/tex]

0.20 M [tex]NH_{4}[/tex]

The objective is to calculate the pH of the buffer using the kb for [tex]NH_3[/tex]

Formula used:

[tex]pOH=pka+log\frac{[salt]}{[base]}\\[/tex]

pH=14-pOH

Solution:

On substituting salt=0.75 and base=0.20 in the formula

[tex]pOH=-log(1.77*10^{-5})+log\frac{0.75}{0.20}\\ =4.75+0.5740\\ =5.324[/tex]

pH=14-pOH

On substituting the pOH value in the above expression,

pH=14-5.324

Therefore,

pH=8.676

A student dropped a piece of nickel metal into a solution of HCl(aq). He observed the formation of gas bubbles and collected the gas into another test tube. The student performed a splint test and observed that the splint flared up when he placed the splint into the test tube of the gas. What can be said about the results of this students experiment?

a. The student performed the splint test incorrectly. He should of observed a popping sound when the splint was placed in the test tube.
b. The experiment was performed incorrectly. Nickel doesn't react with HCl. Therefore, the student picked up the wrong metal when conducting the experiment.
c. The student completed the experiment correctly and there were no errors in the experiment.
d. The student performed the splint test incorrectly. He should of observed the flame being extinguished when the splint was placed in the test tube.

Answers

Answer:

a. The student performed the splint test incorrectly. He should of observed a popping sound when the splint was placed in the test tube.

Explanation:

It is given that a student performed an experiment where he dropped a nickel metal in to HCl solution. He observed the reaction and performed a splint test in the test tube that is filled with a gas which is formed while Nickle is dropped into the solution of HCl.

But the experiment that the student performed was incorrect. He must have observed the popping sound when the splint was placed in the test tube.

When the splint was added to the gas splint flared up. The hydrogen gas pops out when exposed to the flame.

[tex]$Ni + HCl(aq) = NiCl + H_2$[/tex]

Thus the correct option is (a).

How much energy is required to melt 2 kg of gold? Use the table below and this equation.
a. 125.6 kJ
b. 1729 kJ
c. 10.4 kJ
d. 3440kJ

Answers

The equation for the energy required to melt 2 kg of gold is 3440 kJ.

What is energy?

Energy is the ability to do work or cause change. It is an essential part of everyday life and is present in many forms, such as thermal energy, electrical energy, chemical energy, and mechanical energy. Energy can be converted from one form to another in order to do work.


The equation for calculating the energy required to melt a certain mass of material is Q = m x Lf, where Q is the energy required (in joules), m is the mass of the material (in kilograms), and Lf is the latent heat of fusion (in joules per kilogram).
Using the table below, we can see that the latent heat of fusion for gold is 1760 kJ/kg. Therefore, the equation for the energy required to melt 2 kg of gold is: Q = 2 kg x 1760 kJ/kg = 3440 kJ.

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what is the mass of insoluble calcium phosphate produced from .555 grams of calcium chloride​

Answers

Answer:

0.518 g

Explanation:

Step 1: Write the balanced equation

3 CaCl₂ + 2 H₃PO₄ ⇒ Ca₃(PO₄)₂ + 6 HCl

Step 2: Calculate the moles corresponding to 0.555 g of CaCl₂

The molar mass of CaCl₂ is 110.98 g/mol.

0.555 g × 1 mol/110.98 g = 5.00 × 10⁻³ mol

Step 3: Calculate the moles of Ca₃(PO₄)₂ produced

5.00 × 10⁻³ mol CaCl₂ × 1 mol Ca₃(PO₄)₂/3 mol CaCl₂ = 1.67 × 10⁻³ mol Ca₃(PO₄)₂

Step 4: Calculate the mass corresponding to 1.67 × 10⁻³ moles of Ca₃(PO₄)₂

The molar mass of Ca₃(PO₄)₂ is 310.18 g/mol.

1.67 × 10⁻³ mol × 310.18 g/mol = 0.518 g

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