How many oxygen atoms are represented by the formula Fe(CIO4)3? iron(III) chlorate
3
16
12
4

Answer:

the celery would change colour!

Explanation:

Answer:

an energy storing process i think!

Answer:

d

Explanation:

mha

Answer:

100 g of water has the highest number of moles

Explanation:

Recall that the number of moles is obtained as given mass/formula weight

For HCl;

number of moles = 100g/36.5g/mol = 2.7 moles

For H2O;

number of moles = 100g/18g/mol = 5.5 moles

For MgCO3

number of moles = 100g/84.3 g/mol = 1.2 moles

For AlCl3

number of moles = 100g/133.3g/mol = 0.75 moles

For NaCl

number of moles = 100g/58.4 g/mol = 1.7 moles

The enthalpy change of combustion of methane : -892 kJ/mol

The change in enthalpy in the formation of 1 mole of the elements is called enthalpy of formation  

The enthalpy of formation measured in standard conditions (25 ° C, 1 atm) is called the standard enthalpy of formation (ΔHf °)  

Determination of the enthalpy of formation of a reaction can be through a calorimetric experiment, based on the principle of Hess's Law, enthalpy of formation table, or from bond energy data  

The value of ° H ° can be calculated from the change in enthalpy of standard formation:  

∆H ° rxn = ∑n ∆Hf ° (product) - ∑n ∆Hf ° (reactants)  

Reaction for combustion of Methane (CH₄) :

CH₄ + 2O₂ → CO₂ + 2H₂O.

The elements in standard conditions are not included in the enthalpy calculations because the enthalpy of those elements under the standard conditions is zero(∆Hf °O₂ =0)

∆Hf ° CO₂ = -394 kj/mol

∆Hf ° H₂O=-286 kj/mol

∆Hf ° CH₄=-74 kJ/mol

The enthalpy change of combustion of methane :

[tex]\tt \Delta H_{rxn}=\Delta H_f~CO_2+2.\Delta H_fH_2O-(\Delta H_fCH_4)\\\\\Delta H_{rxn}=(-394-2.-286)-(-74)\\\\\Delta H_{rxn}=-892~kJ/mol[/tex]

Answer:

-655

Explanation:

ye

Answer:

the white blood cells are key in fighting foreign invaders.

Answer:

C

Explanation:

Because molecules of solids don't move

while molecules of liquids and gases move.

I Hope this helps!!.

Answer:

Can you attach a picture?

Explanation:

Answer:

the answer is applied force

Answer:

1

Explanation:

The full question is shown in the image attached.

If we look at the structure closely, we will discover that there is only one C=C double bond in the molecule.

It is worthy of note that hydrogen (H2) only reduces the C=C double bond. it does not affect any other bond in the molecule.

Hence, only one molecule of H2 is required to reduce one C=C bond present.

Answer:

Al2(SO4)3

Explanation:

Looking at this carefully, we will discover that Al2(SO4)3 is composed of Al^3+ and SO4^2-.

The aluminum and sulphate ions are ionically bonded. However, the oxygen and sulphur in the sulphate ion are covalently bonded.

Hence, Al2(SO4)3 contains both ionic and covalent bond.

The chemical compound which utilizes both ionic and covalent bonding is: E. [tex]Al_2(SO_4)_3[/tex]

An ionic bond refers to a chemical bond that involves a metallic element transferring one or more electrons to a non-metallic element.

On the other hand, a covalent bond can be defined as a chemical bond that typically exists in chemical compounds and involves the sharing of electrons between the atoms of their chemical element.

In this scenario, we can say that the atoms of aluminum and sulfate ion utilizes an ionic bond because the aluminum atom transfers its electrons to the sulfate ion.

However, there exist is a covalent bond between the atoms of oxygen and sulfur to form sulfate ions.

Read more: https://brainly.com/question/24212500

Answer:

D

Explanation:

Problem #1: If you dilute 175 mL of a 1.6 M solution of LiCl to 1.0 L, determine the new concentration of the solution.

Solution:

M1V1 = M2V2

(1.6 mol/L) (175 mL) = (x) (1000 mL)

x = 0.28 M

Note that 1000 mL was used rather than 1.0 L. Remember to keep the volume units consistent.

Problem #2: You need to make 10.0 L of 1.2 M KNO3. What molarity would the potassium nitrate solution need to be if you were to use only 2.5 L of it?

Solution:

M1V1 = M2V2

(x) (2.5 L) = (1.2 mol/L) (10.0 L)

x = 4.8 M

Please note how I use the molarity unit, mol/L, in the calculation rather than the molarity symbol, M.

Problem #3: How many milliliters of 5.0 M copper(II) sulfate solution must be added to 160 mL of water to achieve a 0.30 M copper(II) sulfate solution?

Solution:

M1V1 = M2V2

(5.00 mol/L) (x) = (0.3 mol/L) (160 + x)

5x = 48 + 0.3x

4.7x = 48

x = 10. mL (to two sig figs)

The solution to this problem assumes that the volumes are additive. That's the '160 + x' that is V2.

Problem #4: What volume of 4.50 M HCl can be made by mixing 5.65 M HCl with 250.0 mL of 3.55 M HCl?

Solution:

Here is the first way to solve this problem:

M1V1 + M2V2 = M3V3

(3.55) (0.250) + (5.65) (x) = (4.50) (0.250 + x)

Where x is volume of 5.65 M HCl that is added

(0.250 + x) is total resultant volume

0.8875 + 5.65x = 1.125 + 4.50 x

1.15x = 0.2375

x= 0.2065 L

Total amount of 4.50 M HCl is then (0.250 + 0.2065) = 0.4565 L

Total amount = 456.5 mL

Here is the second way to solve this problem:

Since the amount of 5.65 M added is not asked for, there is no need to solve for it.

M1V1 + M2V2 = M3V3

(3.55) (250) + (5.65) (x − 250) = (4.50) (x)

That way, x is the answer you want, the final volume of the solution, rather than x being the amount of 5.65 M solution that is added.

Problem #5: A 40.0 mL volume of 1.80 M Fe(NO3)3 is mixed with 21.5 mL of 0.808M Fe(NO3)3 solution. Calculate the molar concentration of the final solution.

Answer:

21.46 kg.

Explanation:

Hello!

In this case, when a variety of measurements are said to contribute to a total, we understand we need to add them all up in order to get that total; in such a way, since the given seafood is composed by 15 kg of salmon, 4.7 kg of crab and 1.76 kg of oysters, the total mass turns out:

[tex]m_T=15kg+4.7kg+1.76kg\\\\m_T=21.46kg[/tex]

Best regards!

Answer:

A. Tertiary alcohol

Explanation:

Secondary alcohols are oxidized to ketones - and that's it. For example, if you heat the secondary alcohol propan-2-ol with sodium or potassium dichromate(VI) solution acidified with dilute Sulphuric acid, you get propanone formed. Playing around with the reaction conditions makes no difference whatsoever to the product.

Answer:

e)  400 mg of a radioactive substance with a half-life of 8 hours after decaying for 8 hours

a) 800 mg of a radioactive substance with a half-life of 2 years after decaying for 4 years.

Explanation:

a) 800 mg of a radioactive substance with a half-life of 2 years after decaying for 4 years.

Number of half lives passed = Time elapsed/ half life

Number of half lives passed = 4 year/2 year

Number of half lives passed = 2

at time zero = 800 mg

At first half life = 800 mg/2 = 400 mg

At 2nd half life = 400 mg/2 = 200 mg

b) 1,000 mg of a radioactive substance with a half-life of 5 years after decaying for 20 years

Number of half lives passed = Time elapsed/ half life

Number of half lives passed = 20 year/ 5 year

Number of half lives passed = 4

at time zero = 1000 mg

At first half life = 1000 mg/2 = 500 mg

At 2nd half life = 500 mg/2 = 250 mg

At 3rd half life = 250 mg/2 =  125 mg

At 4th half life = 125 mg/2 = 62.5 mg

c) 300 mg of a radioactive substance with a half-life of 1 year after decaying for 1 year

Number of half lives passed = Time elapsed/ half life

Number of half lives passed = 1 year/ 1 year

Number of half lives passed = 1

at time zero = 300 mg

At first half life = 300 mg/2 = 150 mg

d) 600 mg of a radioactive substance with a half-life of 6 hours after decaying for 18 hours

Number of half lives passed = Time elapsed/ half life

Number of half lives passed = 18 hours / 6 hours

Number of half lives passed = 3

at time zero = 600 mg

At first half life = 600 mg/2 = 300mg

At 2nd half life = 300 mg/2 = 150 mg

At 3rd half life = 150 mg/2 =  75 mg

e)  400 mg of a radioactive substance with a half-life of 8 hours after decaying for 8 hours

Number of half lives passed = Time elapsed/ half life

Number of half lives passed = 8 hours / 8 hours

Number of half lives passed = 1

at time zero = 400 mg

At first half life = 400 mg/2 = 200 mg

Answer:I think it's in the same column as a sodium on the periodic table because chemical properties are based on their electron structure and the columns called groups are elements with the same shell structure which increases in mass as you go down the column, therefore the elements which are similar to sodium must be in the same group.

Explanation:

Answer:

A carbon atom bound to three atoms (two single bonds, one double bond) is sp2 hybridized and forms a flat trigonal or triangular arrangement with 120° angles between bonds. Notice that acetic acid contains one sp2 carbon atom and one sp3 carbon atom.

Answer:

b no

Explanation:

because it is decomposing into two elements

Answer:

2.5471951319999997

Explanation:

Answer:

C4H9OH + 6O2 → 4CO2 + 5H2O

Explanation:

Answer:

When you fill in 2 reactants, you will have 6 N atoms but on the side of the product there are only 2 N atoms, so the equation cannot be balanced and for the equation to be equal, you need to fill 3 into the part N2 to have 6 N atoms

Explanation:

The given reaction unbalanced because of the number of nitrogen is unequal in both side. Three molecules of N₂ should be there in the product side of this decomposition reaction.Thus option C is correct.

What is decomposition reaction?

A decomposition reaction is the one in which a compound decomposes into its constituent compounds or molecules.It is mostly exothermic thus, release heat energy.

In balanced reaction, each elements have to be equal in  number in both reactant and product side. The imbalanced reaction can be balanced multiplying with simple intergers.

In the given reaction, sodium is balanced and nitrogen is not. There are 6 nitrogen in the reactant side and only 2 in the product side. Hence, the product side have to multiplied by 3. Hence, Three molecules of N₂ should be produced during the decomposition. Thus, option C is correct.

To find more on decomposition reactions, refer here:

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Your question is incomplete.But your complete question is as follows:

An unbalanced chemical equation is shown:

2NaN3 → 2Na + N2

Which of the following statements explains why the equation is not balanced?

A. Four molecules of N2 should be produced during the synthesis reaction.

B. Four molecules of N2 should be produced during the decomposition.

C. Three molecules of N2 should be produced during the decomposition.

D. Three molecules of N2 should be produced during the synthesis reaction.

C. Three molecules of N2 should be produced during the decomposition.

Answer:

Molarity = 12 M

Explanation:

Given data:

Molarity of solution = ?

Number of moles of NaCl = 3.0 mol

Volume of solution = 250 mL (250/1000 = 0.25 L)

Solution:

Molarity is used to describe the concentration of solution. It tells how many moles are dissolve in per litter of solution.

Formula:

Molarity = number of moles of solute / L of solution

Molarity = 3.0 mol / 0.25 L

Molarity = 12 M

Answer:

[tex]\boxed {\boxed {\sf 12 \ M \ NaCl}}[/tex]

Explanation:

Molarity, which tells us the concentration of a solution, is found by dividing the moles of solute by the liters of solution.

[tex]M=\frac{moles \ of \ solute}{liters \ of \ solution}[/tex]

1. Define Values

There are 3.0 Moles of NaCl. This is the moles of solute.

There are 250 milliliters of solution, but we need the liters.

    a. Convert mL to L

1 milliliter is equal to 0.001 liters. We can multiply the given number of              milliliters (250) by 0.001.

250 mL * 0.001 L/mL= 0.25 L

[tex]moles \ of \ solute = 3.0 \ mol \ NaCl \\liters \ of \ solution= 0.25 \ L[/tex]

2. Calculate Molarity

Substitute the values into the formula.

[tex]M=\frac{3.0 \ mol \ NaCl}{0.25 \ L}[/tex]

Divide.

[tex]M= 12 \ mol \ NaCl/L[/tex]

3. Define Units

1 mole per liter is equal to 1 molar.

Our answer of 12 mol NaCl/ L is equal to 12 M NaCl

The molarity is 12 M NaCl

Answer:

Uracil

Explanation:

The base that will NOT combine with 2-deoxyribose to form a nucleic acid is Uracil.

2-deoxyribose is a pentose sugar found in the DNA (Deoxyribonucleic acid). It is devoid of oxygen in its 2' position. The bases found in DNA are Adenine, Guanine, Cytosine and Thymine. Adenine, Guanine, and Cytosine are also found in RNA (Ribonucleic acid). Thymine is not present in RNA, it is only found in DNA. The base found in RNA is Uracil which in turn is not present in DNA. The five carbon sugar present in RNA is ribose which combines with Uracil.

Answer: a: extended or controlled release

Explanation:

Answer:

The correct answer is D) hydration

d

Explanation:

the answer is hydration

Question :

what is matter?

Answer:

In classical physics and general chemistry, matter is any substance that has mass and takes up space by having volume.[1]:21 All everyday objects that can be touched are ultimately composed of atoms, which are made up of interacting subatomic particles, and in everyday as well as scientific usage, "matter" generally includes atoms and anything made up of them, and any particles (or combination of particles) that act as if they have both rest mass and volume. However it does not include massless particles such as photons, or other energy phenomena or waves such as light.[1]:21[2] Matter exists in various states (also known as phases). These include classical everyday phases such as solid, liquid, and gas – for example water exists as ice, liquid water, and gaseous steam – but other states are possible, including plasma, Bose–Einstein condensates, fermionic condensates, and quark–gluon plasma.[3]

Usually atoms can be imagined as a nucleus of protons and neutrons, and a surrounding "cloud" of orbiting electrons which "take up space".[4][5] However this is only somewhat correct, because subatomic particles and their properties are governed by their quantum nature, which means they do not act as everyday objects appear to act – they can act like waves as well as particles and they do not have well-defined sizes or positions. In the Standard Model of particle physics, matter is not a fundamental concept because the elementary constituents of atoms are quantum entities which do not have an inherent "size" or "volume" in any everyday sense of the word. Due to the exclusion principle and other fundamental interactions, some "point particles" known as fermions (quarks, leptons), and many composites and atoms, are effectively forced to keep a distance from other particles under everyday conditions; this creates the property of matter which appears to us as matter taking up space.

For much of the history of the natural sciences people have contemplated the exact nature of matter. The idea that matter was built of discrete building blocks, the so-called particulate theory of matter, independently appeared in ancient Greece and ancient India among Buddhists, Hindus and Jains in 1st-millennium BC.[6] Ancient philosophers who proposed the particulate theory of matter include Kanada (c. 6th–century BC or after),[7] Leucippus (~490 BC) and Democritus (~470–380 BC).[8]