M1 = 3M M2 = 1.25M V1 = V1=.125L unkown What is V2?

A sample of rainwater with a pH of 3.5 is sure to contain a large concentration of [tex]H_3O^+[/tex]

A pH value of 7 downward is said to be acidic. Upward of 7 is said to be basic. A pH of 7 on the dot is said to be neutral.

An acid is defined as a substance that produces  [tex]H_3O^+[/tex] as its only positively charged ion in solution.

Thus, rainwater with a pH of 3.5 is acidic and will contain a high concentration of  [tex]H_3O^+[/tex].

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NO is the limiting reagent and 4.34 g is the amount of the excess reagent that remains after the reaction is complete

The reactant that is entirely used up in a reaction is called as limiting reagent.

The reaction:

[tex]2NO(g) +2H_2(g)[/tex] → [tex]N_2 +2H_2O[/tex]

Moles of nitrogen monoxide

Molecular weight: [tex]M_(_N_O_)[/tex]=30g/mol

[tex]n_(_N_O_) =\frac{mass}{molar \;mass}[/tex]

[tex]n_(_N_O_) =\frac{22.0}{30g/mol}[/tex]

[tex]n_(_N_O_) = 0.73 mol[/tex]

Moles of hydrogen

Molecular weight: [tex]M_(_H_2_)[/tex]=30g/mol

[tex]n_(_H_2_) =\frac{mass}{molar \;mass}[/tex]

[tex]n_(_H_2_) =\frac{5.80g}{2g/mol}[/tex]

[tex]n_(_H_2_) = 2.9 mol[/tex]

Hydrogen gas is in excess.

NO is the limiting reagent.

The amount of the excess reagent remains after the reaction is complete.

[tex]n_(_N_2_) =[/tex] (2.9 mol- 0.73 mol NO x [tex]\frac{1 \;mol \;of \;H_2}{2 \;mole \;of \;NO}[/tex]) x [tex]\frac{2g \;of \;H_2}{mole \;of \;H_2}[/tex]

[tex]n_(_N_2_) =[/tex]4.34 g

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Iron has an atomic mass of 55.845 g/mol, so the amount of iron in 65.0 g is 65.0/55.845 = mol.

Avogadro's law states that in every mole of a substance, there are [tex]6.022 \times 10^{23}[/tex] particles per mole, so in this case, that means there are [tex](1.16393589)(6.022 \times 10^{23})=\boxed{7.01 \times 10^{23}} \text{ atoms (to 3 sf)}}[/tex]

Answer:

A

Explanation:

I'm right I took the test

Answer:

0.8797 J/g oC

Explanation:

use mcAT=mcaT

Answer:

HCO3- + BrO- → HBrO + CO32-

Explanation:

I think this is the answer hope it helps

Answer:

The oxygen atom in water has a negative charge.

Polar molecule: A molecule in which the centroid of the positive charges is different from the centroid of the negative charges.

Oxygen: A colorless, odorless, gaseous element constituting about one-fifth of the volume of the atmosphere and present in a combined state in nature. It is the supporter of combustion in air and was the standard of atomic, combining, and molecular weights until 1961, when carbon 12 became the new standard. Symbol: O; atomic weight: 15.9994; atomic number: 8; density: 1.4290 g/l at 0°C and 760mm pressure.

Water: A transparent, odorless, tasteless liquid, a compound of hydrogen and oxygen, H2O, freezing at 32°F or 0°C and boiling at 212°F or 100°C. that in more or less impure state constitutes rain, oceans, lakes, rivers, etc.: it contains 11.188 percent hydrogen and 88.812 percent oxygen, by weight.

Hydrogen: A colorless, odorless, flammable gas that combines chemically with oxygen to form water: the lightest of the known element. Symbol: H; atomic weight: 1.00797; atomic number: 1; density: 0.0899 g/l at 0°C and 760 mm pressure.

Atom: Am atom is the smallest constituent particle of a chemical element which has the properties of that element. They re comprised of at least an electron and a portion, as is the case for Hydrogen. Atoms of all other elements however, contain at least one neutron.

Proton: A positively charged elementary particle that is a fundamental constituent of all atomic nuclei. It is the lightest and most stable baryon, having a charge equal in magnitude to that of the electron, a spin of 1/2, and a mass of 1.673 × 10-27kg. Symbol: P.

Electron: An elementary particle that is a fundamental constituent of matter, having a negative charge of 1.602 × 10-19 coulombs, ha mass of 9.108 × 10-31 kilograms, and spin of 1/23, and existing independently or as the component outside the nucleus of an atom.

Neutron: An elementary particle having no charge, mass slightly greater than that of a proton, and spin of 1/2: a constituent of the nuclei of all atoms except those of hydrogen. Symbol: n.

A object has a negative charge when it consists of more electrons than protons.

With a partial positive charge on 2 Hs and a partial negative charge on oxygen, water molecules are polar molecules.

Actually, two hydrogen atoms and the core oxygen atom of water are covalently connected. Due to its greater electronegative nature than hydrogen, oxygen pulls the bound electron pair in the middle of the atom toward it.

When illustrating higher electron densities over oxygen, partial negative charge is used, whereas partial positive charge is used to illustrate lower densities over hydrogen atoms.

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Answer:245.088

Explanation: you subtract final temperature with initial temperature = 46 then multiply by grams = 552 then multiplied by J = 245.088

HAZMAT Class 6 Toxic and infectious substances.

Answer:

CLASS 2 Gases

Explanation:

Hope this helps

The standard enthalpy change associated with the given reaction at 25 °C using the following enthalpies of formation will be − 902.4 kJ

The standard enthalpy change of formation of a compound is the enthalpy change which occurs when one mole of the compound is formed from its elements under standard conditions, and with everything in its standard state.

According to the balanced chemical equation ;

CH₄(g) + 2O₂(g) ⟶ CO₂(g) + 2H₂O(l)

Now,

ΔH⁰ = ΔH⁰f(products) − ΔH⁰f(reactants)

​

       = [ΔH⁰(CO₂) +ΔH⁰(H₂O)] − [ΔH⁰(CH₄) + 2ΔH⁰(O₂)]

       = [−393.5 + 2 x (−285.8)] − [−62.7 + 2 x 0]

        = − 902.4 kJ

Hence, The standard enthalpy change associated with the given reaction at 25 °C using the following enthalpies of formation will be − 902.4 kJ

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Answer: B

Explanation:

So mass of sulphur

[tex]\\ \rm\Rrightarrow 0.02\times 757\times 10^{6}[/tex]

[tex]\\ \rm\Rrightarrow 1514\times 10^4[/tex]

Moles =64/32=2

So

So tons of sulphur dioxide

[tex]\\ \rm\Rrightarrow 2(1516)10^4[/tex]

[tex]\\ \rm\Rrightarrow 3032\times 10^4ton[/tex]

Explanation:

2% of 2million tons of sulphur=

2÷100×7.57E9

That is;

[tex] \frac{2}{100} \times 7.57 \times 10 {}^{6} = 0.02 \times 7.57 \times 10 {}^{6} \\ = 2 \times 10 {}^{ - 2} \times7.57 \times 10 {}^{6} = 2 \times 7.57(10 {}^{ - 2 + 6}) \\ = 15.14 \times 10 {}^{4} (for \: sulphur) \\ for \: sulphur \: dioxide = so2 \\ the \: molar \: mass \: of \: so2 = 64g.mol {}^{ -1} \\ if \: 32grams \: of \: sulphur \: weighs \: 15.14 \times 10 {}^{4} tones \\ 64grams \: will \: weigh \: \frac{64}{32} \times 15.14 \times 10 {}^{4} \\ = 2 \times 15.14 \times 10 {}^{4} \\ = 30.28 \times 10 {}^{4} \\ or = 3.028 \times 10 {}^{5} tones[/tex]

The trend for the radii is Mg²⁺ < Na+ < Ne <F⁻

With the increase in number of protons in the nucleus the electrons are more attracted towards nucleus and so the ionic radius decreases.

Ionic size increases with the addition of electrons and ionic size decreases with the loss of electrons.

Higher is the atomic number, higher will be the effective nuclear charge, lower will be the size.

F⁻

no. of proton=9

no of electron = 10

Na+

no. of proton=11

no of electron=10

Mg²⁺

no. of proton=12

no of electron=10

Ne

no. of proton=10

no of electron=10

Mg²⁺ < Na+ < Ne <F⁻

Therefore the trend for the radii is Mg²⁺ < Na+ < Ne <F⁻

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Answer:

0.217 ×2. =9.86×10moles

44

Answer: c) [tex]2\text{Na}_{3}\text{PO}_{4}+3\text{CaCl}_{2} \longrightarrow 6\text{NaCl}+\text{Ca}_{3}\text{(PO}_{4})_{2}[/tex]

Answer: D. The energy needed to change a liquid into a gas

there are five Taste bud present tongue they are -

out of the four type of papillie present on tongue the fungiform circumvallate and folite papillie contain numerous small taste but few such type of Taste but also occur in the mucous membranes of soft palate tonsils and epiglottis all of these days but received taste is normally

each Taste bud is an oval structures formed due to modification of epithelial cells which are situated erect and vertically on the surface long sensory cells and supporting cells are found in this bird is sensory cell is spindle shaped in which a sensory here is found at its free in the century fibre of the facial and vagus cranial nerves are wrapped around the basal and of the gustatory cells each test but open on the epithelial surface through minute taste pore

The volume of the balloon when the mass of oxygen gas is decreased to 50g is 50L.

We will use the ideal gas equation-

PV=nRT

P=Pressure

V=volume

n=no. of moles

T=temperature

In this question temperature and pressure will remains constant then the above equation can be rewritten as-

[tex]V_{1} /n_{1} =V_{2} /n2[/tex]

where V1= initial volume of the balloon

            V2=  volume of the balloon when the mass of oxygen gas is decreased to 50g

now, substitute the values in the above equation-

       100/3.12=V2/1.56

hence, the volume of the balloon when the mass of oxygen gas is decreased to 50g is 50L.

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Answer:

KCl

Explanation:

Ionic compounds are formed between atoms with very different electronegativity (EN) values. The elements with the highest EN values are on the top-right of the periodic table.

Ionic compounds are also generally between a metal and nonmetal. Nonmetals are located in the p-block (the right) of the periodic table and metals are pretty much everywhere else.

That being said, water (H₂O), CO, and CO₂ contain elements that have too similar of EN values to ionically bond.

However, the elements in KCl have different enough EN values to undergo ionic bonding by donating/receiving electrons from each other.

The theoretical yield of NH3 is 10.03 g while the percentage yield of NH3 is 22%.

The theoretical yield is defined as the yield of the product as obtained from the stoichiometry of the reaction.

Given that the reaction equation is; N2 + 3H2 ------> 2NH3

Number of moles of N2 = 9.80 g/28 g/mol = 0.35 moles

Number of moles of H2 = 1.77 g/ 2 g/mol = 0.885 moles

If 1 mole of N2 reacts with 3 moles of H2

0.35 moles reacts with 0.35 moles * 3 moles / 1 mole

= 1.75 moles

Hydrogen is the limiting reactant

Thus;

3 moles of H2 yields 2 moles of NH3

0.885 moles yields  0.885 moles * 2 moles/3 moles

= 0.59 moles

Theoretical yield of NH3 = 0.59 moles * 17 g/mol = 10.03 g

Percent yield = 2.21/10.03 * 100/1

= 22%

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Answer:

A large absorption at 2950 cm⁻¹ indicates that there is likely an sp³C-H functional group somewhere in the molecule. The "sp³" designation means that this is a single bond because these atoms have 4 orbitals available (s + p + p + p) for hybridizing, meaning only single bonds are possible.

The molecular weight of Cr3(AsO4)2 is 433.827.

Furthermore, the primary distinction between each is that molar mass gives the mass of a mole of a selected substance. While molecular weight is the mass of a molecule of a particular substance. At an equal time as the definition and devices are splendid for molar mass and molecular weight, the value is the same.

The sum of the atomic hundreds of all atoms in a molecule is based totally on a scale wherein the atomic loads of hydrogen, carbon, nitrogen, and oxygen are 1, 12, 14, and 16, respectively.

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Answer: There is a very large percentage of C-12.

Explanation:

The atomic mass is must closer to that of C-12 than of C-13, meaning that there is a much larger percentage of C-12.

The statement which is not true is as follows:

Thus, the correct option is C.

An equilibrium may be defined as the condition in which both the reactants and products are present in concentrations that have no additional propensity to alter with time so that there is no noticeable transformation in the possessions of the system.

The causation of repulsive force was thought to be made by the overlap of atomic orbitals and in between the electrons of diverse atoms, it matters a lot.

The equilibrium distance between nuclei of two covalently bonded atoms is called the bond length of that atom. It is the atomic separation at equilibrium.

Equilibrium occurs when the attractive and repulsive forces neglected the effect of each other in a specific state.

Therefore, it is well described above.

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The moles of nitrogen (iv) oxide produced in C and D are 0.284 moles and 68.32 moles respectively.

The mole ratio of the given reaction shows that 2 moles of dinitrogen (v) oxide produces 4 moles of nitrogen (iv) oxide when it decomposes.

B of dinitrogen (v) oxide in 15.4 g is calculated below:

Molar mass of dinitrogen (v) oxide = 108 g/mol

Moles of dinitrogen (v) oxide = 15.4/108 = 0.14 moles

Moles of  nitrogen (iv) oxide produced = 0.142 * 2 = 0.284 moles

Moles of dinitrogen (v) oxide in 3.69 kg is calculated below:

Molar mass of dinitrogen (v) oxide = 108 g/mol

Mass = 3.69 kg = 3690 g

Moles of dinitrogen (v) oxide = 3690/108 = 34.16 moles

Moles of nitrogen (iv) oxide produced = 34.16 * 2 = 68.32 moles

Therefore, the moles of nitrogen (iv) oxide produced in C and D are   0.284 moles and 68.32 moles respectively.

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The initial pH for the weak acid-strong base curve is higher than the initial pH for the strong acid-strong base curve. Hence, option A is correct.

Weak acids are acids that don't completely dissociate in solution.

A weak acid is an acetic acid. It has a of 1.8⋅[tex]10^{-5}[/tex]. Calculate how much it will dissociate in water. Since acetic acid is a weak acid so large part will not dissociate completely.

[tex]k_a=\frac{[CH_3COO^-]H^+]}{[CH_3COOH]}[/tex]

B is not true, it shows the titration curve for weak/strong acid titrated with a strong base. When choosing an indicator for colourimetric titration select one so that the pH jump at the equivalence point contains the interval p±1. Phenolphthalein has a p≈9, so to decide if it is a suitable indicator check if the pH jumps from 8 to 10 at the equivalence point.

C is not true either because of the very slow reaction.

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The percent yield for the chemical  reaction producing sulfur trioxide is 66.51%.

Percent yield is defined as the ratio of actual yield to the theoretical yield multiplied by 100. If the actual and theoretical yield are same then the percent yield is 100%.If actual yield is less than the theoretical yield then the percent yield is less than 100%.Reason of this condition arising is the incompletion of reaction or loss of sample during recovery process.

In cases where percent yield is over 100% it indicates that more sample is recovered than the predicted amount.

For the given reaction, theoretical yield is calculated as,

128.132 g of sulfur dioxide gives 160.12 g of sulfur trioxide

∴705 g of sulfur dioxide gives 705×160.12/128.132= 881 g

Percent yield is, 586/881×100=66.51%

Therefore, the percent yield for the chemical  reaction is 66.51%.

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