Carbon monoxide, a product of combustion, is a toxic gas that has an extremely high affinity for hemoglobin (much higher than that of oxygen for hemoglobin); consequently, as soon as it dissolves in the liquid part of blood at low partial pressure, it diffuses quickly into red blood cells and binds to hemoglobin. In carbon monoxide (CO) poisoning, even with very low partial pressure of inspired CO, CO rapidly binds to hemoglobin (Hgb), leaving a lower fraction of oxygen binding sites on Hgb available to be occupied by oxygen. What would you expect to find if you measure the arterial PO2 of a person with CO poisoning

Answer:

As a giant star.

Explanation:

A

Answer:

C2H2 + Br2 → C2H2Br2

Explanation:

Answer:

90.99 or 91.0

Explanation:

Using the balanced equation, you convert 38.5g of ethanol to moles of water. From there, you plug the values into the Ideal Gas Equation: PV=nRT.

Answer: The volume of oxygen gas is 91.4 L.

Explanation:

The number of moles is defined as the ratio of the mass of a substance to its molar mass. The equation used is:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex] ......(1)

Given mass of ethanol = 38.5 g

Molar mass of ethanol = 46 g/mol

Plugging values in equation 1:

[tex]\text{Moles of ethanol}=\frac{38.5g}{46g/mol}=0.840 mol[/tex]

The given chemical equation follows:

[tex]CH_3CH_2OH(g)+3O_2(g)\rightarrow 2CO_2(g)+3H_2O(g)[/tex]

By stoichiometry of the reaction:

If 1 mole of ethanol produces 3 moles of water

So, 0.840 moles of ethanol will produce = [tex]\frac{3}{1}\times 0.840=2.52mol[/tex] of water

The ideal gas equation is given as:

[tex]PV=nRT[/tex] .......(2)

where

P = pressure = 1.75 atm

V = volume of oxygen gas = ?

n = number of moles= 2.52 moles

R = Gas constant = 0.0821 L.atm/mol.K

T = temperature of the tank = [tex]500^oC=[500+273]K=773K[/tex]

Putting values in equation 2, we get:

[tex]1.75 atm\times V=2.52mol\times 0.0821L.atm/mol.K\times 773K\\\\V=\frac{2.52\times 0.0821\times 773}{1.75}=91.4L[/tex]

Hence, the volume of oxygen gas is 91.4 L.

Answer:

"They established the laws of planetary motion"

Explanation:

Mr. Kepler was the astronomer who came up with the "Laws of Planetary Motion."

Answer:

5g/cm3

Explanation:

firstly convert the litres and kilograms to grams and centimeters.

1l is equivalent to 1000cm3

0.58×1000

580cm3

and 1kg is equivalent to 1000g

2.9×1000

2900

then find the density by using the formula

density=mass/volume

=2900g/580cm3

=5g/cm3

I hope this helps

Answer:

17

Explanation:

Step 1: Calculate the needed concentrations

[A]i = 1.00 mol/5.00 L = 0.200 M

[B]i = 1.80 mol/5.00 L = 0.360 M

[B]e = 1.00 mol/5.00 L = 0.200 M

Step 2: Make an ICE chart

        A(aq) + 2 B(aq) ⇄ C(aq)

I       0.200    0.360        0

C        -x           -2x         +x

E     0.200-x  0.360-2x   x

Then,

[B]e = 0.360-2x = 0.200

x = 0.0800

The concentrations at equilibrium are:

[A]e = 0.200-0.0800 = 0.120 M

[B]e = 0.200 M

[C]e = 0.0800 M

Step 3: Calculate the concentration equilibrium constant (K)

K = [C] / [A] × [B]²

K = 0.0800 / 0.120 × 0.200² = 16.6 ≈ 17

Answer:

Avogadro's number is 1 mol  = 6.02 * 10^23 elements

It means that 1 mol of atoms is 6.02 * 10^23 atoms

1 mol of atoms = 6.02 * 10^23 atoms

From there, if you divide both sides by 1 mol of atoms, you get

1 = 6.02 * 10^23 atoms / 1 mol of atoms.

That means, that to pass from a number of moles of atoms to number of atoms you have to multipby by the conversion factor

         6.02*10^23  atoms Al/ 1 mol Al

That is the second option of the list.

Explanation:

Answer:

79 g/mol

Explanation:

Mass of unknown metal deposited = 3.137 g - 1.4 g = 1.737 g

Number of moles of metal deposited = 0.022 moles

Since;

Number of moles = reacting mass/molar mass

Molar mass = reacting mass/number of moles

Molar mass = 1.737 g/0.022 moles

Molar mass= 79 g/mol

Stoichiometry refers to the relationship between the moles of reactants and products.

This question must be solved using both stoichiometry and the gas laws

The reaction equation is;

2 Al(s) + 6 HCl(aq) --------> 2 AICI3, (aq) + 3 H2(g)

Number of moles of Al = 3g/27g/mol = 0.11 moles

According to the reaction equation;

2 moles of Al yields 3 moles of H2

0.11 moles of Al yields 0.11 * 3/2 = 0.165 moles

From the ideal gas equation;

PV=nRT

P = ?

n= 0.165 moles

V = 0.85 L

T = 297 K

R = 0.082 atmLK-1mol-1

P= nRT/V

P = 0.165 * 0.082 * 297/0.85

P= 4.73 atm

P1 = 4.73 atm

T1 = 297 K

V1 = 0.85 L

P2 = 1 atm

T2 =273 K

V2 =?

P1V1/T1 = P2V2/T2

P1V1T2 = P2V2T1

V2 = P1V1T2/P2T1

V2 =  4.73 * 0.85 * 273/1 * 297

V2 = 3.69 L

P1 = 760 mmHg

T1 = 273 K

V1 = 3.69

P2 = 438 mm Hg

T2 = 284.2 K

V2 =?

P1V1/T1 = P2V2/T2

P1V1T2 = P2V2T1

V2 = P1V1T2/P2T1

V2 = 760 * 3.69 * 284.2/438 *273

V2 = 797010.48/119574

V2= 6.67 L

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Answer:

heat

Explanation:

because it's the cause of change

Answer:

heat

Explanation:

because it is a natural factor that causes the change in Earth's system

the effects are: temperature rises, water shortages, and increased fire threats

Answer:

Answer is B) 20.0 moles

Explanation:

From the equation,

1 mole of Fe2O3 = 2 moles of Al

therefore 10.0 moles of Fe2O3 = 10×2

= 20.0 moles.

Answer:

a. True.

Explanation:

Only primary and secondary alcohols can oxidise to give an aldehyde. But a weak oxidizing agent must be used to prevent formation of a carboxylic acid or ketone.

weak oxidizing agents: Chromyl chloride, silver/oxygen/500°C

take an example of ethanol:

[tex]{ \bf{CH _{3} CH_{2}OH \: \: \frac{Ag/O_{2} }{500 \degree C} > \: \:CH _{3} CHO}}[/tex]

[tex]{ \sf{CH _{3} CHO \: \: is \: ethanal}} [/tex]

By ozonolysis:

Here, reactants are Ozone gas, Carbon tetrachloride at a temperature (<20°C), ethanoic acid, zinc and water.

take an example of propanol:

if it undergoes ozonolysis, it gives ethanal and methanal.

Answer:

A. True

Explanation:

Only primary and secondary alcohols can oxidise to give an aldehyde. But a weak oxidizing agent must be used to prevent formation of a carboxylic acid or ketone.

weak oxidizing agents: Chromyl chloride, silver/oxygen/500°C

take an example of ethanol:

By ozonolysis:

Here, reactants are Ozone gas, Carbon tetrachloride at a temperature (<20°C), ethanoic acid, zinc and water.

take an example of propanol:

if it undergoes ozonolysis, it gives ethanal and methanal.

Answer:

[tex]\boxed {\boxed {\sf 5.56 \times 10^{23} \ molecules \ CH_4}}[/tex]

Explanation:

We are asked to find how many molecules of methane are in 14.8 grams of the substance.

First, we convert grams to moles. We use the molar mass, or the mass of 1 mole of a substance. These values are equivalent to the atomic masses found on the Periodic Table, however the units are grams per mole instead of atomic mass units.

We are given the compound methane, or CH₄. Look up the molar mass of the individual elements (carbon and hydrogen).

Check the formula for subscripts. Hydrogen (H) has a subscript of 4, so there are 4 moles of hydrogen in 1 mole of methane. We must multiply hydrogen's molar mass by 4, then add carbon's molar mass.

Now we use dimensional analysis to convert. To do this, we set up a ratio using the molar mass.

[tex]\frac {16.043 \ g \ CH_4 }{ 1 \ mol \ CH_4}[/tex]

Since we are converting 14.8 grams of methane to moles, we multiply by this value.

[tex]14.8 \ g \ CH_4 *\frac {16.043 \ g \ CH_4 }{ 1 \ mol \ CH_4}[/tex]

Flip the ratio so the units of grams of methane cancel.

[tex]14.8 \ g \ CH_4 *\frac{ 1 \ mol \ CH_4} {16.043 \ g \ CH_4 }[/tex]

[tex]14.8 *\frac{ 1 \ mol \ CH_4} {16.043}[/tex]

[tex]\frac {14.8}{16.043} \ mol \ CH_4= 0.9225207256 \ mol \ CH_4[/tex]

Next, we convert moles to molecules. We use Avogadro's Number or 6.022  × 10²³. This is the number of particles (atoms, molecules, formula units, etc.) in 1 mole of a substance. In this problem, the particles are moles of methane. Set up another ratio using Avogadro's Number.

[tex]\frac { 6.022 \times \ 10^{23} \ molecules \ CH_4}{ 1 \ mol \ CH_4}[/tex]

Multiply by the number of moles we calculated.

[tex]0.9225207256\ mol \ CH_4 * \frac { 6.022 \times \ 10^{23} \ molecules \ CH_4}{ 1 \ mol \ CH_4}[/tex]

The units of moles of methane cancel.

[tex]0.9225207256* \frac { 6.022 \times \ 10^{23} \ molecules \ CH_4}{ 1 }[/tex]

[tex]5.55541981 \times 10^{23} \ molecules \ CH_4[/tex]

The original measurement of grams (14.8) has 3 significant figures, so our answer must have the same. For the number we calculated, that is the hundredth place. The 5 in the thousandth place tells us to round the 5 in the hundredth place up to a 6.

[tex]5.56 \times 10^{23} \ molecules \ CH_4[/tex]

14.8 grams of methane is equal to approximately 5.56 × 10²³ molecules of methane.

Answer:

the number is 7

Explanation:

"Three times" means multiply by 3

"Difference" means subtract

"Sum" means add

3(x - 7) = 23 - (3x + 2)

3x - 21 = 23 - 3x - 2

3x - 21 = 21 - 3x

6x = 42

x = 7

Answer:

Part A

HCOOH(aq) + NaOH(aq) → HCOONa(aq) + H2O(l)

Part B

ClCH2CH2CO2H(aq) + NaOH(aq) ------> ClCH2CH2CO2Na(aq) + H2O(l)

Explanation:

The reaction between an alkanoic acid and a base is a neutralization reaction. The reaction occurs as follows;

RCOOH + NaOH ----> RCOONa + H2O

We have to note the fact that the net ionic reaction still remains;

H^+(aq) + OH^-(aq) ---> H2O(l)

In both cases, the reaction can occur and they actually do occur as written.

Answer:

The lewis structure for NO₃⁻ is shown in the attachment below

For the central nitrogen atom:

The number of lone pairs = 0

The number of single bonds = 2  

The number of double bonds= 1

Explanation:

The lewis structure for NO₃⁻ is shown in the attachment below.

From the Lewis structure

For the central nitrogen atom:

The number of lone pairs = 0

The number of single bonds = 2  

The number of double bonds= 1

Answer:A) V = 16.892 ml

Explanation:

M1 * V1 = M2 * V2

14.8 M * V1 =0.250 M * 1000 ml

V1 = 16.892 ml

a. The volume of 16.89 milliliters of the stock solution of 14.8 M  should be diluted to make 1000.0 mL of 0.250 M.

b. The concentration of the final solution is 0.296 M.

The concentration or the volume of the concentrated or dilute solution can be calculated by using the equation:

M₁V₁ = M₂V₂

where M₁ and V₁ are the concentration and volume of the concentrated solution respectively and M₂ and V₂ are the concentration and volume of the dilute solution.

A stock solution is a solution that has a high concentration and that will be diluted to a low concentration by the addition of water in it.

Given, a stock solution of concentration, M₁ = 14.8 M

The concentration of the diluted solution, M₂ = 0.250 M

The volume of diluted solution, V₂  = 1000ml

Substitute the value of the molarity and volume in equation (1):

(14.8)× (V₁) = (1000) × (0.250)

V₁ = 16.89 ml

Similarly, for part (b): M₁ = 14.8 M, V₁ = 10 ml and V₂  = 0.5L = 500 ml

(14.8)× (10) = (500) × (M₂)

M₂ = 0.296 M

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Answer:

Answer:

D. beta particle

Explanation:

Pu has 94 protons which is only one more than Np (93), that means beta decay. Atomic mass stays the same in beta decay. Beta is very small particle so atomic mass stays at 239.

The given reaction is an example of beta decay where the atomic number of the nuclei increases by one unit and mass number remains unchanged. Therefore, the particle X is a beta particle or an electron.

Heavy unstable isotopes of atoms undergoes nuclear decay by the emission of charged particles such as alpha or beta particle. In alpha decay, the helium nuclei is emitted and in beta decay, electrons are emitted.

In alpha decay, the mass number of the nuclei decreases by 4 units and atomic number by 2 units. In beta decay, the mass number does not change but the atomic number increases by one unit.

Neptunium undergoes beta decay and forms the plutonium nuclei. Thus X is a beta particle. The reaction is written as follows:

[tex]\rm _{93} ^{239}Np \rightarrow _{94}^{239}Pu + _{-1}^{0}e[/tex]

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Answer:

Element

Element : A pure substance composed of the same type of atom throughout. Compound : A substance made of two or more elements that are chemically combined in fixed amounts.

Explanation:

Answer:

C. yes, because it is universally applies to all objects

Answer:

[tex]n_1=4[/tex]

Explanation:

From the question we are told that:

Wavelength [tex]\lambda=489.2 nm =>4.86*10^{-7}[/tex]

nf level= Balmer series

nf level= 2

Generally the equation for Wavelength is mathematically given by

[tex]\frac{1}{\lambda}=R[\frac{1}{nf^2}-\frac{1}{n_1^2}][/tex]

Where

[tex]R=Rydberg Constant[/tex]

[tex]R=1.097*10^7[/tex]

Therefore

[tex]\frac{1}{4.86*10^{-7}}=1.097*10^7[\frac{1}{2^2}-\frac{1}{n_1^2}][/tex]

[tex]n_1=4.0021[/tex]

[tex]n_1=4[/tex]

Nitrogen gained 4 electrons.

Because Nitrogen's redox number went from +6 to +2, it must have gained 4 electrons (-4) in order to achieve this number. Thus, Nitrogen is reduced.

Answer:

[tex]C_t=0.165M[/tex]

Explanation:

From the question we are told that:

Slope [tex]K=0.056 M-1 s -1[/tex]

initial Concentration [tex]C_1=2.2M[/tex]

Time [tex]t=100[/tex]

Generally the equation for Raw law is mathematically given by

[tex]\frac{1}{C}_t=kt+\frac{1}{C}_0[/tex]

[tex]\frac{1}{C}_t=0.056*100+\frac{1}{2.2}_0[/tex]

[tex]C_t=0.165M[/tex]

The second-order reaction is the reaction that depends on the reactants of the first or the second-order reaction. The concentration after 100 seconds will be 0.165 M.

The specific rate constant (k) of the second-order reaction is given in L/mol/s or per M per s. It is the proportionality constant that gives the relation between the concentration and the rate of the reaction.

Given,

Slope (k)= 0.056 per M per s

Initial concentration of the reactant [tex](\rm C_{1})[/tex] = 2.2 M

Time (t) = 100 seconds

The concentration of the reaction after 100 seconds can be given by,

[tex]\rm \dfrac{1}{C_{t}} = kt + \dfrac{1}{C_{1}}[/tex]

Substitute values in the above equation:

[tex]\begin{aligned} \rm \dfrac{1}{C_{t}} &= 0.056 \times 100 + \dfrac{1}{2.02}\\\\&= 0.165 \;\rm M\end{aligned}[/tex]

Therefore, after 100 seconds the concentration is 0.165 M.

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The correct answer is C. square planar

According to the Valence Shell Electron Pair Repulsion Theory(VSEPR), The shape of a molecule depends on the number of electron pairs in the molecule.

VSEPR theory was first coined by Gillespie  and Nyhlom in 1957 as an improvement over the Sidgwick - Powell theory.

According to this theory, the shape of a molecule is determined by the number of electron pairs that surround the valence shell of the central atom in the molecule.  The electron pairs are positioned as far apart in space as possible to minimize repulsion of electron pairs.

However, the presence of lone pairs distorts the shape anticipated for the molecule on the basis of VSEPR.

For a molecule having six electron pairs, an octahedral geometry is expected(electron domain geometry). However, the presence of two lone pairs which are positioned at opposite side of the four single bonds leads to an observed square planar molecular geometry.

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Answer:

square planar

Explanation:

QUESTION:- Manganese-55 has _____neutrons.

OPTIONS :-

A. 55

B. 30

C. 25

ANSWER:- NUMBER OF NEUTRONS IS EQUAL TO THE DIFFERENCE BETWEEN THE MASS IF THE ATOM AND ATOMIC NUMBER

SO DIFFERENCE IS EQUAL TO :- 55-25 = 30 NEUTRONS.

SO THERE IS 30 NEUTRONS IN SINGLE ATOM OF THE MANGANESE-55 ATOM.

Answer:

the mass of an atom is the sum of proton and neutron which are both concentrated in nocleus of an atom. from the question the mass is given as 55 and the proton is 25.

Answer:

0.0347 moles of hydronium ions

Explanation:

The equation of the neutralization reaction between hydroxide and hydronium ions is given below:

H₃O+ (aq) + OH- (aq) ----> 2 H₂O (l)

From the equation above, 1 mole of hydroxide ions will neutralize one mole hydronium ions.

The moles of hydroxide ions present in 1 teaspoon or 5 mL of antacid product is calculated as follows:

Number of moles = mass / molar mass

Molar mass of Magnesium hydroxide, Mg(OH)₂ = 58 g/mol

Molar mass of aluminium hydroxide, Al(OH)₃ = 78 g/mol

Mass of magnesium hydroxide = 450 g = 0.45 g

Mass of aluminium hydroxide = 500 mg = 0.5 g

Moles of magnesium hydroxide = (0.45/58) moles

Moles of aluminium hydroxide = (0.5/78) moles

Equation of the ionization of magnesium hydroxide and aluminium hydroxide is given below:

Mg(OH)₂ (aq) ----> Mg²+ (aq) + 2 OH- (aq)

Al(OH)₃ (aq) ---> Al³+ (aq) + 3 OH- (aq)

Number of moles of hydroxide ions present in (0.45/58) moles of magnesium hydroxide = 2 × (0.45/58) moles = 0.0155 moles

Number of moles of hydroxide ions present in (0.5/78) moles of aluminium hydroxide = 3 × (0.5/78) moles = 0.0192 moles

Total moles of hydroxide ions = 0.0155 + 0.0192 = 0.0347 moles hydroxide ions

Therefore, 0.0347 moles of hydroxide ions will neutralize 0.0347 moles of hydronium ions.

Answer:

Transition temperature is the temperature at which a substance changes from one state to another.

Allotropy is the existence of an element in many forms.