43 mg = [?]g
A. 0.043 g
B. 4.3 g
C. 4300 g
D. 43,000 g​

Answer:2C4H10+2C12+12O2 4CO2+CC14+H20

Answer:

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Answer:

3Ca(OH)2 + 2C6H8O7 → 6H2O + Ca3(C6H5O7)2

And 0.119M is the concentration of the citric acid.

Explanation:

In an acid-base reaction, the proton H+ and the hydroxil ion OH- reacts producing water. The ions of the acid and base (C6H5O7³⁻ and Ca²⁺ ions produce the respective salt) as follows:

Ca(OH)2 + C6H8O7 → H2O + Ca3(C6H5O7)2

To balance the Calcium ions:

3Ca(OH)2 + C6H8O7 → H2O + Ca3(C6H5O7)2

To balance the C6H5O7³⁻ ions:

3Ca(OH)2 + 2C6H8O7 → H2O + Ca3(C6H5O7)2

And to balance the oxygens of water:

3Ca(OH)2 + 2C6H8O7 → 6H2O + Ca3(C6H5O7)2

And this is the balanced reaction.

The moles of Ca(OH)2 that reacts are:

41.27mL = 0.04127L * (0.108mol/L) = 0.004457 moles Ca(OH)2

Moles of citric acid:

0.004457 moles Ca(OH)2 * (2mol C6H8O7 / 3mol Ca(OH)2) = 0.002971 moles C6H8O7

In 25.00mL = 0.02500L:

0.002971 moles C6H8O7 / 0.0250L =

Answer:

Mass CO2 = 3.75 grams

Mass H2O = 2.30 grams

Mass SO2 = 2.73 grams

Explanation:

Step 1: Data given

Density of Liquid ethyl mercaptan, C2H6S = 0.84 g/mL

Volume of ethyl mercaptan = 3.15 mL

Step 2: The reaction

2C2H6S + 9O2 → 4CO2 + 6H2O + 2SO2

Step 3: Calculate mass of ethyl mercaptan

Mass = Volume * density

Mass ethyl mercaptan = 3.15 mL * 0.84 g/mL

Mass ethyl mercaptan = 2.646 grams

Step 4: Calculate moles ethyl mercaptan

Moles = mass / molar mass

Moles ethyl mercaptan = 2.646 grams / 62.13 g/mol

Moles ethyl mercaptan = 0.04259 moles

Step 5: Calculate moles of other products

For 2 moles ethyl mercaptan we need 9 moles O2 to produce 4 moles CO2, 6 moles H2O and 2 moles SO2

For 0.04259 moles we need 0.1917 moles O2 to produce:

2*0.04259 = 0.08518 moles CO2

3*0.04259 = 0.1278 moles H2O

1*0.04259 = 0.04259 moles SO2

Step 6: Calculate mass produced

Mass = moles * molar mass

Mass CO2 = 0.08518 moles * 44.01 g/mol

Mass CO2 = 3.75 grams

Mass H2O = 0.1278 moles * 18.02 g/mol

Mass H2O = 2.30 grams

Mass SO2 = 0.04259 moles * 64.07 g/mol

Mass SO2 = 2.73 grams

Answer:

E = hv

Explanation:

Energy = planck constant × frequency

Answer:

55.44L of the 0.17M NaOH are required

Explanation:

Phosphoric acid, H3PO4, reacts with NaOH as follows:

H3PO4 + 3NaOH → Na3PO4 + 3H2O

Where 1 mole of H3PO4 reacts with 3 moles of NaOH

To solve this question we need to find the moles of H3PO4 in the aliquot. Using the balanced equation we can find the moles of NaOH and its volume with the concentration (0.17M) as follows:

Moles H3PO4:

15.4mL = 0.0154L * (0.204mol/L) = 0.00314 moles H3PO4

Moles NaOH:

0.00314 moles H3PO4 * (3mol NaOH / 1mol H3PO4) = 0.009425moles NaOH

Volume NaOH:

0.009425moles NaOH * (1L/0.17moles NaOH) = 0.05544L 0.17M NaOH =

Answer:

A sample of calcium fluoride was decomposed into the constituent elements. Write a balanced chemical equation for the decomposition reaction. If the sample produced 294 mg of calcium, how many g of fluorine was formed

Explanation:

The balanced chemical equation for the decomposition of calcium fluoride is shown below:

[tex]CaF_2(s)->Ca(s)+F_2(g)[/tex]

The sample produced 294 g of calcium then, how many grams of fluorine is formed?

From the balanced chemical equation,

1 mol of CaF2 forms 1mol of calcium and 1 mol of fluorine.

That is:

40g of calcium and 38.0 g of fluorine are formed.

then,

If 294 g of calcium is formed then how many grams of fluorine is formed?

[tex]294g Ca * 38g F2 / 40g Ca\\=279.3 g F_2[/tex]

Hence, 279.3 g of fluorine will be formed.

Answer: The density of the given vapor is 0.939 g/L.

Explanation:

Given: Pressure = 233 mm Hg (1 mm Hg = 0.00131579 atm) = 0.31 atm

Temperature = [tex]25^{o}C[/tex] = (25 + 273) K = 298 K

According to the ideal gas equation,

[tex]PV = \frac{m}{M}RT[/tex]

where,

P = pressure

V = volume

m = mass

M = molar mass

R = gas constant = 0.0821 L atm/mol K

T = temperature

This formula can be re-written as follows.

[tex]PM = \frac{m}{V}RT[/tex]    (where, [tex]Density = \frac{mass (m)}{Volume (V)}[/tex] )

Hence, formula used to calculate density of diethy ether (molar mass = 74.12 g/mol) vapor is as follows.

[tex]d = \frac{PM}{RT}[/tex]

Substitute values into the above formula as follows.

[tex]d = \frac{PM}{RT}\\= \frac{0.31 atm \times 74.12 g/mol}{0.0821 L atm/mol K \times 298 K}\\= \frac{22.9772}{24.4658}\\= 0.939 g/L[/tex]

Thus, we can conclude that the density of the given vapor is 0.939 g/L.

Answer:

Surely with water

Ok, but how?

There are many Hydrogen Bond between H2O moleculs and london bonds. When fishes take water with their gill,they are broke london bonds. And they can take their needs, Oxygen. Only this.

Good luck :D

Answer:

Yes

Explanation:

=5.7×10-3

=5.7×7=39.9

=0.0007270 greater than 5.7× 10-3

OR

=5.7×10

=57-3

=54

So it is greater number

Answer:

txyxyc8uviciycuyc

Explanation:

gghvj ucy7cyvyfy

Answer:

H2CO

Explanation:

Becuase it has 2 sigma bonds plus one pi bond and one sigma bond that consitute the double bond between C and O.

Answer:

B. H2CO

Explanation:

Answer:

all light waves can be described in terms of their speed, frequency and wavelength

Explanation:

Hope it helps u.....

Answer:

NH3 Ammonia

Explanation:

Ammonia has 4 regions of electron density around the central nitrogen atom (3 bonds and one lone pair). These are arranged in a tetrahedral shape. The resulting molecular shape is trigonal pyramidal with H-N-H angles of 106.7°.

Answer: The type of solvent that should be dried with calcium chloride is esters while magnesium sulfate is diethyl ether

Explanation:

Drying agents are mainly hygroscopic substances that has the ability to absorb water on exposure to the atmosphere but not enough to form solutions. They are used in desiccators. Examples of drying agents include:

--> CALCIUM CHLORIDE: This is a compound of calcium that is found in soil water and sea water. It is prepared by the action of dilute hydrochloric acid on calcium trioxocarbonate(IV). The anhydrous salt is used in drying a wide variety of solvent including esters.

--> MAGNESIUM SULFATE: This is a slightly acidic drying agent. It works well in solvents like diethyl ether. It is a fast drying agent because it comes as a fine powder with a large surface area.

answer; 1/ 20_2[2-] +2e - ->0.

Answer: The fractional abundance of lighter isotope is 0.518

Explanation:

Average atomic weight is the sum of the masses of the individual isotopes each multiplied by its fractional abundance. The equation used is:[tex]\text{Average atomic weight}=\sum_{i=1}^{n}\text{(Atomic mass of isotope)}_i\times \text{(Fractional abundance)}_i[/tex]           ......(1)

Let the fractional abundance of Ag-107 isotope be 'x'

Atomic mass = 106.90509 amu

Fractional abundance = x

Atomic mass = 108.9047 amu

Fractional abundance = (1 - x)

Average atomic mass of silver = 107.8682 amu

Plugging values in equation 1:

[tex]107.8682=(106.90509 \times x) + (108.9047 \times (1-x))\\\\107.8682=106.90509x+108.9047-108.9047x\\\\1.99961x=1.0365\\\\x=0.518[/tex]

Fractional abundance of Ag-107 isotope (lighter) = x = 0.518

Hence, the fractional abundance of lighter isotope is 0.518

Answer:

The formula of Al³⁺ and SO₄²⁻ is aluminum sulfate.

Explanation:

The formula for aluminum sulfate is Al₂(SO₄)₃. If we say in terms of ions. The ions are Al³⁺. It is a positive ion or the cation. Other ion is SO₄²⁻. It is sulfate ion. It is anion.

Aluminum sulphate is used in water purification and as a mordant in dyeing and printing textiles.

Hence, the formula of Al³⁺ and SO₄²⁻ is aluminum sulfate.

Answer:

C₃H₆O₃

Explanation:

To solve this question we need to find, as first, the moles of each atom in order to find empirical formula (Simplest whole-number ratio of atoms present in a molecule).

With the molar mass of the substance and the empirical formula we can find the molecular formula as follows:

Moles C -Molar mass:12.0g/mol-

0.24g * (1mol/12.0g) = 0.020 moles C

Moles H = Mass H because molar mass = 1g/mol:

0.040 moles H

Moles O -Molar mass: 16g/mol-

Mass O: 0.60g - 0.24g - 0.040g = 0.32g O

0.32g O * (1mol/16g) = 0.020 moles O

Ratio of atoms (Dividing in moles of C: Lower number of moles):

C = 0.020 moles C / 0.020 moles C = 1

H = 0.040 moles H / 0.020 moles C = 2

O = 0.020 moles O / 0.020 moles C = 1

Empirical formula:

CH₂O.

Molar mass CH2O:

12g/mol + 2*1g/mol + 16g/mol = 30g/mol

As molecular formula has a molar mass 3 times higher than empirical formula, the molecular formula is 3 times empirical formula:

The molecular formula of the organic acid would be C3H6O3

Molecular formula = [empirical formula]n

Where n = molar mass/mass of empirical formula

Empirical formula

C = 0.24/12 = 0.02

H = 0.040/1 = 0.04

O = 0.6 - (0.24+0.04) = 0.32/16 = 0.02

Divide by the smallest

C = 1

H = 2

O = 1

Empirical formula = CH2O

Empirical formula mass = 12 + 2 + 16 = 30

n = 90/30 = 3

Molecular formula = [CH2O]3

                               = C3H6O3

More on molecular formula can be found here: https://brainly.com/question/1247523

Answer:

The Experimental Molar Volume in L/mol of the Hydrogen gas=23.36L/mol

Explanation:

We are given that

Volume of H2 at STP=52.8mL

Mass of magnesium metal ,M(Mg)=0.055g

We have to find the Experimental Molar Volume in L/mol of the Hydrogen gas.

Molar mass of Mg=24.305 g/mol

Number of moles=[tex]\frac{given\;mass}{molar\;mass}[/tex]

Using the formula

Number of moles of Mg=[tex]\frac{0.055}{24.305}[/tex]moles

Number of moles of Mg=0.00226moles

Number of moles of Mg=Number of moles of H2

Number of moles of H2=0.00226moles

Molar volume of Hydrogen gas (H2)=[tex]\frac{volume\;at\;STP}{No\;of\;moles\;H_2}[/tex]

Molar volume of Hydrogen gas (H2)=[tex]\frac{52.8}{0.00226}mL/mol[/tex]

Molar volume of Hydrogen gas (H2)=[tex]\frac{52.8}{0.00226}\times 10^{-3}L/mol[/tex]

[tex]1L=1000mL[/tex]

Molar volume of Hydrogen gas (H2)=23.36L/mol

Hence, the Experimental Molar Volume in L/mol of the Hydrogen gas=23.36L/mol

Answer:

0.605 g

Explanation:

First we calculate how many Cl₂ moles need to be produced, using the PV=nRT formula:

Inputting the data:

Then we convert 0.00696 moles of Cl₂ to MnO₂ moles:

Finally we convert 0.00696 moles of MnO₂ to grams, using its molar mass:

Answer:

hexane

I hope it's helps you

Answer:

yes is very possible to be

Answer:

Cl

Explanation:

Electronegativity is the ability of an electron to attract electrons.

Now, due to the fact that halogens need just one more electron to become stable in their outermost shell, it means all halogens are electronegative.

However, the smaller the atomic number, the bigger the charge density and thus the more electronegative.

Thus, it is the halogen element with the highest atomic number further down the periodic table that will have the least electro negativity and thus require highest amount of energy to attract other electrons.

Thus, since chlorine (Cl) has the least atomic number of 17, then it means that it will be the one that will easily accept the electrons the most from other elements. Therefore the process of transferring electrons from potassium to chlorine will take the least amount of energy.

Answer:

second law of thermodynamics

Explanation:

According to the second law of thermodynamics, the total entropy of a system and its surroundings always increases for a spontaneous process.

Entropy is defined as the degree of disorderliness of a system. The entropy of a system never remains constant. It either increases or decreases in a process. The total entropy is the sum of the entropy of the system and its surrounding. The total entropy must increase in a spontaneous process.

Thus, the implication of this law is that even, if the entropy of a system decreases, this must be compensated for by increase in entropy of the surroundings in order for the process to be spontaneous.

Answer:

[tex]\mathbf{\lambda \simeq 3.039 \times 10^{-8} \ m}[/tex]

Explanation:

For a hydrogen-like atom, the spectral line wavelength can be computed by using the formula:

[tex]\bar v = Z^2 R_H \Big(\dfrac{1}{n_f^2}-\dfrac{1}{n_i^2}\Big)[/tex]

where:

emitted radiation of the wavenumber  [tex]\bar v[/tex] = ???

atomic no of helium Z = 2

Rydberg's constant [tex]R_H = 1.097*10^7 \ m^{-1}[/tex]

the initial energy of  the principal quantum [tex]n_1[/tex] = 2

the initial energy of  the principal quantum [tex]n_1[/tex] = 2

Now, the emitted radiation of the wavenumber can be computed as:

[tex]\bar v = (2)^2 (1.097*10^7 \ m^{-1} ) \Big(\dfrac{1}{1^2}-\dfrac{1}{2^2}\Big)[/tex]

[tex]\bar v = 3.291 \times 10^ 7/m[/tex]

Now, the wavelength for the transition can be computed by using the relation between the wavelength λ and the emitted radiation of the wavenumber  [tex]\bar v[/tex], which is:

[tex]\bar v = \dfrac{1}{\lambda}[/tex]

[tex]\lambda = \dfrac{1}{\bar v}[/tex]

[tex]\lambda = \dfrac{1}{3.291 \times 10^{7}}\times \dfrac{m}{1}[/tex]

[tex]\mathbf{\lambda =3.03859 \times 10^{-8} \ m}[/tex]

[tex]\mathbf{\lambda \simeq 3.039 \times 10^{-8} \ m}[/tex]

Answer:

6.40x10^-7

Explanation:

answer with work is attached.

Answer:

–0.09

Explanation:

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

Measured value = 5.88 g

Actual value = 5.97 g

Error =?

The error in the student's measurement can be obtained as follow:

Error = Measured value – Actual value

Error = 5.88 – 5.97

Error = –0.09

Therefore, the error in the student's measurement is –0.09

Answer:

O B. Alkaline earth metals

Explanation:

Noble gases → 8th column.

Alkali metal → first column.

Halogen → 7th

Answer:

B. Alkaline earth metals

Explanation:

Alkaline-earth metals: The alkaline-earth metals make up Group 2 of the periodic table, from beryllium (Be) through radium (Ra). Each of these elements has two electrons in its outermost energy level, which makes the alkaline earths reactive enough that they're rarely found alone in nature. But they're not as reactive as the alkali metals. Their chemical reactions typically occur more slowly and produce less heat compared to the alkali metals.