what is the molar masses for C8H10N2O4

Answer:

S + 2CO = SO2 + 2C

First, look for the amount of substance of sulfur:

n(S) = m / M

n(S) = 14.8 g/32 g / mol = 0.4625 mol

n(CO) = m (CO) / M (CO)

M(CO) = 12 + 16 = 28 g/mol

n(CO) = 19.9 g/28 g/mol = 0.71 mol

S in excess, so for calculating we take CO:

n(SO2) = n(CO)/2 = 0.71 mol/2 = 0.355 mol

m(SO2) = M(SO2)*n(SO2)

M(SO2) = 32 + 16*2 = 64 g/mol

m(SO2) = 64 g/mol * 0.355 mol = 22.74 g

Answer:

7.44 C is the answer of the question.

Explanation:

Answer:

C2H4O3

Explanation:

Empirical formula is defined as the simplest whole number ratio of atoms present in a molecule.

To solve this question we need to find the moles of carbon = Moles of CO2, the moles of hydrogen (Using moles of H2O) and the moles of oxygen (Finding the mass of the mass of each atom) as follows:

Moles Carbon -Molar mass CO2: 44.01g/mol-:

33.86g CO2 * (1mol/44.01g) = 0.769 moles CO2 = Moles C * (12g/mol) =

9.23g C

Moles Hydrogen -Molar mass H2O: 18.01g/mol-

13.86g H2O * (1mol/18.01g) = 0.770 moles H2O * (2mol H / 1mol H2O) = 1.54 moles H * (1g/mol) = 1.54g H

Moles Oxygen:

Mass: 29.26g - 9.23g C - 1.54g H = 18.49g O * (1mol/16g) = 1.156 moles O

Dividing each number of moles in the moles of C (Lowe number of moles):

C = 0.769 moles C / 0.769 moles C  = 1

H = 1.54 moles H / 0.769 moles C = 2

O = 1.156 moles O / 0.769 moles C = 1.5

As the number must be a whole number each ratio twice:

C = 2

H = 4

O = 3

Empirical formula is:

Answer:

243.75 mL

Explanation:

In 1min 32.5 mL of oxygen

In 7.5 min electrolysis of water produces

(32.5mL × 7.5 min)/ 1min

= 243.75mL

Answer:

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Explanation:

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Answer:

The movement of the earth around the sun in a fixed path or orbit is called Revolution. The axis of the earth which is an imaginary line, makes an angle of 66½° with its orbital plane. ... The earth takes about 24 hours to complete one rotation around its axis.

Answer:

I expect to observe a change in colour from reddish brown to a colourless solution

Explanation:

Bromine (Br2) attacks the electron rich carbon-carbon triple bond in but-2-yne, an alkyne to form an initial product 2, 3 dibromobut-2-ene; which reacts with excess bromine to form a final product 2,2,3,3 -tetrabromobutane.

The reaction occurs in two steps. On approaching but-2-yne, bromine molecule becomes polarised forming an induced dipole containing a bromonium ion.

Br - Br → Br+ - Br-

The bromonium ion (Br+) formed then attacks the carbon - carbon triple bond to form the initial product

2,3- dibromobut-2-ene

CH2-C≡C-CH2 + Br+ →

CH2 - CBr =CBr-CH2

(2,3- dibromobut-2-ene)

Which in the presence of excess bromine gives the final product

2,2,3,3 - tetrabromobutane.

CH2 - CBr =CBr-CH2 + Br2 →

CH3 -CBr2-CBr2 - CH3

2,2,3,3 - tetrabromobutane.

A visible change in colour from the reddish-brown colour of Bromine to a colourless solution is observed during the reaction.

Explanation:

here is the answer to your question.

Answer:

B. False

Explanation:

Water does NOT react too copper. Copper does not react with water because the oxygen in water is locked into a compound with one part oxygen and two parts hydrogen. Copper oxide is a compound from the two elements copper and oxygen. Everything else listed does but since water is on this list it is false.

Answer:

When this equation is solved, the two values of the unknown are 0.0643 and -0.082

Explanation:

Given

[tex]1.77 * 10^{-2} = \frac{x^2}{0.298 - x}[/tex] --- the actual equation

Required

The values of x

We have:

[tex]1.77 * 10^{-2} = \frac{x^2}{0.298 - x}[/tex]

Cross Multiply

[tex]1.77 * 10^{-2} * (0.298 - x)= x^2[/tex]

Multiply both sides by 100

[tex]1.77 * (0.298 - x)= 100x^2[/tex]

Open bracket

[tex]0.52746 - 1.77x= 100x^2[/tex]

Rewrite as:

[tex]100x^2 + 1.77x - 0.52746 =0[/tex]

Using quadratic formula:

[tex]x = \frac{-b \± \sqrt{b^2 - 4ac}}{2a}[/tex]

Where:

[tex]a = 100; b = 1.77; c = -0.52746[/tex]

So, we have:

[tex]x = \frac{-1.77 \± \sqrt{1.77^2 - 4*100*- 0.52746 }}{2*100}[/tex]

[tex]x = \frac{-1.77 \± \sqrt{214.1169}}{2*100}[/tex]

[tex]x = \frac{-1.77 \± 14.63}{200}[/tex]

Split

[tex]x = \frac{-1.77 + 14.63}{200}\ or\ x = \frac{-1.77 - 14.63}{200}[/tex]

[tex]x = \frac{12.86}{200}\ or\ x = \frac{-16.40}{200}[/tex]

[tex]x = 0.0643\ or\ x = -0.082[/tex]

Answer:

The mass of bromine liquid that participates in  a chemical reaction=12.8 g

Explanation:

We are given that

Total number of moles of bromine liquid participate in chemical reaction=0.0800 moles

We have to find the mass of bromine liquid that participates.

Atomic mass of Br=79.9 g

1  mole of bromine liquid=2 atomic mass of bromine (Br)

1 mole of bromine liquid ([tex]Br_2[/tex]) =[tex]2\times 79.9=159.8 g[/tex]

0.0800 moles of  bromine liquid=[tex]159.8\times 0.0800[/tex] g

0.0800 moles of  bromine liquid=12.784 g

0.0800 moles of  bromine liquid[tex]\approx 12.8[/tex] g

Hence, the mass of bromine liquid that participates in  a chemical reaction=12.8 g

Answer:

3 > 2> 1

Explanation:

Aromatic compounds undergo electrophilic substitution reaction which passes through a positively charged intermediate to yield the product.

Substituted benzenes may be more or less reactive towards electrophilic aromatic substitution than benzene molecule depending on the nature of the substituent.

Certain substituents increase the ease of reaction of benzene towards aromatic substitution.

If we look at the compounds closely, we will notice that toluene reacts readily with CH3Cl / AlCl3. This is because, the methyl group is electron donating hence it stabilizes the positively charged intermediate produced in the reaction.

Carbonyl compounds are electron withdrawing substituents hence they decrease the magnitude of the positive charge and hence decrease the rate of electrophilic aromatic substitution.

gasoline is the chemical that is coming out of the air

Answer:

Multiply the number of moles of butane by its molar mass, 58.12g/mol, to produce the mass of butane. Mass of butane = 18.8g.

Explanation:

Part B:

The mass of water produced when 4.86 g of butane(C4H10) react with excess oxygen is calculated as below

calculate the moles of C4H10 used = mass/molar mass

moles = 4.86g/58 g/mol =0.0838 moles

write a balanced equation for reaction

2 C4H10 + 13 O2 = 8 CO2 + 10 H2O

by use of mole ratio between C4H10 to H2O which is 2:10 the moles of

H20= 0.0838 x10/2 = 0.419 moles of H2O

mass = moles x molar mass

=0.419 molx 18 g/mol = 7.542 grams of water is formed

i can not understand the question. There seems to be missing info

Answer:

The average mass of one atom of arsenic is 74.92 amu, and the mass of one mole of arsenic atoms is 74.92 grams.

The mass of one mole of arsenic atoms is 74.92 grams.

Calculation,

Mass of one mole of arsenic = Mass of 1 arsenic atom × [tex]N_{A}[/tex]

Where, [tex]N_{A}[/tex] = Avogadro's number = 6.022 × [tex]10^{23}[/tex]

Mass of one arsenic atom = 74.921 u

One u =  1.66 × [tex]10^{-27}[/tex] kg

And,

Mass of one mole of arsenic atom = 6.022 ×[tex]10^{23}[/tex] × 74.921 × 1.66 × [tex]10^{-27}[/tex]×[tex]10^{3}[/tex] kg

Mass of one mole of arsenic atom = 74.92 [tex]gmol^{-1}[/tex]

To learn more about molar mass,

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Answer: The total partial pressure of the solution is 131.37 torr.

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 glucose = 46.8 g

Molar mass of glucose = 180 g/mol

Plugging values in equation 1:

[tex]\text{Moles of glucose}=\frac{46.8g}{180g/mol}=0.26 mol[/tex]

Given mass of methanol = 117 g

Molar mass of methanol = 32 g/mol

Plugging values in equation 1:

[tex]\text{Moles of methanol}=\frac{117g}{32g/mol}=3.66 mol[/tex]

Mole fraction is defined as the moles of a component present in the total moles of a solution. It is given by the equation:

[tex]\chi_A=\frac{n_A}{n_A+n_B}[/tex] .....(2)

where n is the number of moles

Putting values in equation 2:

[tex]\chi_{methanol}=\frac{3.66}{0.26+3.66}=0.934[/tex]

Raoult's law is the law used to calculate the partial pressure of the individual gases present in the mixture. The equation for Raoult's law follows:

[tex]p_A=\chi_A\times p_T[/tex] .....(3)

where [tex]p_A[/tex] is the partial pressure of component A in the mixture and [tex]p_T[/tex] is the total partial pressure of the mixture

We are given:

[tex]p_{methanol}=122.7torr\\\chi_{methanol}=0.934[/tex]

Putting values in equation 3, we get:

[tex]122.7torr=0.066\times p_T\\\\p_T=\frac{122.7torr}{0.934}=131.37torr[/tex]

Hence, the total partial pressure of the solution is 131.37 torr.

Answer:

Theory

Explanation:

We are given the statement;

"A helium filled balloon floats."

Now, this statement is an attempt to explain to us why balloons float. And the reason is because they are filled with helium. This is no prediction or law or observation because to find out if the balloons contain helium, there must have been experiments to confirm that.

Thus, the statement in the question is a theory.

Complete question:

What mass of steam initially at 120 ⁰C is needed to warm 200g of water in a 100 g glass container from 20.0 oC to 50.0 ⁰C

Answer:

the initial mass of the steam is 10.82 g

Explanation:

Given;

mass of water, m₁ = 200 g

mass of the glass, m₂ = 100 g

temperature of the steam = 120 ⁰C

initial temperature of the water, 20⁰ C

final temperature of the water, = 50⁰ C

let the mass of the steam = m

specific heat capacity of water c = 1 cal/g ⁰ C

specific heat capacity of glass c₂ = 0.2 cal/g ⁰ C

laten heat of vaporization of steam L = 540 cal/g

Apply principle of conservation energy;

Heat given off by the steam = Heat absorbed by water + heat absorbed by glass

[tex]mc\Delta T_1 + mL + mc\Delta T_2 = m_1c\Delta T_3 + m_2c_2\Delta T_3\\\\mc\Delta T_1 + mL + mc\Delta T_2 = [m_1c + m_2c_2]\Delta T_3[/tex]

m(1) (120 - 100)  +  m(540)  + m(1) (100 - 50) = [200(1)  +  100(0.2)] (50 - 20)

20m + 540m + 50m = 6600

610 m = 6600

m = 6600 / 610

m = 10.82 g

Therefore, the initial mass of the steam is 10.82 g

Answer: The first law of thermodynamics is a statement of conservation of energy.

Explanation:

According to the first law of thermodynamics, heat provided to a system is actually the sum of internal energy and work done by the system or on the system.

Mathematically, [tex]\Delta Q = \Delta U + \Delta W[/tex]

The first law of thermodynamics also means that energy can neither be created nor it can be destroyed. Hence, energy is conserved.

Thus, we can conclude that the first law of thermodynamics is a statement of conservation of energy.

Answer:

Multiply the number of moles of butane by its molar mass, 58.12g/mol, to produce the mass of butane. Mass of butane = 18.8g.

Answer:

The wavelength of a microwave is LONGER than the wavelength if visible light.

Answer:

1, 1, 2, 3

Explanation:

The numbers 1 and 8 both have 1 sig. fig.

The number 13 has 2 sig. figs.

The number 104 has 3 sig. figs.

Answer: There are 8 protons in oxygen-18 and 36 neutrons in copper-65.

Explanation:

An atom contains three sub-atomic particles, that is, protons, neutrons and electrons.

The atomic number is the total number of protons present in an atom. For oxygen-18, the atomic mass is 18.

Atomic mass is the sum of total number of protons and electrons present in an atom. As the atomic number of an oxygen atom is 8 so the number of protons present in oxygen-18 is 8.

The atomic mass of copper is 65 and for a copper atom, the atomic number is 29. Hence, the number of neutrons for copper-65 is as follows.

Atomic mass = no. of protons + no. of neutrons

65 = 29 + no. of neutrons

no. of neutrons = 65 - 29 = 36

Thus, we can conclude that there are 8 protons in oxygen-18 and 36 neutrons in copper-65.

Answer:

e. The lipid bilayer is a permeability barrier for integral membrane proteins.

Explanation:

Integral membrane proteins enter the lipid bilayer by translocon channel which gives nascent polypeptide segment a chance to partition itself into lipid bilayer hydrophobic core.

Answer:

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ASAP PROJECT

Answer:

The kinds of intermolecular forces that are present in each element Kr-Kr.

Explanation:

Since Kr is an inert gas and in atomic form only it is highly stable.

So, Kr gas does not form molecules.

Between the atoms of inert gas, there exist London dispersion forces.

Hence, the intermolecular forces that are present between Kr-Kr atoms is London dispersion forces.

Answer:

Nuclear

Solar

Wind

Hydro

Geo-Thermal

Explanation:

hope it helps

☛ 299,792,458 meters per second.

Answer:

C₂H₃O

Explanation:

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

Carbon (C) = 48.8383%

Hydrogen (H) = 8.1636%

Oxygen (O) = 43.1981%

Empirical formula =?

The empirical formula of the compound can be obtained as follow:

C = 48.8383%

H = 8.1636%

O = 43.1981%

Divide by their molar mass

C = 48.8383 / 12 = 4.07

H = 8.1636 / 1 = 8.1636

O = 43.1981 / 16 = 2.7

Divide by the smallest

C = 4.07 / 2.7 = 2

H = 8.1636 / 2.7 = 3

O = 2.7 /2.7 = 1

Thus, the empirical formula of the compound is C₂H₃O

Answer:

[tex]Q\approx6.4~kJ[/tex]

Explanation:

Quantity of heat required by 10 gram of ice initially warm it from -5°C to 0°C:

[tex]Q_1=m.C_s.\Delta T[/tex]

here;

mass, m = 10 g

specific heat capacity of ice, [tex]C_s=2.09~J.g^{-1}.^{\circ}C^{-1}[/tex]

change in temperature, [tex]\Delta T=(5-0)=5^{o}C[/tex]

[tex]Q_1=10\times2.09\times 5[/tex]

[tex]Q_1=104.5~J[/tex]

Amount of heat required to melt the ice at 0°C:

[tex]Q_2=m.\Delta H_{fus}[/tex]

where, [tex]\Delta H_{fus}=6020~J/mol[/tex]

we know that no. of moles is = (wt. in gram) [tex]\div[/tex] (molecular mass)

[tex]Q_2=\frac{10}{18} \times 6020[/tex]

[tex]Q_2=3344.44~J[/tex]

Now, the heat required to bring the water to 70°C from 0°C:

[tex]Q_3=m.C_L.\Delta T[/tex]

specific heat of water, [tex]C_L=4.18~J/g/^oC[/tex]

change in temperature, [tex]\Delta T=(70-0)=70^oC[/tex]

[tex]Q_3=10\times 4.18\times 70[/tex]

[tex]Q_3=2926~J[/tex]

Therefore the total heat required to warm 10.0 grams of ice at -5.0°C to a temperature of 70.0°C:

[tex]Q=Q_1+Q_2+Q_3[/tex]

[tex]Q=104.5+3344.44+2926[/tex]

[tex]Q=6374.94~J[/tex]

[tex]Q\approx6.4~kJ[/tex]