Glycerol (C3H8O3), also called glycerine, is widely used in the food and pharmaceutical industries. Glycerol is polar and dissolves readily in water and polar organic solvents like ethanol. Calculate the mole fraction of the solvent in a solution that contains 2.51 g glycerol dissolved in 21.10 mL ethanol (CH3CH2OH; density

Answer:

b. temperature difference

Answer:

It uses this term to represent the economic use of land and nature, with minimal degradation and maximum environmental preservation.

Explanation:

Capitalism is an economic system that promotes the maximum economic exploitation of terrestrial resources. It is common for this exploitation to be strongly associated with environmental degradation, destruction of natural habitats, deforestation and death of fauna and flora. With this, we can consider that capitalism is one of the main formulators of environmental degradation, however many professionals have considered the term "capitalism mated with conservation," where terrestrial resources are exploited consciously and with the objective of reducing degradation to the maximum. environmental impact and maximize conservation.

Answer:

0.591 g of magnesium phosphate is the theoretical yield.

Magnesium nitrate is the limiting reactant.

Explanation:

Hello!

In this case, since the balanced reaction turns out:

[tex]3Mg(NO_3)_2+2Na_3PO_4\rightarrow Mg_3(PO_4)_2+6NaNO_3[/tex]

Next, we compute the grams of magnesium phosphate yielded by each reactant, considering the present mole ratios and molar masses:

[tex]m_{Mg_3(PO_4)_2}^{by\ Mg(NO_3)_2}=1.00gMg(NO_3)_2*\frac{1molMg(NO_3)_2}{148.31gMg(NO_3)_2}*\frac{1molMg_3(PO_4)_2}{3molMg(NO_3)_2} *\frac{262.86gMg_3(PO_4)_2}{1molMg_3(PO_4)_2} \\\\m_{Mg_3(PO_4)_2}^{by\ Mg(NO_3)_2}= 0.591gMg_3(PO_4)_2\\\\m_{Mg_3(PO_4)_2}^{by\ Na_3PO_4}=1.00gNa_3PO_4*\frac{1molNa_3PO_4}{163.94gNa_3PO_4}*\frac{1molMg_3(PO_4)_2}{2molNa_3PO_4} *\frac{262.86gMg_3(PO_4)_2}{1molMg_3(PO_4)_2} \\\\m_{Mg_3(PO_4)_2}^{by\ Na_3PO_4} = 0.802gMg_3(PO_4)_2[/tex]

Thus, we infer that the correct theoretical yielded mass is 0.591 g as magnesium nitrate is the limiting reactant for which it produces the fewest grams of product.

However, is not possible to compute the percent yield since no actual yield is given, and must be provided or indicated by the problem or an experiment and it not here, nevertheless, you may compute the percent yield by dividing the actual yield by the theoretical and then multiplying by 100:

[tex]Y=\frac{actual}{0.591g}*100\%[/tex]

Best regards!

Answer:

4.82 moles of Ag.

Explanation:

We'll begin by calculating the number of mole in 450.98 g of Cu(NO₃)₂. This can be obtained as follow:

Molar mass of Cu(NO₃)₂ = 63.5 + 2[14 + (16×3)]

= 63.5 + 2[14 + 48]

= 63.5 + 2[62]

= 63.5 + 124

= 187.5 g/mol

Mass of Cu(NO₃)₂ = 450.98 g

Mole of Cu(NO₃)₂ =?

Mole = mass /Molar mass

Mole of Cu(NO₃)₂ = 450.98 / 187.5

Mole of Cu(NO₃)₂ = 2.41 moles

Next, we shall determine the number of mole of Cu needed to produce 450.98 g (i.e 2.41 moles) of Cu(NO₃)₂. This can be obtained as follow:

Cu + 2AgNO₃ —> Cu(NO₃)₂ + 2Ag

From the balanced equation above,

1 mole of Cu reacted to produce 1 mole of Cu(NO₃)₂.

Therefore, 2.41 moles of Cu will also react to produce 2.41 moles of Cu(NO₃)₂.

Thus, 2.41 moles of Cu is needed for the reaction.

Finally, we shall determine the number of mole of Ag produced from the reaction. This can be obtained as follow:

From the balanced equation above,

1 mole of Cu reacted to produce 2 moles of Ag.

Therefore, 2.41 moles of Cu will react to produce = 2× 2.41 = 4.82 moles of Ag.

Thus, 4.82 moles of Ag were obtained from the reaction.

Answer:

I'm pretty sure its the one that says very little at the beginning but if I get it wrong I'm sorry

The mass allowance of Vitamin C for children aged 4-8 years is equal to 0.025 grams.

A mole can be defined as a standard unit that can be utilized to evaluate the number of entities such as atoms, molecules, ions, or other particular particles in a particular amount of the substance.

The number of elementary entities present in one mole of any chemical substance was found to be equal to 6.023 × 10²³ which is also known as the Avogadro number.

Given, the number of moles of vitamin C = 0.000142 moles

Given, the molecular formula of Vitamin C is C₆H₈O₆.

The mass of one mole of C₆H₈O₆ = 176 g

One mole of Vitamin C has mass = 176 g

0.000142 mol of Vitamin C has mass  = 0.000142×176 = 0.025 g

Therefore, the mass of 0.000142 mol of Vitamin C is 0.025 g.

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Answer:

nice handwrtting

Explanation:

Answer:

C8H16O4

Explanation:

C2H4O= 24+4+16

44

n=molar mass/empirical formula

n=176.21/44

=4

Therefore

Molar Formula= (C2H4O)4=C8H16O4

Answer:

0.73g/cm^3

Explanation:

d=m/v

d=11/15

d=0.73

Answer:

323.22 ml

Explanation:

Given that :

Diameter, d = 6.8cm

Height, h = 8.9cm

V = arh

Recall :

Volume, V = πr²h

Radius, r = diameter / 2 = 6.8 / 2 = 3.4cm

V = π * 3.4^2 * 8.9

V = 323.21961 cm³

Recall:

1ml = 1cm³

Hence,

323.21961 cm³ = 323.21961 ml

Volume = 323.22 ml

Answer:

Kp = Kc (RT) ^(-2)

Explanation:

For the reaction;

N2 (g) + 3H2 ⇄ 2NH3(g)

We can write;

Kc = [NH3]^2/[N2] [H2]^3

But

Kp = pNH3^2/pN2 . PH2^3

To convert from Kc to Kp

Kp = Kc (RT) ^Δn

where Δn is the change in number of moles going from reactants

to products.

For this reaction;

Δn = 2- (3+1) = -2

Kp = Kc (RT) ^(-2)

Answer:

no, and next time take it right

Explanation:

Answer:

44.45 g of tetrahydrofuran.

Explanation:

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

Volume of tetrahydrofuran = 50 mL

Density of tetrahydrofuran = 0.889 g/mL

Mass of tetrahydrofuran =?

Density of a substance is simply defined as the mass of the substance per unit volume of the substance. Mathematically, density is expressed as shown below:

Density = mass / volume

With the above formula, we shall determine the mass of tetrahydrofuran needed. This can be obtained as follow:

Volume of tetrahydrofuran = 50 mL

Density of tetrahydrofuran = 0.889 g/mL

Mass of tetrahydrofuran =?

Density = mass / volume

0.889 = mass / 50

Cross multiply

Mass = 0.889 × 50

Mass of tetrahydrofuran = 44.45 g

Therefore, the student should weigh out 44.45 g of tetrahydrofuran.

Answer:

50 g of S are needed

Explanation:

To star this, we begin from the reaction:

S(s) + O₂ (g) →  SO₂ (g)

If we burn 1 mol of sulfur with 1 mol of oxygen, we can produce 1 mol of sulfur dioxide. In conclussion, ratio is 1:1.

According to stoichiometry, we can determine the moles of sulfur dioxide produced.

100 g. 1mol / 64.06g = 1.56 moles

This 1.56 moles were orginated by the same amount of S, according to stoichiometry.

Let's convert the moles to mass

1.56 mol . 32.06g / mol = 50 g

Answer:

298. 7 K.

Explanation:

Hello!

In this case, since equation we use to compute the heat in a cooling or heating process is:

[tex]Q=mC(T_f-T_i)[/tex]

Whereas we are given the heat, mass, specific heat and initial temperature. Thus, we infer that we need to solve for the final temperature just as shown below:

[tex]T_f=T_i+\frac{Q}{mC}\\\\T_f=1000 K+\frac{-270000J}{1000g*0.385\frac{J}{g*K} } \\\\T_f=298.7 K[/tex]

It is important to notice that the iron release heat as water absorbs it, that is why it is taken negative.

Best regards!

Answer:

84.8 mL

Explanation:

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

Mass of CuNO₃ = 3.53 g

Molarity of CuNO₃ = 0.330 M

Volume of solution =?

Next, we shall determine the number of mole in 3.53 g of CuNO₃. This can be obtained as follow:

Mass of CuNO₃ = 3.53 g

Molar mass of CuNO₃ = 63.5 + 14 + (16×3)

= 63.5 + 14 + 48

= 125.5 g/mol

Mole of CuNO₃ =?

Mole = mass / Molar mass

Mole of CuNO₃ = 3.53 / 125.5

Mole of CuNO₃ = 0.028 moles

Next, we shall determine the volume of the solution. This can be obtained as follow:

Molarity of CuNO₃ = 0.330 M

Mole of CuNO₃ = 0.028 moles

Volume of solution =?

Molarity = mole /Volume

0.330 = 0.028 / Volume

Cross multiply

0.330 × Volume = 0.028

Divide both side by 0.330

Volume = 0.028 / 0.330

Volume = 0.0848 L

Finally, we shall convert 0.0848 L to millilitres (mL). This can be obtained as follow:

1 L = 1000 mL

Therefore,

0.0848 L = 0.0848 L × 1000 mL / 1 L

0.0848 L = 84.8 mL

Therefore, the volume of the solution is 84.8 mL.

Answer:

There are no options provided dude,

But i guess the  answer will be a metal with valency 2 for sure as the subscript given for N in 'M3N2' is 2 so the valency of the metal u need to select will be 2 for sure

It can be magnesium or some other if the provided options in real question has Mg then its the answer

Answer:

The correct answer is A :))

Answer:

definite volume

Explanation:

Both solids and liquids have a definite volume: The difference is that solids have definite shape while liquids do not -liquids take the shape of their containers-.

Having a definite volume is why liquids cannot be compressed, a property that is used in hydraulic systems.

Answer:

Formulas

3.2 Determining Empirical and Molecular Formulas

Learning Objectives

By the end of this section, you will be able to:

Compute the percent composition of a compound

Determine the empirical formula of a compound

Determine the molecular formula of a compound

The previous section discussed the relationship between the bulk mass of a substance and the number of atoms or molecules it contains (moles). Given the chemical formula of the substance, one may determine the amount of the substance (moles) from its mass, and vice versa. But what if the chemical formula of a substance is unknown? In this section, these same principles will be applied to derive the chemical formulas of unknown substances from experimental mass measurements.

Percent Composition

The elemental makeup of a compound defines its chemical identity, and chemical formulas are the most succinct way of representing this elemental makeup. When a compound’s formula is unknown, measuring the mass of each of its constituent elements is often the first step in the process of determining the formula experimentally. The results of these measurements permit the calculation of the compound’s percent composition, defined as the percentage by mass of each element in the compound. For example, consider a gaseous compound composed solely of carbon and hydrogen. The percent composition of this compound could be represented as follows:

%H=mass Hmass compound×100%

%C=mass Cmass compound×100%

If analysis of a 10.0-g sample of this gas showed it to contain 2.5 g H and 7.5 g C, the percent composition would be calculated to be 25% H and 75% C:

%H=2.5g H10.0g compound×100%=25%

%C=7.5g C10.0g compound×100%=75%

EXAMPLE 3.9

Calculation of Percent Composition

Analysis of a 12.04-g sample of a liquid compound composed of carbon, hydrogen, and nitrogen showed it to contain 7.34 g C, 1.85 g H, and 2.85 g N. What is the percent composition of this compound?

Solution

To calculate percent composition, divide the experimentally derived mass of each element by the overall mass of the compound, and then convert to a percentage:

%C=7.34g C12.04g compound×100%=61.0%%H=1.85g H12.04g compound×100%=15.4%%N=2.85g N12.04g compound×100%=23.7%

The analysis results indicate that the compound is 61.0% C, 15.4% H, and 23.7% N by mass.

Check Your Learning

A 24.81-g sample of a gaseous compound containing only carbon, oxygen, and chlorine is determined to contain 3.01 g C, 4.00 g O, and 17.81 g Cl. What is this compound’s percent composition?

ANSWER:

12.1% C, 16.1% O, 71.8% Cl

Determining Percent Composition from Molecular or Empirical Formulas

Percent composition is also useful for evaluating the relative abundance of a given element in different compounds of known formulas. As one example, consider the common nitrogen-containing fertilizers ammonia (NH3), ammonium nitrate (NH4NO3), and urea (CH4N2O). The element nitrogen is the active ingredient for agricultural purposes, so the mass percentage of nitrogen in the compound is a practical and economic concern for consumers choosing among these fertilizers. For these sorts of applications, the percent composition of a compound is easily derived from its formula mass and the atomic masses of its constituent elements. A molecule of NH3 contains one N atom weighing 14.01 amu and three H atoms weighing a total of (3 × 1.008 amu) = 3.024 amu. The formula mass of ammonia is therefore (14.01 amu + 3.024 amu) = 17.03 amu, and its percent composition is:

%N=14.01amu N17.03amuNH3×100%=82.27%%H=3.024amu H17.03amuNH3×100%=17.76%

This same approach may be taken considering a pair of molecules, a dozen molecules, or a mole of molecules, etc. The latter amount is most convenient and would simply involve the use of molar masses instead of atomic and formula masses, as demonstrated Example 3.10. As long as the molecular or empirical formula of the compound in question is known, the percent composition may be derived from the atomic or molar masses of the

458 g H₂SO₄

Math

Pre-Algebra

Order of Operations: BPEMDAS

Chemistry

Atomic Structure

Stoichiometry

Step 1: Define

4.67 mol H₂SO₄

Step 2: Identify Conversions

[PT] Molar Mass of H - 1.01 g/mol

[PT] Molar Mass of S - 32.07 g/mol

[PT] Molar Mass of O - 16.00 g/mol

Molar Mass of H₂SO₄ - 2(1.01) + 32.07 + 4(16.00) = 98.09 g/mol

Step 3: Convert

Step 4: Check

Follow sig fig rules and round. We are given 3 sig figs.

458.08 g H₂SO₄ ≈ 458 g H₂SO₄

Answer:

a period

Explanation:

Answer: The row of elements across the periodic table is called "periods".

Explanation:

In the Periodic Table, there are seven rows of elements, which is called periods.

Answer: 1and 4

Explanation: iron is an element not an alloy. An ionic lattice is not bonded covalently.

Answer:

The conditions are

1) Small enough

2) Electronegative atom

3) highly electronegative

4) lone pair of electrons

The correct statement therefore is

It needs to be small enough not to bump into other atoms when approaching the 1s orbital of the hydrogen, it needs to carry at least one electronegative atom, it needs to be highly electronegative enough to create a delta on the connected hydrogen, and it needs to have at least one lone pair of electrons.

Explanation:

Hydrogen bonding is a type of intermolecular bond that occurs between the partial positive charge (delta) on a hydrogen atom bonded to a small highly electronegative element (like nitrogen, oxygen or fluorine) and the free electrons on another electronegative element of another molecule.

The hydrogen atom with the partial positive charge (delta) is known as the hydrogen bond donor, while the electronegative element, carrying lone electrons is called the hydrogen bond acceptor.

Let's take a deeper look at these terms:

1) Hydrogen bond donor

Using water (H₂O) as an example, the high electronegativity of the oxygen atom covalently bonded to the hydrogen atom draws the lone electron in the 1s orbital of the hydrogen atom, creating a partial positive charge (d⁺) on the hydrogen atom. This is what happens within one water molecule

2) Hydrogen bond acceptor

When two or more molecules of water interact, the partial positive charge (d⁺) on the hydrogen atom of one molecule, is attracted to the valence or free electrons on the oxygen atom of a nearby molecule of water thus creating a dipole-dipole intermolecular bond known as a hydrogen bond.

For the hydrogen bond to be effective, the electronegative atom bonded to the hydrogen acting as the hydrogen bond donor in the first water molecule needs to be small enough so as not to disrupt the 1s orbital of the hydrogen atom. The smaller the size of the electronegative atom, the stronger the partial negative charge created on the hydrogen atom.

The valence or free pair of electrons on the electronegative (oxygen) atom of the second molecule of water (hydrogen bond acceptor) is what attracts the partial positive charge on the hydrogen atom to create the hydrogen bond

Answer:

Empirical formula

Explanation:

The empirical formula of a compound is the simplest whole number ratio of atoms of each element in the compound. It is determined using data from experiments and therefore empirical.

Answer: Thermal Equilibrium

Explanation:

Answer:

temperature  

Explanation:

I think that's it i'm sorry if i'm wrong

Answer:

[H2]2[S2][H2S]2Kc=[H2]2[S2][H2S]2

Explanation:

2H2S(g)⇋2H2(g)+S2(g)2H2S(g)⇋2H2(g)+S2(g)

The equilibrium constant expression in terms of concentrations is:

Kc=[H2]2[S2][H2S]2Kc=[H2]2[S2][H2S]2.

The equilibrium expression for the given reaction can be written in terms of equilibrium constant which is the ratio of power of molar concentration of the product to the product of power of molar concentration of the reactants.

Equilibrium is a state for a reversible reaction where, the rate of forward reaction is equal to the rate of backward reaction. The rate of a reaction is the rate of decrease in the concentration of reactants or the rate of increase in the concentration of the products.

The given reaction at equilibrium state is written as:

[tex]\rm 2H_{2}S (g)\leftrightharpoons 2H_{2} (g)+ S_{2}(g)[/tex]

The equilibrium constant Kb is ratio of power of molar concentration of the product to the product of power of molar concentration of the reactants.

[tex]Kb = \rm \frac{[H_{2}S]^{2}}{[H_{2}]^{2} [S_{2}]}[/tex]

The rate of the reaction will be r = Kb [H₂]² [S₂].

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