La is element 57 on the periodic table a sample contains 2.82 * 10€25 power atoms of La calculate the amount of LA

Answer:

B

Explanation:

the concentration will be the same disabled

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Answer:

0.2 M

Explanation:

Step 1: Given data

There are different ways to express the concentration of a solution. We will calculate molarity, which is one of the most used.

Step 2: Calculate the moles of sucrose

The molar mass of sucrose is 342.3 g/mol.

15 g × 1 mol/342.3 g = 0.044 mol

Step 3: Calculate the molarity of the solution

Molarity is equal to the moles of solute divided by the liters of solution.

M = 0.044 mol/0.2 L = 0.2 M

Answer:

THE ANSWER IS: copper(I) sulfide.

hope this helped <3

Explanation:

Answer:

MgO

Explanation:

Answer:

sorry i cant give the answer but you can gi end check in answer sheet of this becoz i had same question in exam in chemistry so i revised then i checked the answer if you want to check answer go to www.coachscotchemistry.com there you xan find the answer

Answer: A volume of 20.49 milliliters of sodium formate do you need to measure to make this buffer (assuming the rest is formic acid).

Explanation:

Given: Total volume of the buffer = 21.0 mL

[tex]\frac{[HCOONa]}{[HCOOH]} = 4[/tex] ... (1)

It is assumed that the volume of HCOONa is x. Hence, volume of HCOOH is (21.0 - x) mL.

Hence,

[HCOONa] = Molarity [tex]\times[/tex] Volume

= 0.10 [tex]\times[/tex] x

= 0.1x mmol

Similarly, [HCOOH] = Molarity [tex]\times[/tex] Volume

= 0.10 [tex]\times[/tex] (21.0 - x) mmol

Using equation (1),

[tex]\frac{[HCOONa]}{[HCOOH]} = 4\\\frac{0.1x}{(21.0 - x)} = 4\\0.1x = 84.0 - 4x\\4.1x = 84.0\\x = 20.49 mL[/tex]

As x is the volume of sodium formate. Hence, 20.49 mL of sodium formate is required to make the buffer.

Thus, we can conclude that a volume of 20.49 milliliters of sodium formate do you need to measure to make this buffer (assuming the rest is formic acid).

Answer:

5 atoms form one formula unit of Fe2O3

Explanation:

2 atoms of Fe (Iron)

and 3 atoms of O ( Oxygen)

so total = 3 + 2

= 5

Answer:

fe203 the right answer is thus

Answer:

14.9 g

Explanation:

Step 1: Write the balanced equation

C₄H₁₀ + 6.5 O₂ ⇒ 4 CO₂ + 5 H₂O

Step 2: Calculate the moles corresponding to 9.57 g of C₄H₁₀

The molar mass of C₄H₁₀ is 58.12 g/mol.

9.57 g × 1 mol/58.12 g = 0.165 mol

Step 3: Calculate the moles of H₂O produced from 0.165 moles of C₄H₁₀

0.165 mol C₄H₁₀ × 5 mol H₂O/1 mol C₄H₁₀ = 0.825 mol H₂O

Step 4: Calculate the mass corresponding to 0.825 mol of H₂O

The molar mass of H₂O is 18.02 g/mol.

0.825 mol × 18.02 g/mol = 14.9 g

Answer:

it's B

Explanation:

Write the balanced equation: H2SO4 + 2KOH → K2SO4 +2H2O. So 2(moles KOH) = (moles H2SO4); 2(volume KOH)(concentration KOH) = (volume H2SO4)(concentration H2SO4); 2(40ml)(0.2M) = (30ml)(x); x = 0.53M

The concentration of  H₂SO₄ solution is equal to 0.133 M.

A neutralization reaction can be described as a chemical reaction in which an acid and base react together to form respective salt and water. When a strong acid such as HCl will react with a strong base such as NaOH the salt can be neither acidic nor basic.

When H₂SO₄ (a strong acid) reacts with KOH, the resulting salt will be  K₂SO₄ and water.

H₂SO₄    +  2KOH   → K₂SO₄  +  2H₂O

Given, the concentration of KOH solution = 0.2 M

The volume of the KOH solution = 40 ml = 0.040 ml

The number of moles of KOH, n = M × V = 0.2 × 0.04 = 0.008 mol

The volume of the H₂SO₄ = 30 ml = 0.03 L

The number of moles of H₂SO₄, n = 0.008/2 = 0.004 mol

The concentration of  H₂SO₄ solution = 0.004/0.03 = 0.133 M

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Answer:

The open system evaporates the solvent in the solution

Explanation:

An open system is a system in which exchange of materials and energy can occur. If a TLC set up is left open, then the set up constitutes an open system.

During TLC, the sample is dotted on the plate and inserted into a suitable solvent. The solvent moves up the plate and achieves the required separation of the mixture.

Most of these solvents used used TLC are volatile organic compounds. Therefore, if the TLC set up is left open, the solvent will evaporate leading to poor results after running the TLC experiment.

Answer:

[tex]4.67\times 10^{-4}[/tex]

Explanation:

Given that,

The pH of a certain orange juice is 3.33.

We need to find the +ion concentration.

We know that,

[tex]pH=-log[H^+][/tex]

So,

[tex]3.33=-log[H^+]\\\\\[H^+=10^{-3.33}\\\\=4.67\times 10^{-4}[/tex]

So, the +ion concentraion is equal to [tex]4.67\times 10^{-4}[/tex].

Answer:

Ionic compounds conduct electricity when molten (liquid) or in aqueous solution (dissolved in water), because their ions are free to move from place to place. Ionic compounds cannot conduct electricity when solid, as their ions are held in fixed positions and cannot move.

Explanation:

because their ions are free to move from place to place.

Answer:

AlCl3 + 3NaOH ———>Al(OH)3 + 3NaCl

Answer: The empirical and molecular formula for the given organic compound is [tex]C_9H_{13}O_3N[/tex]

Explanation:

Let the mass of the compound be 100 g

Given values:

% of C = 59.0%

% of H = 7.15%

% of O = 26.20%

% of N = 7.65%

Mass of C = 59.0 g

Mass of H = 7.15 g

Mass of O = 26.20 g

Mass of N = 7.65 g

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)

To formulate the empirical formula, we need to follow some steps:

Molar mass of C = 12 g/mol

Molar mass of H = 1 g/mol

Molar mass of O = 16 g/mol

Molar mass of N = 14 g/mol

Putting values in equation 1, we get:

[tex]\text{Moles of C}=\frac{59.0g}{12g/mol}=4.917 mol[/tex]

[tex]\text{Moles of H}=\frac{7.15g}{1g/mol}=7.15 mol[/tex]

[tex]\text{Moles of O}=\frac{26.20g}{16g/mol}=1.6375 mol[/tex]

[tex]\text{Moles of N}=\frac{7.65g}{14g/mol}=0.546 mol[/tex]

Calculating the mole fraction of each element by dividing the calculated moles by the least calculated number of moles that is 0.546 moles

[tex]\text{Mole fraction of C}=\frac{4.917}{0.546 }=9[/tex]

[tex]\text{Mole fraction of H}=\frac{7.15}{0.546 }=13[/tex]

[tex]\text{Mole fraction of O}=\frac{1.6375}{0.546 }=2.99\approx 3[/tex]

[tex]\text{Mole fraction of N}=\frac{0.546}{0.546 }=1[/tex]

The ratio of C : H : O : N = 9 : 13 : 3 : 1

The empirical formula of the compound becomes [tex]C_9H_{13}O_3N_1=C_9H_{13}O_3N[/tex]

To calculate the molecular formula, the number of atoms of the empirical formula is multiplied by a factor known as valency that is represented by the symbol, 'n'.

[tex]n =\frac{\text{Molecular mass}}{\text{Empirical mass}}[/tex] .....(2)

We are given:

Mass of molecular formula = 183 g/mol

Mass of empirical formula = 183 g/mol

Putting values in equation 2, we get:

[tex]n=\frac{183g/mol}{183g/mol}=1[/tex]

Multiplying this valency by the subscript of every element of empirical formula, we get:

[tex]C_{1\times 9}H_{1\times 13}O_{1\times 3}N_{1\times 1}=C_9H_{13}O_3N[/tex]

Hence, the empirical and molecular formula for the given organic compound is [tex]C_9H_{13}O_3N[/tex]

Answer:

A

Explanation:

Answer:

INCREASE in the difference between the melting point measured and the true melting temperature.

Explanation:

Melting point of a compound is defined as the temperature at which the soils compound changes into liquid at the atmospheric pressure. There are different circumstances that can lead to inaccurate melting point. These include:

--> presence of impurities in the compound,

--> Molecular composition,

--> Force of attraction, and

--> Rapid heating of the compound.

Under the circumstances of rapid heating of the compound, there would be an increase in the melting point range when compared with the true melting point range of the compound.

The higher the heating rate, the more rapid the rise in oven temperature, increasing the difference between the melting point measured and the true melting temperature.

Answer:

1 +-0.05cm

Explanation:

because this is more suitable

Answer:

Aluminum tribromide: AlBr₃, however, it should be just aluminum bromide.

Sulfur dioxide: SO₂.

Beryllium hydride: BeH₂

Magnesium(II) oxide: MgO; however the roman numeral is not used in Mg as it just has one oxidation number.

Copper(II) oxide: CuO.

Calcium sulfate: CaSO₄

Nitric acid: HNO₃.

Explanation:

Hello there!

In this case, it seems that the formulas were not given, however, we can write the correct one for each given compound according to the widely used nomenclature rules as shown below:

Aluminum tribromide: AlBr₃, however, it should be just aluminum bromide.

Sulfur dioxide: SO₂.

Beryllium hydride: BeH₂

Magnesium(II) oxide: MgO; however the roman numeral is not used in Mg as it just has one oxidation number.

Copper(II) oxide: CuO.

Calcium sulfate: CaSO₄

Nitric acid: HNO₃.

Regards!

Answer:

1.207 ××10⁻¹⁷ J

Explanation:

Step 1: Given and data

Step 2: Calculate the energy (E) of the photon

We will use the Planck-Einstein's relation.

E = h × ν

E = (6.626 × 10⁻³⁴ J.s) × ( 1.821 × 10¹⁶ s⁻¹) = 1.207 ××10⁻¹⁷ J

The question is incomplete, the complete question is:

A certain liquid X has a normal freezing point of [tex]0.80^oC[/tex] and a freezing point depression constant [tex]K_f=7.82^oC.kg/mol[/tex] . Calculate the freezing point of a solution made of 81.1 g of iron(III) chloride () dissolved in 850. g of X. Round your answer to significant digits.

Answer: The freezing point of the solution is [tex]-17.6^oC[/tex]

Explanation:

Depression in the freezing point is defined as the difference between the freezing point of the pure solvent and the freezing point of the solution.

The expression for the calculation of depression in freezing point is:

[tex]\text{Freezing point of pure solvent}-\text{freezing point of solution}=i\times K_f\times m[/tex]

OR

[tex]\text{Freezing point of pure solvent}-\text{Freezing point of solution}=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times w_{solvent}\text{(in g)}}[/tex] ......(1)

where,

Freezing point of pure solvent = [tex]0.80^oC[/tex]

Freezing point of solution = [tex]?^oC[/tex]

i = Vant Hoff factor = 4 (for iron (III) chloride as 4 ions are produced in the reaction)

[tex]K_f[/tex] = freezing point depression constant = [tex]7.82^oC/m[/tex]

[tex]m_{solute}[/tex] = Given mass of solute (iron (III) chloride) = 81.1 g

[tex]M_{solute}[/tex] = Molar mass of solute (iron (III) chloride) = 162.2 g/mol

[tex]w_{solvent}[/tex] = Mass of solvent (X) = 850. g

Putting values in equation 1, we get:

[tex]0.8-(\text{Freezing point of solution})=4\times 7.82\times \frac{81.1\times 1000}{162.2\times 850}\\\\\text{Freezing point of solution}=[0.8-18.4]^oC\\\\\text{Freezing point of solution}=-17.6^oC[/tex]

Hence, the freezing point of the solution is [tex]-17.6^oC[/tex]

Answer:

Gains

Explanation:

It gets more electrons

Answer:

The mole is defined as a collection of 6.022 × 1023 particles.

The atomic mass given on a periodic table that is given in grams is the mass of

one mole (6.022 × 1023 particles) of that element

Explanation:

Answer:

Neutron has no charge while electron has a negative charge and proton has a positive charge

Answer:

Identify the indicated protons in the following molecules as unrelated, homotopic, enantiotopic, or diastereotopic. a) Methyls a & b: _________ b) Ha & Hc: ________

Explanation:

Homotopic hydrogens:

Consider two hydrogens in the given molecule and replace one by one with a different atom say for example deuterium, then if the two molecules formed by replacing hydrogens are the same then the two hydrogens are called homotopic hydrogens.

After replacing the two hydrogens with a different atom then, enantiomers are formed then, the two hydrogens are called enantiotopic hydrogens.

After replacing the two hydrogens with a different atom then, diastereomers are formed then, the two hydrogens are called diastereotopic hydrogens.

In the methyl group, select two hydrogens and replace one hydrogen atom with a D-atom name the compound.

Again replace another hydrogen atom with D-atom.

Name it.

If both are the same then, the hydrogens are homotopic and they are shown below:

Hence, they are homotopic protons.

The density (g/mL) for a 20.0-mL sample of a salt solution has a mass of 24.0 g is 1.2 g/ml.

Density is the mass per unit volume. Density is a scalar quantity. It is denoted by d and the symbol for density is given as rho, a Greek symbol. Density is calculated as mass divided by volume.

Mass is the quantity of matter in a physical body. The product of the molar mass of the compound and the moles of the substance are defined as mass.

Volume is the space occupied by a three-dimensional object. Volume is calculated by dividing mass by density.

Given, the 20.0-mL sample of a salt solution, which is the volume.

The mass of the solution is 24.0 g

To calculate the density

mass/volume

24.0 / 20.0 = 1.2 g/ml

Thus, the density of the given salt solution is 1.2 g/ml.

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Answer:

LIBs have had a huge impact on our society. They enabled modern portable electronics such as laptops and mobile phones. And they are now enabling clean and low-carbon transport, be it via electric cars or even flying taxis, and grid-scale storage of renewable energy

Explanation:

Answer:

H2O.....................