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3 years ago

Please solve this for me <3

Chemistry

1 answer:

d1i1m1o1n [39]3 years ago

3 0

Answer: Brainliest Plz

Democritus and Dalton = invisble sphere

Thomson = pudding model

Rutherford = central nucleus with electron son the outside

Explanation:

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When of a certain molecular compound X are dissolved in of dibenzyl ether , the freezing point of the solution is measured to be

kondaur [170]

This question is incomplete, the complete question is;

When 4.28 g of a certain molecular compound X are dissolved in 60.0 g of dibenzyl ether [(C₆H₅CH₂)₂0] , the freezing point of the solution is measured to be -3.2°C . Calculate the molar mass of X.

If you need any additional information on dibenzyl ether, use only what you find in the ALEKS Data resource. Also, be sure your answer has a unit symbol, and is rounded to significant digit.

Answer: molar mass of solute (X) is 88.03 g/mol

Explanation:

Given that;

mass of solute = 4.28 g

mass of solvent = 60.0 g = 0.060 kg (Dibenzyl ether)

depression constant kf = 6.17 °CKg/mol

Freezing Point of solvent T₀ = 1.80°C (Dibenzyl ether)

freezing point of solution Tsol = -3.20°C

Now we know that

Depression in freezing point ΔTf = depression constant kf × molaity m

and (ΔTf = T₀-Tsol)

so T₀ - Tsol = kf × m

we substitute

1.80 - (-3.20) = 6.17 × m

5 = 6.17 × m

m = 5 / 6.17

m = 0.8103 kg/mol

so molaity m = 0.8103 kg/mol

we know that

Molaity of solute m = (mass of solute / M.wt of solute) × ( 1 / mass of solvent in Kg)

solve for molar mass of solute

molar mass of solute = (mass of solute / molaity) × ( 1 / mass of solvent in Kg)

now we substitute

molar mass = (4.28g / 0.8103 kg/mol) × (1 / 0.060kg)

molar mass = ( 5.2839 × 16.66 ) g/mol

molar mass = 88.0297 g/mol ≈ 88.03 g/mol

Therefore molar mass of solute (X) is 88.03 g/mol

3 0

3 years ago

How many atoms are in 10.0 moles of titanium

Tema [17]

The answer should be 479 i could possibly be wrong but that’s what i got because one mole is 47.90 grams

8 0

3 years ago

The three particles that make up atoms are

Anni [7]

A. Protons neutrons and electrons.

Haha those three make up a simple Atom.

6 0

3 years ago

Read 2 more answers

A 50.0 mL sample of 12.0 M HCl is diluted to 200 mL. What is true about the diluted solution?

kobusy [5.1K]

The concentration of the solution reduces and the number of moles of solute isn't affected.

Data;

V1 = 50mL
C1 = 12.0M
V2 = 200mL
C2 = ?

<h3>Facts about the diluted solution</h3>

1. When the solution is diluted, the concentration changes and this time, the concentration reduces.

Using dilution formula

$c_1 v_1 = c_2 v_2\\12 * 50 = c_2 * 200\\c_2 = \frac{600}{200} \\c_2 = 3M$

The concentration of the solution reduces.

2. The number of moles remains the same.

When a solution is diluted, the number of moles remains the same because there's no change in the mass of the solute.

Learn more on concentration of a solution here;

brainly.com/question/2201903

7 0

3 years ago

II. Ionic Equations

mario62 [17]

Answer:

Complete ionic: $\begin{aligned}& \rm 2\, Ag^{+}\, (aq) + 2\, {NO_3}^{-} \, (aq) + Ca^{2+}\, (aq) + 2\, Cl^{-}\, (aq) \\ & \rm \to 2\, AgCl\, (s) + Ca^{2+}\, (aq) + 2\, {NO_3}^{-}\, (aq)\end{aligned}$ .

Net ionic: $\begin{aligned}& \rm Ag^{+}\, (aq) + Cl^{-}\, (aq) \to AgCl\, (s)\end{aligned}$ .

Explanation:

Start by identifying species that exist as ions. In general, such species include:

Soluble salts.
Strong acids and strong bases.

All four species in this particular question are salts. However, only three of them are generally soluble in water: $\rm AgNO_3$ , $\rm CaCl_2$ , and $\rm Ca(NO_3)_2$ . These three salts will exist as ions:

Each $\rm AgNO_3\, (aq)$ formula unit will exist as one $\rm Ag^{+}$ ion and one $\rm {NO_3}^{-}$ ion.
Each $\rm CaCl_2$ formula unit will exist as one $\rm Ca^{2+}$ ion and two $\rm Cl^{-}$ ions (note the subscript in the formula $\rm CaCl_2\!$ .)
Each $\rm Ca(NO_3)_2$ formula unit will exist as one $\rm Ca^{2+}$ and two $\rm {NO_3}^{-}$ ions.

On the other hand, $\rm AgCl$ is generally insoluble in water. This salt will not form ions.

Rewrite the original chemical equation to get the corresponding ionic equation. In this question, rewrite $\rm AgNO_3$ , $\rm CaCl_2$ , and $\rm Ca(NO_3)_2$ (three soluble salts) as the corresponding ions.

Pay attention to the coefficient of each species. For example, indeed each $\rm AgNO_3\, (aq)$ formula unit will exist as only one $\rm Ag^{+}$ ion and one $\rm {NO_3}^{-}$ ion. However, because the coefficient of $\rm AgNO_3\, (aq)\!$ in the original equation is two, $\!\rm AgNO_3\, (aq)$ alone should correspond to two $\rm Ag^{+}\!$ ions and two $\rm {NO_3}^{-}\!$ ions.

Do not rewrite the salt $\rm AgCl$ because it is insoluble.

$\begin{aligned}& \rm 2\, Ag^{+}\, (aq) + 2\, {NO_3}^{-} \, (aq) + Ca^{2+}\, (aq) + 2\, Cl^{-}\, (aq) \\ & \rm \to 2\, AgCl\, (s) + Ca^{2+}\, (aq) + 2\, {NO_3}^{-}\, (aq)\end{aligned}$ .

Eliminate ions that are present on both sides of this ionic equation. In this question, such ions include one unit of $\rm Ca^{2+}$ and two units of $\rm {NO_3}^{-}$ . Doing so will give:

$\begin{aligned}& \rm 2\, Ag^{+}\, (aq) + 2\, Cl^{-}\, (aq) \to 2\, AgCl\, (s)\end{aligned}$ .

Simplify the coefficients:

$\begin{aligned}& \rm Ag^{+}\, (aq) + Cl^{-}\, (aq) \to AgCl\, (s)\end{aligned}$ .

7 0

3 years ago