explain observation made when anhydrous calcium chloride and anhydrous copper (ii) sulphate are separately exposed to the atmosphere for about two days​

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Answer:

a neutral solution has a pH of 7.

ph=-log(H+)

Explanation:

hence (H+)

=10⁷mol/dm³

The molar mass of the unknown nonelectrolyte compound is

Using the formula;

∆T = K m i

Where;

K = freezing point depression constant

m = molality of the solution

i = Van't Hoft factor

Note that i = 1 since the compound is a nonelectrolyte.

To find molality;

Number of moles = 0.520 g/Molar mass

Let the molar mass of the unknown compound be MM

Number of moles = 0.520 g/MM

Number of kilograms of solvent = 4.12 g/1000 = 0.00412 Kg

Molality = 0.520 g/MM * 1/0.00412 Kg

Freezing point depression is 4.20 °C

To find the molar mass of the compound;

4.20 °C = 3.90 °C/m *  0.520 g/MM * 1/0.00412 Kg

4.20  = 492.23/MM

MM = 492.23/4.20

MM = 117.19 g/mol

Learn more; https://brainly.com/question/14762341

Answer:

2. 181.25 K.

3. 0.04 atm.

Explanation:

2. Determination of the temperature.

Number of mole (n) = 2.1 moles

Pressure (P) = 1.25 atm

Volume (V) = 25 L

Gas constant (R) = 0.0821 atm.L/Kmol

Temperature (T) =?

The temperature can be obtained by using the ideal gas equation as illustrated below:

PV = nRT

1.25 × 25 = 2.1 × 0.0821 × T

31.25 = 0.17241 × T

Divide both side by 0.17241

T = 31.25 / 0.17241

T = 181.25 K

Thus, the temperature is 181.25 K.

3. Determination of the pressure.

Number of mole (n) = 10 moles

Volume (V) = 5000 L

Temperature (T) = –10 °C = –10 °C + 273 = 263 K

Gas constant (R) = 0.0821 atm.L/Kmol

Pressure (P) =?

The pressure can be obtained by using the ideal gas equation as illustrated below:

PV = nRT

P × 5000 = 10 × 0.0821 × 263

P × 5000 = 215.923

Divide both side by 5000

P = 215.923 / 5000

P = 0.04 atm

Thus, the pressure is 0.04 atm

Answer:

Examples of physical change include changes in the size or shape of matter. Changes of state—for example, from solid to liquid or from liquid to gas—are also physical changes. Some of the processes that cause physical changes include cutting, bending, dissolving, freezing, boiling, and melting.

Explanation:

Answer:

205.12 atm

Explanation:

Using the ideal gas law equation:

PV = nRT

Where;

P = pressure (atm)

V = volume (L)

R = 0.0821 Latm/perK)

T = temperature (K)

n = number of moles (mol)

According to the information in this question;

P = ?

V = 34.25 mL = 34.25 ÷ 1000 = 0.03425L

n = 0.215 mol

T = 125.0°C = 125 + 273 = 398K

Using PV = nRT

P = nRT ÷ V

P = (0.215 × 0.0821 × 398) ÷ (0.03425)

P = 7.025 ÷ 0.03425

P = 205.12 atm

Answer:

CH₂

Explanation:

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

Mass of compound = 1 g

Mass of CO₂ = 3.14 g

Mass of H₂O = 1.29 g

Empirical formula =?

Next, we shall determine the mass of Carbon and hydrogen present in the compound. This can be obtained as follow:

For Carbon, C:

Mass of CO₂ = 3.14 g

Molar mass of CO₂ = 12 + (2×16)

= 12 + 32

= 44 g/mol

Molar mass of C = 12 g/mol

Mass of C =?

Mass of C = molar mass of C/ Molar mass of CO₂ × Mass of CO₂

Mass of C = 12/44 × 3.14

Mass of C = 0.86 g

For hydrogen, H:

Mass of C = 0.86 g

Mass of compound = 1 g

Mass of H =?

Mass of H = (Mass of compound) – (mass of C)

Mass of H = 1 – 0.86

Mass of H = 0.14 g

Finally, we shall determine the empirical formula of the cyclopropane. This can be obtained as follow:

Mass of C = 0.86 g

Mass of H = 0.14 g

Divide by their molar mass

C = 0.86 / 12 = 0.07

H = 0.14 / 1 = 0.14

Divide by the smallest

C = 0.07 / 0.07 = 1

H = 0.14 / 0.07 = 2

Thus, the empirical formula of cyclopropane is CH₂

[tex]mass \: of \: the \: car \: = 1250 \: kg \\ \\ velocity \: of \: the \: car \: is \: = 2.3m/s \\ \\ formula \: to calculate \: momentum \: \\ \\ \: \: \: \: p = mv \: \: \: (m = mass \: and \: v = velocity) \\ \\ momentum \: = 1250 \times 2.3 = 2875m/s[/tex]

Answer:

Part (1a): The number of moles are 2.14 moles.

Part (1b): The number of moles are 12.2 moles.

Part (1c): The number of moles are 0.545 moles.

Part 2: The number of molecules are 1.38 × 10²⁵.

Part 3: The number of moles are 5.65 moles.

Part 4: The number of atoms are 1.81 × 10²⁴.

Part 5: The number of molecules are 1.20 × 10²⁶.

Explanation:

According to the mole concept:

1 mole of substance contains 6.022 × 10²³ number of atoms or molecules.

Part (1a): 1.29 × 10²⁴ hydrogen atoms in H

As, 6.022 × 10²³ number of hydrogen atoms present in 1 mole

So, 1.29 × 10²⁴ number of hydrogen atoms present in [tex]\frac{1.29\times 10^{24}}{6.022\times 10^{23}}=2.14[/tex] mole

Part (1b): 7.36 × 10²⁴ oxygen atoms

As, 6.022 × 10²³ number of oxygen atoms present in 1 mole

So, 7.36 × 10²⁴ number of oxygen atoms present in [tex]\frac{7.36\times 10^{24}}{6.022\times 10^{23}}=12.2[/tex] mole

Part (1c): 3.28 × 10²³ Na atoms

As, 6.022 × 10²³ number of Na atoms present in 1 mole

So, 3.28 × 10²³ number of Na atoms present in [tex]\frac{3.28\times 10^{23}}{6.022\times 10^{23}}=0.545[/tex] mole

Part 2:

As, 1 mole of oxygen contains 6.022 × 10²³ number of molecules

So, 23.0 moles of oxygen contains 23.0 × 6.022 × 10²³ = 1.38 × 10²⁵ number of molecules

Part 3:

As, 6.022 × 10²³ number of H₂SO₄ molecules present in 1 mole

So, 3.4 × 10 number of H₂SO₄ molecules present in [tex]\frac{3.4\times 10}{6.022\times 10^{23}}=5.65[/tex] mole

Part 4:

As, 1 mole of Na contains 6.022 × 10²³ number of atoms

So, 3.0 moles of Na contains 3.0 × 6.022 × 10²³ = 1.81 × 10²⁴ number of atoms

Part 5:

As, 1 mole of O₃ contains 6.022 × 10²³ number of molecules

So, 200 moles of O₃ contains 200 × 6.022 × 10²³ = 1.20 × 10²⁶ number of molecules

Answer:

approximately 1.772 grams

Explanation:

molecular mass of AlCl3 is 132 g per mole and of Al(OH)3 is 78 g per mole

the reaction is

AlCl3 + 3 NaOH ---> Al(OH)3 + 3 NaCl

from the reaction it is clear that 1 mole AlCl3 makes 1 mole Al(OH)3

implies 132g AlCl3 gives 78g Al(OH)3

Implies 3g AlCl3 gives

[tex]3 \times \frac{132}{78} = 1.772 \: g \: al(oh)3[/tex]

Answer:

To produce a solution that frees at -50°C we need 1668.5 g of ethylene glycol

Explanation:

Formula for freezing point depression is:

Freezing T° of pure solvent - Freezing T° of solution = Kf . m . i

Our solute is the ethylene glycol, then water is our solvent.

As ethylene glycol is a non ionic compound, i = 1 (Van't Hoff factor, numbers of ions dissolved)

Kf = Cryoscopic constant. For water is 1.86 °C/m

We replace data:

0°C - (-50°C) = 1.86 °C/m . m . 1

To determine grams of ethylene glycol needed, we need to find m (molality)

50°C / 1.86 m/°C = m → 26.8 mol/kg

As this moles of solute are contained in 1kg, definetely we need 26.8 moles of ethylene glycol.

To find the answer, we convert moles to mass:

26.8 mol . 62.07g / mol = 1668.5 g

Answer:

All ionic compounds with alkali metals are soluble.

Explanation:

Certain elements always form soluble substances. Most of those elements are alkali metals/group 1 metals.

Answer:

This statement is true.........

Answer:

i think true

Explanation:

Answer:

A- 286.3L

Explanation:

Using the ideal gas law equation;

PV = nRT

Where;

P = pressure (atm)

V = volume (L)

n = number of moles (mol)

R = gas law constant (0.0821 Latm/molK)

T = temperature (K)

According to the information given in this question;

P = 1.14atm

V = ?

n = 15moles

T = 265K

Using PV = nRT

V = nRT ÷ P

V = (15 × 0.0821 × 265) ÷ (1.14)

V = 326.35 ÷ 1.14

V = 286.27 L

V = 286.3L

Answer:

use grahamns law to get the answer

Answer:

it may also become impure,I think.

Answer:

22

Explanation:

3x4+  5x2

Answer:

0.0286M

Explanation:

Using the formula as follows:

CaVa = CbVb

Where;

Ca = concentration of HCl acid (M)

Cb = concentration of NaOH base (M)

Va = volume of HCl acid (mL)

Vb = volume of NaOH base (mL)

According to the information provided in this question;

Ca = 0.10 M

Cb = ?

Va = 10mL

Vb = 35mL

Using CaVa = CbVb

Cb = CaVa/Vb

Cb = (0.10 × 10) ÷ 35

Cb = 1 ÷ 35

Cb = 0.0286M

Answer:

Both Arrhenius Base and Bronsted Lowry Base

Both Arrhenius Acid and Bronsted Lowry Acid

Arrhenius Acid Only

Arrhenius Base Only

Bronsted Lowry Acid Only

Bronsted Lowry Base Only

Explanation:

Answer:

Metilaminas

Explanation:

Las metilaminas a menudo están presentes en los tejidos corporales de organismos marinos como crustáceos, moluscos y en todos los peces marinos como N-óxido de trimetilamina, donde actúan para estabilizar la proteína que se desestabiliza por la presión y también se cree que actúa como un depresor del punto de congelación. en peces polares

Las metilaminas tienen olor a pescado podrido, basura o huevos podridos, de modo que una acumulación de metilamina en el cuerpo da como resultado la liberación de un olor a pescado en la amplitud, el sudor y la orina de un individuo

Answer:

this is a required answer. look it once.

Answer:

See explanation

Explanation:

The elements in group form univalent positive ions and element in group 17 form univalent negative ions. Hence, when a group 1 element reacts with a group 17 element, a compound of the sort MX is formed. Hence, when a group 1 element reacts with bromine, a salt is formed with the general formula MBr.

Elements of group 1 are highly electro positive metals. They react with water to form the metal hydroxide and release hydrogen gas. Hence, when group 1 elements react with water, hydrogen gas is released.

A group 1 element forms a univalent positive ion while a group 16 element forms a divalent negative ion. Hence, when a groups 1 element reacts with oxygen, the compound formed must have the general formula M2O.

The reactivity of group 1 metal increases down the group hence Cs is the most reactive group 1 element.

Lithium displays a slightly different chemistry from other group 1 elements because of its small size.

Answer:

Heterogeneous mixture...is the answer

The dude above me is right. I thought he was wrong, so I put A but it turns out he was right!

Answer:

is your answe b?

Explanation:

Im really asking

Answer:

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Explanation:

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Answer:

Explanation:

Ultimately, milk can be described as both. The actual description depends on the individual and/or company involved. For example, in many places milk if extracted from cows, bottled, and delivered to the end consumer. In this case, milk would be considered as natural because nothing has been done to change its nutritional composition. This changes when companies begin to pasteurise and homogenize the milk so that it has a longer shelf life and create different variations. The milk in this case no longer has the same nutritional composition and has been altered, and is therefore considered as processed.