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2 years ago

When you combine chocolate powder and milk, the result is best describe as?

Chemistry

2 answers:

Butoxors [25]2 years ago

4 0

Answer:

suspension solution

Explanation:

Suspension solution -

The mixture of two or more substance , to obtain a solution , where the constituents can be observed with naked eye .

Hence , milk and chocolate powder leads a homogeneous solution , but after keeping it for some time , the chocolate particles will settle down , making a suspension solution .

Hence , the answer is suspension solution .

NeX [460]2 years ago

3 0

Answer:

The answer to your question would be substance, but chocolate power mixed into milk would be more of a suspension.

Explanation:

Neither chocolate powder nor milk are elements. They are both complex molecules. Their mixture will not result in the formation of a compound since no chemical reaction will take place.

The molecules of the chocolate powder will simply intermingle with the fatty molecules of the milk to form the substance.

When thoroughly mixed the solution will become homogeneous so there will be no lumps of chocolate power visible. But after time, the chocolate will become visible at the bottom of the clear container in which we asked you to prepare the mixture.

HOPE THIS HELPS :)

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describes a type of reaction in which negative ions in two compounds switch places , forming two new compounds

Nadusha1986 [10]

Answer:

A double displacement reaction or double replacement reaction.

Explanation:

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3 years ago

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Novay_Z [31]

The correct answer is letter C

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2 years ago

CO2(g)+CCl4(g)⇌2COCl2(g) Calculate ΔG for this reaction at 25 ∘C under these conditions: PCO2PCCl4PCOCl2===0.140 atm0.185 atm0.7

padilas [110]

Answer: The $\Delta G$ for the reaction is 54.425 kJ/mol

Explanation:

For the given balanced chemical equation:

$CO_2(g)+CCl_4(g)\rightleftharpoons 2COCl_2(g)$

We are given:

$\Delta G^o_f_{CO_2}=-394.4kJ/mol\\\Delta G^o_f_{CCl_4}=-62.3kJ/mol\\\Delta G^o_f_{COCl_2}=-204.9kJ/mol$

To calculate $\Delta G^o_{rxn}$ for the reaction, we use the equation:

$\Delta G^o_{rxn}=\sum [n\times \Delta G_f(product)]-\sum [n\times \Delta G_f(reactant)]$

For the given equation:

$\Delta G^o_{rxn}=[(2\times \Delta G^o_f_{(COCl_2)})]-[(1\times \Delta G^o_f_{(CO_2)})+(1\times \Delta G^o_f_{(CCl_4)})]$

Putting values in above equation, we get:

$\Delta G^o_{rxn}=[(2\times (-204.9))-((1\times (-394.4))+(1\times (-62.3)))]\\\Delta G^o_{rxn}=46.9kJ=46900J$

Conversion factor used = 1 kJ = 1000 J

The expression of $K_p$ for the given reaction:

$K_p=\frac{(p_{COCl_2})^2}{p_{CO_2}\times p_{CCl_4}}$

We are given:

$p_{COCl_2}=0.735atm\\p_{CO_2}=0.140atm\\p_{CCl_4}=0.185atm$

Putting values in above equation, we get:

$K_p=\frac{(0.735)^2}{0.410\times 0.185}\\\\K_p=20.85$

To calculate the gibbs free energy of the reaction, we use the equation:

$\Delta G=\Delta G^o+RT\ln K_p$

where,

$\Delta G$ = Gibbs' free energy of the reaction = ?

$\Delta G^o$ = Standard gibbs' free energy change of the reaction = 46900 J

R = Gas constant = $8.314J/K mol$

T = Temperature = $25^oC=[25+273]K=298K$

$K_p$ = equilibrium constant in terms of partial pressure = 20.85

Putting values in above equation, we get:

$\Delta G=46900J+(8.314J/K.mol\times 298K\times \ln(20.85))\\\\\Delta G=54425.26J/mol=54.425kJ/mol$

Hence, the $\Delta G$ for the reaction is 54.425 kJ/mol

7 0

3 years ago

What is the molarity of a solution that contains 0.082 mol KI in a 2.03 L solution?

astraxan [27]

Answer:

molarity of the KI solution = 0.04 mol/L

Explanation:

Molarity (M) is the concentration of a solution expressed as the number of moles of solute per liter of solution.

The law that we can applied to calculate the M is:

M = n / V

n - number of moles

V- volume of the solution (liters)

Then insert in the equation the values from the question;

M = 0.082 mol / 2.03 L = 0.04 mol/L

4 0

3 years ago

What is the volume of 40.0 grams of argon gas at STP ?

MrRa [10]

Answer:

24.9 L Ar

General Formulas and Concepts:

Atomic Structure

Reading a Periodic Table
Moles
STP (Standard Conditions for Temperature and Pressure) = 22.4 L per mole at 1 atm, 273 K

Aqueous Solutions

States of Matter

Stoichiometry

Using Dimensional Analysis

Explanation:

Step 1: Define

[Given] 40.0 g Ar

[Solve] L Ar

Step 2: Identify Conversions

[PT] Molar Mass of Ar - 39.95 g/mol

[STP] 22.4 L = 1 mol

Step 3: Convert

[DA] Set up: $\displaystyle 40.0 \ g \ Ar(\frac{1 \ mol \ Ar}{39.95 \ g \ Ar})(\frac{22.4 \ L \ Ar}{1 \ mol \ Ar})$
[DA] Divide/Multiply [Cancel out units]: $\displaystyle 24.9235 \ L \ Ar$

Step 4: Check

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

24.9235 L Ar ≈ 24.9 L Ar

5 0

2 years ago