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nasty-shy [4]
3 years ago
8

Classify the following mixturesin their proper column​

Chemistry
1 answer:
PolarNik [594]3 years ago
8 0

Answer:

homogeneous mixtures: iron,alcohol,zonrox,wine.

heterogeneous mixtures: smoke,batchoy,spaghetti,halo halo,book,clothes.

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A piece of metal with a mass of 150 g is placed in a 50 mL graduated cylinder. The water level rises from 20 mL to 45mL. What is
Flauer [41]

Answer:

25 mL

Explanation:

The water level increased from 20mL to 45mL

That is the volume of the metal

45 - 20 = 25

The volume of the metal was 25 mL

Density = mass / volume

Density = 150 / 25

Density = 6 grams / mL

7 0
3 years ago
Questions:<br> 1. Why are atomic masses on the periodic table decimal numbers?
fomenos

Answer:

The atomic mass on the periodic table is a decimal number because it is an average of the different isotopes of an element.

P.S.

An isotope is each of two or more forms of the same element that contain equal numbers of protons but different numbers of neutrons in their nuclei, and differ in relative atomic mass but not in chemical properties; in particular, a radioactive form of an element.

4 0
2 years ago
Suppose you have just added 200.0 ml of a solution containing 0.5000 moles of acetic acid per liter to 100.0 ml of 0.5000 M NaOH
uranmaximum [27]

Answer:

The final pH is 3.80

Explanation:

Step 1: Data given

Volume of acetic acid = 200.0 mL = 0.200 L

Number of moles acetic acid = 0.5000 moles

Volume of NaOH = 100.0 mL = 0.100 L

Molarity of NaOH = 0.500 M

Ka of acetic acid = 1.770 * 10^-5

Step 2: The balanced equation

CH3COOH + NaOH → CH3COONa + H2O

Step 3: Calculate moles

moles = molarity * volume

Moles NaOH = 0.500 M * 0.100 L

Moles NaOH = 0.0500 moles

Step 4: Calculate the limiting reactant

For 1 mol CH3COOH we need 1 mol NaOH to produce 1 mol CH3COONa and 2 moles H2O

NaOH is the limiting reactant. It will completely be consumed (0.0500 moles). CH3COOH is in excess. There will react 0.0500 moles . There will remain 0.500 - 0.0500 = 0.450 moles

There will be produced 0.0500 moles CH3COONa

Step 5: Calculate the total volume

Total volume = 200.0 mL + 100.0 mL = 300.0 mL

Total volume = 0.300 L

Step 6: Calculate molarity

Molarity = moles / volume

[CH3COOH] = 0.450 moles / 0.300 L

[CH3COOH] = 1.5 M

[CH3COONa] = 0.0500 moles / 0.300 L

[CH3COONa]= 0.167 M

Step 7: Calculate pH

pH = pKa + log[A-]/ [HA]

pH = -log(1.77*10^-5) + log (0.167/ 1.5)

pH = 4.75 + log (0.167/1.5)

pH = 3.80

The final pH is 3.80

7 0
3 years ago
What would happen if an alkali metal was combined with a halogen?explain ​
RideAnS [48]
All alkail metals react with halogens vigorously to produce salt. The alkali metal would loose an electron to form and ion with the halogen
7 0
3 years ago
Which of the following reactions have a positive ΔSrxn? Check all that apply.
PolarNik [594]

Answer:

The reactions that have a <em>positive ΔS rxn </em>are the first and the fourth choices:

  • <em>2A(g) + B(s) → 3C(g)</em>

  • <em>2A(g) + 2B(g) → 5C(g)</em>

Explanation:

<em>ΔS rxn </em>is the change of entropy of the chemical reaction.

ΔS rxn = S after reaction - S before reaction.

Therefore, a positive ΔS rxn  means that the entropy after the reaction is greater than the entropy before the reaction.

You may use some assumptions to predict whether a reaction will lead an increase or decrease of the entropy.

First, assume that all the non-shown conditions, such as temperature and pressure, are constant.

Under that assumption, and from the meaning of entropy as a measure of the disorder or randomness of a system you can predict the sign of the change of entropy.

  • <em><u>2A(g) + B(s) → 3C(g)</u></em>

        1)  The solid compounds, B(s) in this case, are very ordered and so they have low entropy.

        2) Gas molecules are highly disordered (scattered), and the greater the number of molecules of the gas the larger the entropy, S).

Hence, since the product side shows 3 gas molecules and the reactant side shows 2 gas molecules and 1 solid molecule, you predict that the products have a larger entropy than the reactants, meaning an increase in entropy: <em>ΔS rxn is positive.</em>

  • <em><u>2A(g) + B(g) → C(g)</u></em>

Using the same reasoning, 3 gas molecules in the  reactant side have more entropy than 1 molecule in the product side, and so the reaction leads to a decrease in the entropy: ΔS rxn is negative

  • <u><em>A(g) + B(g) → C(g)</em></u>

Again, 2 gas molecules in the  reactant side have more entropy than 1 molecule in the product side, and so the reaction leads to a decrease in the entropy: ΔS rxn is negative

  • <u><em>2A(g) + 2B(g) → 5C(g)</em></u>

With the same reasoing, 5 molecules in the product side, lets you predict that will have more entropy than 4 molecules in the reactant side, and, the entropy will increase: <em>ΔS rxn is positive.</em>

6 0
3 years ago
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