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Butoxors [25]
3 years ago
5

Explain why thinner containers are more effective for measuring the volume of liquids than shorter or wide containers?

Chemistry
2 answers:
Westkost [7]3 years ago
7 0

Taller and Thinner containers can give a much more accurate reading on what it's holding, while shorter and wider containers leave less space between each marking, and the level of water might look close to one measurement, but could actually be a lot more or less! It's like zooming out on a graph, the volume is thinned out to a more generalized area instead of an accurate way of reading it.

Dmitry_Shevchenko [17]3 years ago
4 0

Answer:THINNER contains more effective for measurin

Explanation:

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You are holding four identical balloons each containing 10.0g of a different gas. The balloon containing which gas is the larges
Vaselesa [24]

Answer:

Hydrogen, H_2

Explanation:

mass of each gas is 10.0 g

number of mole =  mass/ molar mass

number of moles is directly proportional to volume at constant temp and pressure

this implies that the  volume is inversely proportional to molar mass. And Among all the gases in periodic table the molar mass of Hydrogen is the least.

molar mass of H2=2 g/mol

Since, H2 has minimum molar mass then for the same mass of the gases Hydrogen will have maximum volume.

6 0
3 years ago
Arrange the following elements in order of increasing ionization energy
Gnoma [55]

Ionization Trend: First ionization energy will increase left to right across a period and increase bottom to top of a family (column).

A) Sr, Be, Mg are all in column 2 of the periodic table. Based on the first ionization rule above, from increasing to decreasing energy, the order is: Be, Mg, Sr

B) Bi, Cs, Ba are all in the same row of the periodic table. Based on the first ionization rule above, from increasing to decreasing energy, the order is: Bi, Ba, Cs

C) Same rule as above. Order is: Na, Al, S

8 0
3 years ago
2. Classify the following solutions as acidic, basic, or neutral at 25OC.
PilotLPTM [1.2K]

Answer: a) pH = 13.00 : basic

b) [H_3O^+]=1.0\times 10^{-12}: basic

c) pOH = 5.00 : basic

d) [OH^-]=1.0\times 10^{-9}: acidic

Explanation:

pH or pOH is the measure of acidity or alkalinity of a solution.

pH is calculated by taking negative logarithm of hydrogen ion concentration.

pH=-\log [H_3O^+]

Acids have pH ranging from 1 to 6.9 and bases have pH ranging from 7.1 to 14.Neutral substances have pH of 7.

a) pH = 13.00

As pH is more than 7, the solution is basic.

b)  [H_3O^+]=1.0\times 10^{-12}

Putting in the values:

pH=-\log[1.0\times 10^{-12}]

pH=12

As pH is more than 7, the solution is basic.

c) pOH = 5.00

pH+pOH=14.0

pH=14.0-5.00=9.00

As pH is more than 7, the solution is basic.

d) [OH^-]=1.0\times 10^{-9}

Putting in the values:

pOH=-\log[1.0\times 10^{-9}]

pOH=9.00

pH+pOH=14.0

pH=14.0-9.00=5.00

As pH is less than 7, the solution is acidic.

3 0
3 years ago
Suppose that 25,0 mL of a gas at 725 mmHg and 298K is converted to
posledela

The new volume : 21.85 ml

<h3>Further explanation</h3>

Given

V1=25,0 ml

P1=725 mmHg

T1=298K is converted to

T2=273'K

P2=760 mmHg atm

Required

V2

Solution

Combined gas law :

\tt \dfrac{P_1.V_1}{T_1}=\dfrac{P_2.V_2}{T_2}

Input the value :

V2=(P1.V1.T2)/(P2.T1)

V2=(725 x 25 ml x 273)/(760 x 298)

V2=21.85 ml

5 0
3 years ago
For the following electron-transfer reaction:
creativ13 [48]

Answer:

1. The oxidation half-reaction is: Mn(s) ⇄ Mn²⁺(aq) + 2e⁻

2. The reduction half-reaction is: Ag⁺(aq) + 1e⁻ ⇄ Ag(s)  

Explanation:

Main reaction: 2Ag⁺(aq) + Mn(s) ⇄ 2Ag(s) + Mn²⁺(aq)

In the oxidation half reaction, the oxidation number increases:

Mn changes from 0, in the ground state to Mn²⁺.

The reduction half reaction occurs where the element decrease the oxidation number, because it is gaining electrons.

Silver changes from Ag⁺ to Ag.

1. The oxidation half-reaction is: Mn(s) ⇄ Mn²⁺(aq) + 2e⁻

2. The reduction half-reaction is: Ag⁺(aq) + 1e⁻ ⇄ Ag(s)  

To balance the hole reaction, we need to multiply by 2, the second half reaction:

Mn(s) ⇄ Mn²⁺(aq) + 2e⁻

(Ag⁺(aq) + 1e⁻ ⇄ Ag(s)) . 2

2Ag⁺(aq) + 2e⁻ ⇄ 2Ag(s)  

Now we sum, and we can cancel the electrons:

2Ag⁺(aq) + Mn(s) + 2e⁻ ⇄ 2Ag(s) + Mn²⁺(aq) + 2e⁻

4 0
3 years ago
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