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alexandr1967 [171]
1 year ago
13

A substance occupies one half of an open container. The atoms of the substance are closely packed but are still able to slide pa

st each other.
What is most likely the phase of the substance?

O: Gas
O: liquid
O: Solid and gas
O: liquid and solid

Pls help ;-;
Chemistry
1 answer:
SCORPION-xisa [38]1 year ago
8 0

For a substance to occupy one half of an open container with the atoms being able to slide past each other, the most likely phase of the substance would be liquid.

Atoms of gases would not occupy just one-half of an open container because atoms of gases will diffuse and spread out from an open container.

Atoms of solids are fixed about a particular position and will most likely not be able to slide past each other in an open container.

Only the atoms of liquid take the shapes of their containers and do not have the capacity to diffuse out ordinarily. They are also able to flow despite the closeness of the atoms and would occasionally slide past each other.

More on states of matter can be found here: brainly.com/question/9402776

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What was the original volume of a gas if heating it from 22k to 85k produced a volume of 3.8 liters?
mel-nik [20]

Answer: the original volume was 0.98L

Explanation:

V1 =?

T1 = 22k

V2 = 3.8L

T2 = 85k

V1 /T1 = V2 /T2

VI/22 = 3.8/85

V1 = 22 x (3.8/85)

V1 = 0.98L

3 0
1 year ago
Upon heating with acid salicylic acid can form a polymer. what is its structure likely to be
Gnesinka [82]
Take a look at the attached picture. This is not the reaction of polymerization of salicylic acid. It just shows the structure of salicylic acid and the structure when it's polymerized. The polymerization is done by connecting the salicylic acids where you produce one molecule of water.

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2 years ago
A solution was made in a 200.00 mL volumetric flask consisting of 18.00 mL of 2.125 M HCl and 0.4104 grams of CaCO3, followed by
____ [38]

Answer:

16.2mL of 0.1111M NaOH are used

Explanation:

<em>volumetric flask of 250.0mL</em>

As first, the HCl reacts with CaCO3 as follows:

2HCl + CaCO3 → H2O + CaCl2 + CO2

<em>Where 2 moles of HCl react with 1 mole of CaCO3</em>

To solve this question we must find the moles of each reactant in order to determine the moles of HCl that remains:

<em>Moles HCl:</em>

0.01800L * (2.125mol / L) = 0.03825 moles HCl

<em>Moles CaCO3 -Molar mass: 100.09g/mol-</em>

0.4104g * (1mol / 100.09g) = 0.00410 moles CaCO3

Moles HCl that reacts:

0.00410 moles CaCO3 * (2 mol HCl / 1 mol CaCO3) =

0.00820 moles HCl

Moles HCl that remains:

0.03825 moles HCl - 0.00820 moles HCl:

<em>0.03005 moles HCl remains </em>

The HCl reacts with NaOH as follows:

HCl + NaOH → H2O + NaCl

That means, to reach the equivalence point, the moles of NaOH that must be added = Moles HCl, in a sample of 15.00mL, the moles of HCl are:

0.03005 moles HCl * (15.0mL / 250.0mL) = 0.001803 moles HCl = Moles NaOH. The volume of 0.1111M NaOH is:

0.001803 moles NaOH * (1L / 0.1111moles) = 0.016L 0.1111M NaOH are required, that is:

<h3>16.2mL of 0.1111M NaOH are used</h3>
6 0
1 year ago
A flask of fixed volume contains 1.0 mole of gaseous carbon dioxide and 88 g of solid carbon dioxide. The original pressure and
lianna [129]

Answer:

The correct option is: C. 250 K

Explanation:

Given: <em><u>Before Sublimation-</u></em>

Initial Temperature: T₁ = 300 K, Initial Pressure: P₁ = 1 atm, Initial number of moles of gas: n₁ = 1 mol, given mass of solid Carbon dioxide: w = 88 g    

<u><em>After Sublimation- </em></u>      

Final Pressure: P₂ = 2.5 atm, Final number of moles of gas: n₂ = ? mol

Final Temperature: T₂ = ? K,            

Also, Volume is constant, Molar mass of Carbon dioxide: m = 44 g/mol

As we know,

<em>The number of moles:</em>

n = \frac {given\: mass\: (w)} {Molar\: mass\: (m)}

<em>So the number of moles of carbon dioxide sublimed:</em>

n = \frac {w}{m} = \frac {88\: g} {44\: g/mol} = 2 mol

<em><u>Therefore, the final number of moles of gas after sublimation:</u></em>

n_{2} = n_{1} + n = 1\: mol + 2\: mol = 3\: mol

<u><em>According to the </em></u><u><em>Ideal gas equation</em></u><u><em>:</em></u>

P.V = n.R.T

or, \frac {P_{1}.V_{1}}{n_{1}.T_{1}} = \frac {P_{2}.V_{2}}{n_{2}.T_{2}} \: \: \: \: \: \: ....equation\: (1)

<em>Since the volume is constant, so the equation (1) can be written as:</em>

\frac {P_{1}}{n_{1}.T_{1}} = \frac {P_{2}}{n_{2}.T_{2}}

\Rightarrow \frac {1\:atm}{1\:mol \times 300\:K} = \frac {2.5\:atm}{3\:mol \times T_{2}}

\therefore T_{2} = \frac {2.5\:atm \times 300\:K \times 1\:mol}{3\:mol \times 1\:atm}

\Rightarrow T_{2} = 250\:K

<u>Therefore, the final temperature: T₂ = 250 K</u>

6 0
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