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Pani-rosa [81]
2 years ago
7

1. Imagine you have a sealed 20.0 L balloon filled with helium gas at 750 mmHg in the

Chemistry
1 answer:
SVETLANKA909090 [29]2 years ago
6 0

Answer:

17.6510 L

Explanation:

First we should get the number of moles of helium here by Boyle's law

PV=nRT

P=750/760= 0.9868 atm

T=25+273=298 kelvin

R= 0.08206

V= 20L

so

n=PV/RT

n=0.9868×20/0.08206×298

n=0.80707 mol

Then use the same law

V=0.80707×0.08206×263/0.9869=

17.6510L

SO THE VOLUME WILL BE 17.6510 L

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What volume (in L) of a 1.25 M
sleet_krkn [62]

Answer:

V1= 0.305L

Explanation:

To find the initial volume of 1.25M potassium fluoride needed to make tge dilution specified in the question, we can use: C1 × V1 = C2 × V2

Since the question wants the volume in litres, convert 455 mL to L

455/ 1000

= 0.455 L

Now make the substitution

1.25 × V1 = 0.838 × 0.455

Rearrange to make V1 the subject

V1=

\frac{0.838 \times 0.455}{1.25 }  = 0.305

4 0
3 years ago
A science fair volcano bubbles and fizzes. what is taking place?
vaieri [72.5K]
Answer is B. gas formation
7 0
3 years ago
Standard reduction half-cell potentials at 25∘c half-reaction e∘ (v) half-reaction e∘ (v) au3 (aq) 3e−→au(s) 1.50 fe2 (aq) 2e−→f
Zanzabum

<em>K</em> = 5.0 × 10^25

<h2>Part (a). Calculate <em>E</em>° for the reaction </h2>

<em>Step 1.</em> Write the equations for the two half-reactions

2H^(+)(aq) + 2e^(-) → H2(g); _0.00 V

Zn^(2+)(aq) + 2e^(-) → Zn(s); -0.76 V

<em>Step 2.</em> Identify the cathode and the anode

The half-cell with the more negative <em>E</em>° (Zn) is the anode.

<em>Step 3.</em> Calculate <em>E</em>°

Zn(s) → Zn^(2+)(aq) + 2e^(-); _________+0.76 V

2H^(+)(aq) + 2e^(-) → H2(g); __________0.00 V

Zn(s) + 2H^(+)(aq) → Zn^(2+)(aq) + H2(g); +0.76 V

<em>E</em>° = +0.76 V

<h2>Part (b). Calculate <em>K</em> for the reaction </h2>

The relation between <em>E</em>° and <em>K</em> is

<em>E</em>° = (<em>RT</em>)/(<em>nF</em>)ln<em>K </em>

where

<em>R</em> = the universal gas constant: 8.314 J·K^(-1)mol^(-1)

<em>T</em> = the Kelvin temperature

<em>n</em> = the moles of electrons transferred

<em>F</em> = the Faraday constant: 96 485 J·V^(-1)mol^(-1)

Then

0.76 V = [8.314 J·K^(-1)mol^(-1) × 298.15 K]/[2 × 96 485 J·V^(-1)mol^(-1)]ln<em>K</em>

0.76 = 0.012 85 ln<em>K</em>

ln<em>K</em> = 0.76/0.012 85 = 59.16

<em>K</em> =e^59.16 = 5.0 × 10^25

4 0
3 years ago
Find the number of moles of water that can be formed if you have 182 mol of hydrogen gas and 86 mol of oxygen gas.
frutty [35]
Water has a chemical formula of H2O. This means that for every 2 moles of hydrogen and 1 mole of oxygen, one mole of water will be formed.

Note that hydrogen gas and oxygen gas are both biatomic molecules. 

 (1)                   (182 mol H2) x (1 mol H2O/ 1 mol H2) = 182 mol H2O
 (2)                   (86 mol O2) x (2 mol H2O / 1 mol O2) = 172 mol H2O

We choose the smaller number of the two as the answer to this item. Thus, the answer to this question is 172 mol of H2O can be formed out of the given quantities. 
3 0
3 years ago
Which elements are in the same period?
worty [1.4K]

Answer:

Elements that are in the same period have chemical properties that are not all that similar. Consider the first two members of period 3: sodium (Na) and magnesium (Mg). In reactions, they both tend to lose electrons (after all, they are metals), but sodium loses one electron, while magnesium loses two.

Explanation:

(Hoped this helped! :D)

5 0
3 years ago
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