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Lina20 [59]
2 years ago
13

Is the sweetened iced tea an electrolyte mixture?

Chemistry
2 answers:
klio [65]2 years ago
5 0
Yes it is an electrolyte mixture
Arada [10]2 years ago
3 0

Answer:

Yes it is

Explanation:

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nataly862011 [7]

Scientific Notation Is Basically The Shorthand For Writing Numbers. Scientists Use It When They Want To Write Numbers That Are Very Big Or Very Small.  

4 0
3 years ago
Read 2 more answers
A sample of table sugar (sucrose, C12H22O11) has a mass of 1.202 g.
Archy [21]

Answer:

a) 0.003512 moles

b) Moles C= 0.04214 moles carbon

Moles H = 0.07726 moles hydrogen

Moles O = 0.03863 moles of oxygen

c) C atoms = 2.54 *10^22 carbon atoms

H atoms = 4.65 *10^22 hydrogen atoms

O atoms = 2.33 *10^22 oxygen atom

Explanation:

Step 1: Data given

Mass of sucrose = 1.202 grams

Molar mass of sucrose = 342.3 g/mol

Step 2: Calculate moles of sucrose

Moles sucrose = Mass sucrose / molar mass sucrose

Moles sucrose = 1.202 grams / 342.3 g/mol

Moles sucrose = 0.003512 moles

Step 3: Calculate moles of each element

For 1 mol of C12H22O11 we have 12 moles of carbon, 22 moles of hydrogen and 11 moles of oxygen

Moles C: 12*0.003512 = 0.04214 moles carbon

Moles H: 22* 0.003512 = 0.07726 moles hydrogen

Moles O: 11* 0.003512 = 0.03863 moles of oxygen

Step 4: Calculate the number of atoms

C atoms = 6.022 *10^23 / mol * 0.04214 moles = 2.54 *10^22 atoms carbon

H atoms = 6.022 * 10^23 / mol * 0.07726 moles = 4.65 *10^22 atoms H

O atoms = 6.022 * 10^23 / mol * 0.03863 moles = 2.33 *10^22 atoms O

0.003512 moles of sucrose containse 6.022 *10^23 * 0.003512 = 2.11 * 10^21 sucrose molecules

7 0
2 years ago
What's the smallest part of a compound
tigry1 [53]
I believe an atom. may be wrong, science is not my strong suit
6 0
3 years ago
Calculate the values of ΔU, ΔH, and ΔS for the following process:
ladessa [460]

Answer:

ΔU = 45.814 KJ

ΔH = 46.4375 KJ

ΔS = 18.76 J/K

Explanation:

            H2O(l)        →          H2O(l)                →              H2O(steam)

   298.15K, 1atm   ΔHp     373.15K,1atm       ΔHv         373.15K,1 atm

∴ ΔHp = Qp = nCpΔT

∴ n H2O = 1 mol

∴ Cp,n = 75.3 J/mol.K

∴ ΔT = 373.25 - 298.15 = 75 K

⇒ Qp = (1 mol)*(75.3 J/mol.K)*(75K) = 5647.5 J

⇒ ΔHp = 5647.5 J = 5.6475 KJ

⇒ ΔH = ΔHp + ΔHv

∴ ΔHv = 40.79 KJ/mol * 1 mol = 40.79 KJ  

⇒ ΔH = 5.6475 KJ + 40.79 KJ = 46.4375 KJ

ideal gas:

∴ ΔH = ΔU + PΔV

∴ V1 = nRT1/P1 = ((1)*(0.082)*(298.15))/1 = 24.45 L

∴ V2 = nRT2/P2 = ((1)*(0.082)*(373.15))/ 1 = 30.59 L

⇒ ΔV = V2 - V1 = 6.15 L * (m³/1000L) = 6.15 E-3 m³

∴ P = 1 atm * (Pa/ 9.86923 E-6 atm) = 101325.027 Pa

⇒ ΔU = ΔH - PΔV = 46.4375 KJ - ((101325.027 Pa*6.15 E-3m³)*(KJ/1000J))

⇒ ΔU = 46.4375 KJ - 0.623 KJ

⇒ ΔU = 45.814 KJ

∴ ΔS = Cv,n Ln (T2/T1) + nR Ln (V2/V1)

⇒ ΔS = (75.3) Ln(373.15/298.15) + (1)*(8.314) Ln (30.59/24.45)

⇒ ΔS = 16.896 J/K + 1.863 J/K

⇒ ΔS = 18.76 J/K

3 0
3 years ago
What is the final volume of a gas (in liters) when the initial volume is 14.00 L at
8_murik_8 [283]

Answer:

V_2=1344L

Explanation:

Hello there!

In this case, since we have a problem about volume-pressure relationship, the idea here is to use the Boyle's law to calculate the final volume as shown below:

P_2V_2=P_1V_1\\\\V_2=\frac{P_2V_2}{P_1}\\

Then, we plug in the initial and final pressures and the initial volume to obtain:

V_2=\frac{14.00L*0.9600atm}{0.01000atm}\\\\V_2=1344L

Regards!

4 0
2 years ago
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