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Mrrafil [7]
3 years ago
11

What are the chemical elements in iced tea?

Chemistry
1 answer:
Natali5045456 [20]3 years ago
6 0

Explanation:

I hope you interested about the chemical what they add in tea

You might be interested in
A helium-filled balloon has a volume of 2.48 L and a pressure of 150 kPa. The volume of the balloon increases to 2.98 L after yo
Tema [17]

Answer:

THE NEW PRESSURE OF THE HELIUM GAS AT 2.98 L VOLUME IS 124.8 kPa.

AT AN INCREASE ALTITUDE, THERE IS A LOWER PRESSURE ENVIRONMENT AND THE HELIUM GAS PRESSURE DECREASES AND HENCE AN INCREASE IN VOLUME.

Explanation:

The question above follows Boyle's law of the gas law as the temperature is kept constant.

Boyle's law states that the pressure of a fixed mass of gas is inversely proportional to the volume, provided the temperature remains constant.

Mathematically, P1 V1 = P2 V2

P1 = 150 kPa = 150 *10^3 Pa

V1 = 2.48 L

V2 = 2.98 L

P2 = ?

Rearranging the equation, we obtain;

P2 = P1 V1 / V2

P2 = 150 kPa * 2.48 / 2.98

P2 = 372 *10 ^3 / 2.98

P2 = 124.8 kPa.

The new pressure of the gas when at a height which increases the volume of the helium gas to 2.98 L is 124.8 kPa.

6 0
3 years ago
Unit: Chemical Quantities
Vaselesa [24]

Answer:

(See explanation for further details)

Explanation:

1) The quantity of moles of sulfur is:

n = \frac{1.20\times 10^{24}\,atoms}{6.022\times 10^{23}\,\frac{atoms}{mol} }

n = 1.993\,moles

2) The number of atoms of helium is:

x = (1.5\,moles)\cdot \left(6.022\times 10^{23}\,\frac{atoms}{mole} \right)

x = 9.033\times 10^{23}\,atoms

3) The quantity of moles of carbon monoxide is:

n = \frac{4.15\times 10^{23}\,molecules}{6.022\times 10^{23}\,\frac{molecules}{mol} }

n = 0.689\,moles

4) The number of molecules of sulfur dioxide is:

x = (2.25\,moles)\cdot \left(6.022\times 10^{23}\,\frac{molecules}{mole} \right)

x = 1.355\times 10^{24}\,molecules

5) The quantity of moles of sodium chloride is:

n = \frac{2.4\times 10^{23}\,molecules}{6.022\times 10^{23}\,\frac{molecules}{mol} }

n = 0.399\,moles

6) The number of formula units of magnesium iodide is:

x = (1.8\,moles)\cdot \left(6.022\times 10^{23}\,\frac{f.u.}{mole} \right)

x = 1.084\times 10^{24}\,f.u.

7) The quantity of moles of potassium permanganate is:

n = \frac{3.67\times 10^{23}\,f.u.}{6.022\times 10^{23}\,\frac{f.u.}{mol} }

n = 1.214\,moles

8) The number of molecules of carbon tetrachloride is:

x = (0.25\,moles)\cdot \left(6.022\times 10^{23}\,\frac{molecules}{mole} \right)

x = 1.506\times 10^{23}\,molecules

9) The quantity of moles of aluminium is:

n = \frac{3.67\times 10^{23}\,atoms}{6.022\times 10^{23}\,\frac{atoms}{mol} }

n = 0.609\,moles

10) The number of molecules of oxygen difluoride is:

x = (3.52\,moles)\cdot \left(6.022\times 10^{23}\,\frac{molecules}{mole} \right)

x = 2.120\times 10^{24}\,molecules

3 0
3 years ago
What substance do you place in the burette during a titration
strojnjashka [21]

Answer:

Acid solution

Explanation:

In acid-base titrations carried out in school and college labs, many of the older generation, well-informed teachers told their students that the acid solution should be taken in the burette.

5 0
3 years ago
why is whenever one raised to the power of any exponent 1 and why is 0 raised to the power of any exponent 1
IceJOKER [234]
When you use exponents, think of it like this. 1 squared is 1 x 1. 1 cubed is 1 x 1 x 1. And 1 to the power of 4 is 1 x 1 x 1 x 1. And so on. You basically just multiply them by themselves. 0 to the power of any exponent is 1, well that's just a rule. It doesn't make much sense but it's easy to remember and I wouldn't worry about it.
5 0
3 years ago
The normal freezing point of water (H2O) is 0.00 oC and its Kf value is 1.86 oC/m. A nonvolatile, nonelectrolyte that dissolves
adelina 88 [10]

Answer:

3.74g of ethylene glycol must be added to decrease the freezing point by 0.400°C

Explanation:

One colligative property is the freezing point depression due the addition of a solute. The equation is:

ΔT=Kf*m*i

<em>Where ΔT is change in temperature = 0.400°C</em>

<em>Kf is freezing point constant of the solvent = 1.86°C/m</em>

<em>m is molality of the solution (Moles of solute / kg of solvent)</em>

<em>And i is Van't Hoff constant (1 for a nonelectrolyte)</em>

Replacing:

0.400°C =1.86°C/m*m*1

0.400°C / 1.86°C/m*1 = 0.215m

As mass of solvent is 280.0g = 0.2800kg, the moles of the solute are:

0.2800kg * (0.215moles / 1kg) = 0.0602 moles of solute must be added.

The mass of ethylene glycol must be added is:

0.0602 moles * (62.10g / mol) =

3.74g of ethylene glycol must be added to decrease the freezing point by 0.400°C

<em />

6 0
3 years ago
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