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madam [21]
3 years ago
5

What would happen if you play a note in a cold day on a flute?

Chemistry
1 answer:
Maurinko [17]3 years ago
7 0
I don't know if this is the answer you are looking for but it would be flat unless the player pushed the tuning slide in.
You might be interested in
When each piece of ice is placed on a piece of oil it is observed that ice piece 2 melts faster than ice plece 1
solniwko [45]

We cannot see it its so small

Also yes it matches your hypothesis

4 0
3 years ago
Calculate the mass in micrograms of zinc oxalate the chemist has added to the flask. be sure your answer has the correct number
Vera_Pavlovna [14]
<span>A chemist adds 155.0ml of a 4.10 X 10^-5 mmol/L of a zinc oxalate (ZnC2O4)solution to a reaction flask. Calculate the mass in micrograms of zinc oxalate the chemist has added to the flask. 1mmol = 10^-3 mol Therefore 4.10*10^-5mmol = 4.10*10^-8mol molar mass ZnC2O4 = 65.39+(2*12.011)+(4*15.99) = 153.372g/mol You have 4.10*10^-8 mol/litre =153.372 * 4.10*10^-8 = 6.29*10^-6 grams / litre (* see below) But you have 155ml. Mass of ZnC2O4 = 155/1000*6.29*10^-6 g Mass is = 9.75*10^-7 grams 1µg = 10^-6 g You then have 9.75*10^-7/10^-6 = 0.975µg ZnC2O4 (*see below) at this point you could have said: 1µg = 10^-6 g therefore you have a solution of 6.29µg per litre, 155ml = 6.29*155/1000 = 0.975µg ZnC2O4</span>
3 0
3 years ago
Calculate the energy E of a sample of 3.50 mol of ideal oxygen gas (O2) molecules at a temperature of 310 K. Assume that the mol
love history [14]

Answer:

13.53 kJ

Explanation:

The energy of a gas can be calculated by the equation:

E = (3/2)*n*R*T

Where n is the number of moles, R is the gas constant (8.314 J/mol.K), and T is the temperature.

E = (3/2)*3.5*8.314*310

E = 13,531.035 J

E = 13.53 kJ

7 0
2 years ago
How many moles of oxygen gas, o2, are in a storage tank with a volume of 1.000×105 l at stp?
Mumz [18]
STP means standard temperature and pressure which is equivalent to 273 K and 1 atm, respectively. Assuming ideal gas behavior, the solution for this problem is as follows:

PV = nRT
Solve for n,
n = RT/PV
n = (0.0821 L-atm/mol-K)(273 K)/(1 atm)(1×10⁵ L)
<em>n = 2.24×10⁻⁴ moles</em>
6 0
3 years ago
Read 2 more answers
A. At STP, what is the volume of 708 mol of nitrogen gas? 708 mol = 708 mol X L B. A sample of hydrogen gas occupies 14.1 L at S
elena-14-01-66 [18.8K]

Answer:

A. 15859.2 L or 15900 L

B. 0.629 mol

Explanation:

At STP, one mole is equal to approximately 22.4 L

L or mL is volume, so you are attempting to solve for L or mL.

A.

708 mol x (22.4 L/1 mol) = 15859.2 L (w/ significant figures included - 15900 L)

B.

(14.1 L) x (1 mole/ 22.4 L) = 0.629 mol.

5 0
2 years ago
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