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3 years ago

The cure for polio is a(n) ________science. basic applied

1 answer:

4 0

The cure for polio is an applied science. The vaccination as the cure of the disease uses applied science<span> that applies existing knowledge in science to develop practical technology or inventions like the vaccine. Answer to this problem is applied/</span>

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You place a balloon in a closed chamber at standard temperature and pressure (STP) . You double the chamber temperature while le

Likurg_2 [28]

Answer:

Volume of balloon will be doubled

Explanation:

We are asked to state what happens to a balloon which was placed in a closed chamber at STP if the temperature is doubled and the pressure is kept constant.

We have three parameters here, the temperature and pressure of the chamber, and the volume of the balloon.

Now, since the pressure of the chamber is kept constant, we can say that the gas law we need to use is the Charles’ law and it posits that volume and temperature are directly proportional:

Given;

V1 =? , V2 = ?, T1 = 273.15K(at STP) and T2 = 273.15 * 2 = 546.3K

We substitute these values in the Charles’ law equation:

V1/T1 = V2/T2

V2 = V1T2/T1

V2 = (V1 * 546.3)/273.15

V2 = 2V1

This means that the balloon will double in volume

4 0

4 years ago

How many moles are in a 63.60 g sample of carbon? (the molar mass of carbon is 12.01 g/mol.)

skelet666 [1.2K]

63.60/12.01=5.296mole of carbon

5 0

3 years ago

Read 2 more answers

A gas at 155 kilopascals and 25°C has an initial volume of 1.00 liter. The pressure of the gas increases to 605 kilopascals as t

monitta

I think the answer will be A) 0.34

5 0

4 years ago

It is difficult to break the ionic bonds in a compound because of the...

mestny [16]

The answer to the first one is C. Strong attraction between ions

The answer to the second one is A. High melting points

Hope this helps :)

6 0

4 years ago

Read 2 more answers

PLEASE SOMEONE! The density of the acetic acid solution is 1.05 g/mL. Calculate the %(m/m) of the acetic acid solution (convert

tigry1 [53]

Answer:

The Percentage concentration of acetic acid = 1.39 %

Explanation:

Density of acetic acid solution = 1.05 g/mL

Volume of acetic acid solution = 0.1 L = 100 mL

From the formula, Density = mass / volume; mass = density × volume

Mass of acetic acid solution = 1.05 g/mL × 100 mL = 105 g

Molar concentration of acetic acid solution = 0.243 mol/L

Molar mass of acetic acid, CH₃COOH = (12 × 2 + 1 ×4 + 16 ×2) = 60 g/mol

From the formula, mass concentration = molar concentration × molar mass

Mass concentration of acetic acid, CH₃COOH = 0.243 mol/L × 60 g/mol = 14.58 g/L

In one liter of acetic acid solution, there are 14.58 g of acetic acid. Therefore, in 0.1 L, there will be 14.58 × 0.1 = 1.458 g of acetic acid.

Percentage concentration of acetic acid = mass of acetic acid / mass of acetic acid solution × 100%

Percentage concentration of acetic acid = (1.458 / 105) × 100% = 1.39 %

The Percentage concentration of acetic acid = 1.39 %

8 0

3 years ago