Physical or chemical ?

For some jobs, “good enough” is good enough. That’s sometimes true in chemistry. More often, though, careful planning, calculati

insens350 [35]

When carrying out experiment in the laboratory to investigate physical changes, absolute measurement is usually not needed. For instance, one can just take a little sample of sodium chloride and dissolve it in water to demonstrate physical change. But for other experiments, such as the measurement of the catalase enzyme activity in a liver tissue. Careful measurements have to be made when preparing the chemicals that will be used in the experiment.

6 0

3 years ago

Read the descriptions below of two substances and an experiment on each. Decide whether the result of the experiment tells you t

Lelu [443]

Answer:

Sample A - mixture

Sample B- Mixture

Explanation:

Looking at sample A, we can see that as water was poured over sample A, the sample was separated into its components as the powder disappeared leaving behind a gritty black sand-like material on the surface of the paper. A separation of the mixture has taken place.

In sample B, we can clearly see that it is a mixture because the amount of solid recovered is much less than the total mass of the solid put into the beaker. The sample must have been separated into its components.

8 0

3 years ago

I’ve been stuck on these 5 questions!? Can you guys help?!

Varvara68 [4.7K]

Answer:

1. 0.224 moles of oxygen

3. 143.36 L oxygen gas

5. 0.059 atm

10. 5.14 atm

11. 307 K

Explanation:

1. You have to use the ideal gas law: PV=nRT where P is pressure in atm, V is volume in liters, n is number of moles, R is the constant 0.08206 L atm mol^-1 K^-1, and T is the temperature in Kelvins where K=degrees celsius+273.15 . By rearranging the equation, you solve for n, which is n=(PV)/(RT)

P= 28.3 atm

V=0.193 L

R= 0.08206 L atm mol^-1 K^-1

T= 24.5+273.15= 297.65 K

Plugging the values in,

n=(28.3 atm x 0.193 L)/(0.08206 L atm mol^-1 K^-1 x 297.65 K)

n= 0.224 moles of oxygen

3. At STP, there are 22.4 L of gas for every mole of gas present. So 6.4 moles of oxygen would mean that there are:

6.4 mol x 22.4 L= 143.36 L oxygen gas

5. You have to use the ideal gas law: PV=nRT where P is pressure in atm, V is volume in liters, n is number of moles, R is the constant 0.08206 L atm mol^-1 K^-1, and T is the temperature in Kelvins where K=degrees celsius+273.15 . By rearranging the equation, you solve for P, which is P=(nRT)/V

n= 0.72 g converting to moles, divide by molar mass of oxygen gas: 0.72 g/32g= 0.0225 moles

V=9.3 L

R= 0.08206 L atm mol^-1 K^-1

T= 23.0+273.15= 296.15 K

Plugging the values in,

P=(0.0225 moles x 0.08206 L atm mol^-1 K^-1 x 296.15 K)/ 9.3 L

P= 0.059 atm

10. Ideal gas law again using the same equation as 5 above: You have to use the ideal gas law: PV=nRT where P is pressure in atm, V is volume in liters, n is number of moles, R is the constant 0.08206 L atm mol^-1 K^-1, and T is the temperature in Kelvins where K=degrees celsius+273.15 . By rearranging the equation, you solve for P, which is P=(nRT)/V

n= 0.108 mol

R=0.08206 L atm mol^-1 K^-1

T=20.0+273.15= 293.15 K

V= 0.505 L

Plugging the values in,

P=(0.108 mol x 0.08206 L atm mol^-1 K^-1 x 293.15 K)/0.505 L

P= 5.14 atm

11. You have to use the ideal gas law: PV=nRT where P is pressure in atm, V is volume in liters, n is number of moles, R is the constant 0.08206 L atm mol^-1 K^-1, and T is the temperature in Kelvins where K=degrees celsius+273.15 . By rearranging the equation, you solve for Y, which is T=(PV)/(nR)

P= 0.988 atm

V= 1.20 L

n= 0.0470 mol

R=0.08206 L atm mol^-1 K^-1

Plugging the numbers in,

T=(0.988 atm x 1.20 L)/(0.0470 mol x 0.08206 L atm mol^-1 K^-1)

T= 307 K

4 0

3 years ago

I have to show my work after I find the atomic weight of each element please help !!

wariber [46]

Explanation: