All categories

1 year ago

Question 3 of 25

1 answer:

Colt1911 [192]1 year ago

8 0

The statement that describes a question that can guide the design of a scientific investigation could be about the necessity of a controlled variable. The correct answer is A.

<h3>What is a controlled variable?</h3>

It is a variable whose values are controlled by the researcher during the course of experimental investigations.

When designing an experiment, it is necessary to identify the different variables that would be involved for a successful investigation.

Variables can be independent, dependent, controlled, or constant. Identifying the necessary variables will help in designing the experiment.

More on variables can be found here: brainly.com/question/15740935

#SPJ1

You might be interested in

1. Your friend has a balloon that can hold a volume of 5 L. If she added a 3 mol of CO2 gas at a pressure of 0.90 atm, what is t

masha68 [24]

Answer:

I don't <u>understand</u><u> </u><u>your</u><u> </u><u>question</u><u> </u>

6 0

2 years ago

7. A 2.0 L container had 0.40 mol of He(g) and 0.60 mol of Ar(g) at 25°C.

Finger [1]

Answer:

a) Ek Ar > Ek He

b) v Ar < v He

c) If v Ar = 431 m/s ⇒ v He = 1710.44 m/s

d) Pt = 12.218 atm

e) P He = 4.887 atm and P Ar = 7.33 atm

Explanation:

container:

∴ V = 2.0 L

∴ n He = 0.4 mol

∴ n Ar = 0.6 mol

∴ T = 25°C ≅ 298 K

a) Internal energy (U) :

∴ U = Ek + Ep = kinetic energy + potential energy

∴ Ep: the potential interaction energy is neglected, assuming ideal gas mixture

⇒ U = Ek = N(1/2mv²)= 3/2 NKT

∴ N = nNo ....number of moleculas

∴ K = 1.380 E-23 J/K....Boltzmann's constant

∴ No = 6.022 E23 molec/mol....Avogadro's number

for He:

⇒ N = (0.4)(6.022 E23) = 2.4088 E23 molec

⇒ Ek = (3/2)(2.4088 E23)(1.380 E-23 J/K)(298) = 1485.892 J

for Ar:

⇒ N = (0.6)(6.022 E23) = 3.6132 E 23 molec

⇒ Ek = (3/2)(3.6132 E23)(1.380 E-23 J/K)(298) = 2228.838 J

** Ar gas has a greater average kinetic energy

b) He:

∴ N(1/2)mv² = (3/2)NKT

⇒ mv² = 3KT

⇒ v² = 3KT/m

⇒ v = √3KT/m

∴ m He = (0.4 mol)(4.0026 g/mol) = 1.601 g He = 1.601 E-3 Kg He

⇒ v = √(3(1.380 E-23)(298)/(1.601 E-3)) = 2.776 E-9 m/s He

Ar:

∴ m Ar = (0.6)(39.948 g/mol) = 23.969 g = 0.0239 Kg Ar

⇒ v = 6.99 E-10 m/s

** v Ar < v He

c) r = V Ar / v He = (6.99 E-10 m/s)/(2.776 E-9 m/s) = 0.252

∴ If v Ar = 431 m/s

⇒ v He = v Ar/0.252 = 431 m/s / 0.252 = 1710.44 m/s

d) Pt = ntRT / V

∴ nt = 0.4 + 0.6 = 1 mol

⇒ Pt = (1mol)(0.082 atm.L/K.mol)(298 K)/(2.00 L) = 12.218 atm

e) P He = nRT/V = (0.4)(0.082)(298)/2 = 4.8872 atm

⇒ P Ar = Pt - PHe = 12.218 - 4.8872 = 7.33 atm

3 0

3 years ago

An object has a volume of 2 milliliters and a mass of 10 grams, calculate the density of the object

MArishka [77]

5000 kilogram/cubic meter

5 0

3 years ago

How to balance NH3+NO=N2+H2O

Kryger [21]

Explanation:

NH₃+NO=N₂+H₂O;

NH₃+NO=N₂+6H₂O;

NH₃+6NO=N₂+6H₂O;

4NH₃+6NO=N₂+6H₂O;

4NH₃+6NO=5N₂+6H₂O.

3 0

1 year ago

When a mercury-202 nucleus is bombarded with a neutron, a proton is ejected. What element is formed?

Galina-37 [17]

That will make a gold-202 nucleus.

<h3>Explanation</h3>

Refer to a periodic table. The atomic number of mercury Hg is 80.

Step One: Bombard the $\displaystyle ^{202}_{\phantom{2}80}\text{Hg}$ with a neutron $^{1}_{0}n$ . The neutron will add 1 to the mass number 202 of $^{202}_{\phantom{2}80}\text{Hg}$ . However, the atomic number will stay the same.

New mass number: 202 + 1 = 203.
Atomic number is still 80.

$^{202}_{\phantom{2}80}\text{Hg} + ^{1}_{0}n \to ^{203}_{\phantom{2}80}\text{Hg}$ .

Double check the equation:

Sum of mass number on the left-hand side = 202 + 1 = 203 = Sum of mass number on the right-hand side.
Sum of atomic number on the left-hand side = 80 = Sum of atomic number on the right-hand side.

Step Two: The $^{203}_{\phantom{2}80}\text{Hg}$ nucleus loses a proton $^{1}_{1}p$ . Both the mass number 203 and the atomic number will decrease by 1.

New mass number: 203 - 1 = 202.
New atomic number: 80 - 1 = 79.

Refer to a periodic table. What's the element with atomic number 79? Gold Au.

$^{203}_{\phantom{2}80}\text{Hg} \to ^{202}_{\phantom{2}79}\text{Au} + ^{1}_{1}p$ .

Double check the equation:

Sum of mass number on the left-hand side = 203 = 202 + 1 = Sum of mass number on the right-hand side.
Sum of atomic number on the left-hand side = 80 = 79 + 1 = Sum of atomic number on the right-hand side.

A gold-202 nucleus is formed.

6 0

3 years ago