The given information in the question is :
Frequency = 170 Hz
1 Hz = 
Frequency = 170
Wavelength given = 2 m
Speed is unknown which we need to calculate.
Speed can be calculated by using the formula : Speed = wavelength X frequency
By plugging the value of wavelength and frequency in the above formula we get :
Speed = 
Speed = 340
Trait-descriptive adjectives.
adjectives that can be used to describe characteristics of people are called.
Answer:
2.40 M
Explanation:
The molarity of a solution tells you how many moles of solute you get per liter of solution.
Notice that the problem provides you with the volume of the solution expressed in milliliters,
mL
. Right from the start, you should remember that you must convert this volume to liters by using the conversion factor
1 L
=
10
3
mL
Now, in order to get the number of moles of solute, you must use its molar mass. Now, molar masses are listed in grams per mol,
g mol
−
1
, which means that you're going to have to convert the mass of the sample from milligrams to grams
1 g
=
10
3
mg
Sodium chloride,
NaCl
, has a molar mass of
58.44 g mol
−
1
, which means that your sample will contain
unit conversion
280.0
mg
⋅
1
g
10
3
mg
⋅
molar mass
1 mole NaCl
58.44
g
=
0.004791 moles NaCl
This means that the molarity of the solution will be
c
=
n
solute
V
solution
c
=
0.004791 moles
2.00
⋅
10
−
3
L
=
2.40 M
The answer is rounded to three sig figs, the number of sig figs you have for the volume of the solution.
Answer:
A model or simulation is only as good as the rules used to create it. It is very difficult to create an entirely realistic model or simulation because the rules are based on research and past events. The main disadvantage of simulations is that they aren't the real thing.
Explanation:
Answer is: c. CuSO4 + 2NaOH yields Cu(OH)2 + Na2SO4.
Copper atom has oxidation number +2, sulfur atom oxidation number +6, oxygen has oxidation number-2, sodium has oxidation number +1 and hydrogen has +1 on both side of chemical reaction, so elements did not change their oxidation numbers.
