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defon
3 years ago
7

True or False: Sound waves are a type of transverse wave.

Chemistry
1 answer:
vazorg [7]3 years ago
8 0
<span>The answer is false. Transverse waves are waves that oscillate perpendicular to the direction of the wave. Examples of transverse waves is that of electromagnetic radiation and seismic S waves. The other type of wave is longitudinal waves that oscillates in the same plane as the direction of the wave. An example, of longitudinal waves, is sound waves and seismic P-waves</span>




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How many kilojoules of heat are released when 32.0 g of NaOH are dissolved in water? (The molar heat of solution of NaOH is –445
Katen [24]
The energy release when dissolving 1 mol of NaOH in water is 445.1 kJ
the mass of NaOH to be dissolved is 32.0 g
The number of NaOH moles in 32.0 g - 32.0 g / 40 g/mol =  0.8 mol
the energy released whilst dissolving 1 mol of NaOH - 445.1 kJ
when dissolving 0.8 mol - the energy released is 445.1 kJ/mol x 0.8 mol
therefore heat released is - 356.08 kJ
answer is -356.08 kJ
6 0
3 years ago
Read 2 more answers
The equilibrium constant Kc for the reaction below is 0.00584 at a certain temperature. Br2(g) ⇌ 2Br(g) If the initial concentra
Vinil7 [7]

Answer:

Explanation:

Given that:

The chemical equation for the reaction is:

             Br2(g)    ⇌  2Br(g)

Initially  0.0345M  0.0416M

Q_C = \dfrac{[Br]^2}{[Br_2]} = \dfrac{(0.0416)^2}{(0.0345)}= 0.05016

Q_C =0.05016 >>> K_c(0.00584)

Thus, the given reaction will proceed in the  backward direction

 The I.C.E table is as follows:

                       Br2(g)    ⇌  2Br(g)

I               0.0345                 0.0416

C                 +x                        -2x

E             (0.0345+x)            (0.0416 -2x)

K_c = \dfrac{[Br]^2}{[Br_2]} = \dfrac{(0.0416-2x)^2}{(0.0345+x)} = 0.00584

= 0.00173056 - 0.0832x - 0.0832x + 4x² = 0.00584 (0.0345 +x)

= 0.00173056 - 0.166x + 4x² = 2.0148× 10⁻⁴ + 0.00584x

= 0.00173056 - 2.0148× 10⁻⁴ - 0.166x - 0.00584x + 4x²

= 0.00152908  - 0.17184x + 4x²

Solving by using Quadratic formula

x = 0.03038 or 0.0126

For x = 0.03038

At equilibrium

[Br₂] = (0.0345 + 0.03038) = 0.06488 M

[Br] =  (0.0416 -2(0.03038)) = - 0.01916 M

Since we have a negative value for [Br], we discard the value for x

For x = 0.0126

At equilibrium

[Br₂] = (0.0345 + 0.0126) = 0.0471 M

[Br] =  (0.0416 -2(0.0126)) = 0.0164 M

4 0
3 years ago
Draw a mechanism for the reaction of methylamine with 2-methylpropanoic acid. Draw any necessary curved arrows. Show the product
Nitella [24]

Answer:

See figure 1

Explanation:

On this case we have a <u>base</u> (methylamine) and an <u>acid</u> (2-methyl propanoic acid). When we have an acid and a base an <u>acid-base reaction </u>will take place, on this specific case we will produce an <u>ammonium carboxylate salt.</u>

Now the question is: <u>¿These compounds can react by a nucleophile acyl substitution reaction?</u> in other words <u>¿These compounds can produce an amide? </u>

Due to the nature of the compounds (base and acid), <u>the nucleophile</u> (methylamine) <u>doesn't have the ability to attack the carbon</u> of the carbonyl group due to his basicity. The methylamine will react with the acid-<u>producing a positive charge</u> on the nitrogen and with this charge, the methylamine <u>loses all his nucleophilicity.</u>

I hope it helps!

5 0
3 years ago
The combustion of glucose (c6h12o6) with oxygen gas produces carbon dioxide and water. this process releases 2803 kj per mole of
V125BC [204]

Answer:- 335 kcal of heat energy is produced.

Solution:- The balanced equation for the combustion of glucose in presence of oxygen to give carbon dioxide and water is:

C_6H_1_2O_6+6O_2\rightarrow 6CO_2+6H_2O

From given info, 2803 kJ of heat is released bu the combustion of 1 mol of glucose. We need to calculate the energy produced when 3.00 moles of oxygen react with excess of glucose.

We could solve this using dimensional analysis as:

3.00mol O_2(\frac{1mol glucose}{6mol O_2})(\frac{2803 kJ}{1mol glucose})

= 1401.5 kJ

Now, let's convert kJ to kcal.

We know that, 1kcal = 4.184kJ

So, 1401.5kJ(\frac{1kcal}{4.184kJ})

= 335 kcal

Hence, 335 kcal of heat energy is produced by the use of 3.00 moles of oxygen gas.


8 0
3 years ago
Identify the number of each type of atom in this formula 5C12H22O11<br><br>wdym
Lady bird [3.3K]
There are 12 carbon atoms, 22 hydrogen atoms, and 11 oxygen atoms in one molecule of C12H22O11. 
In five molecules there are 60 carbon atoms, 110 hydrogen atoms, and 55 oxygen atoms.
6 0
3 years ago
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