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meriva
2 years ago
12

In your second paragraph, describe properties of sound waves. Then make a claim about how loud an explosion in space should be.

Provide evidence and reasoning to support your claim.
Chemistry
1 answer:
morpeh [17]2 years ago
8 0

Answer:

Here are five important characteristics: wavelength, amplitude, frequency, time period, and velocity. The wavelength of a sound wave can tell the distance that wave travels before it repeats itself. The wavelength itself is a longitudinal wave that show compression and rarefactions of sound waves.

Explanation:

Hope this helped please pick me brainliest

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A 4.36 g sample of an unknown alkali metal hydroxide is dissolved in 100.0 ml of water. an acid-base indicator is added and the
Fofino [41]

Answer:

(a) 102.6g/mol

(b) Rubidium

Explanation:

Hello,

This titration is carried out by assuming that the volume of base doesn't have a significant change when the mass is added, thus, we state the following data a apply the down below formula to compute the molarity of the base solution:

V_{base}=0.1L; M_{acid}=2.5M, V_{acid}=0.017L\\V_{base}M_{base}=V_{acid}M_{acid}

Solving for the molarity of base we've got:

M_{base}=\frac{M_{acid}*V_{acid}}{V_{base}}=\frac{2.50M*0.017L}{0.1L} =0.425M=0.425mol/L

Now, we can compute the moles of the base as:

n_{base}=0.425mol/L*0.1L=0.0425mol

(a) Now, one divides the provided mass over the previously computed moles to get the molecular mass of the unknown base:

\frac{4.36g}{0.0425mol} =102.6g/mol

(b) Subtracting the atomic mass of oxygen and hydrogen, the metal's atomic mass turns out into:

102.6g/mol-16g/mol-1g/mol=85.6g/mol

So, that atomic mass dovetails to the Rubidium's atomic mass.

Best regards.

8 0
3 years ago
A 1.25g sample of copper (cCu=0.386Jg∘C) is initially at a temperature of 25.0∘C. If the sample absorbs 87.4J of heat, what is i
professor190 [17]

Answer: Final temperature = 206∘C

Explanation:

Heat Energy is given as  

q= mCΔT

ehere

q= Heat energy = 87.4J

m= mass=1.25g

C=specific heat c= 0.386Jg∘C) ,

ΔT =  Change in temperate of which the final temperature= 25.0∘C

 q= mCΔT

ΔT = q/mC

ΔT = 87.4/ 1.25 X 0.386=181.14∘C

But,

T final- T initial = ΔT

T final = T initial + ΔT

T final = 25.0∘C +181.14∘C=206.14∘C rounded to 206∘C

4 0
3 years ago
Scientific endeavors is sometimes accompanied by negative trade offs true or false?
elena-14-01-66 [18.8K]
Idk bro i havent learned that yet
8 0
3 years ago
The element in Period 4 and Group 14 of the Periodic Table would be classified as a
Paul [167]

Answer:

Chalcogen

Explanation:

7 0
2 years ago
MnCl₂ (aq) + (NH₄)₂CO₃ (aq) --> MnCO₃ (s) + 2 NH₄Cl (aq)
LuckyWell [14K]

this equation is balanced.

8 0
3 years ago
Read 2 more answers
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