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natka813 [3]
3 years ago
8

According to the law of conservation of matter, what number must be the same on each side of a chemical equation?

Physics
1 answer:
olga2289 [7]3 years ago
7 0

Answer:

4. The number of atoms of each element

Explanation:

<em><u>Why is 1 wrong?</u></em>

The number of substances (i.e. number of mols of substances) do not have to be the same. For instance, 2 mol of water reacts with 1 mol of oxygen to form 2 mol of water. Obviously the numer of substances is not conserved. Thus, it isn't a requirement.

<em><u>Why is 2 wrong?</u></em>

The number of chemical symbols do not mean anything. They do not tell us anything about the quantity of matter, they only tell us about the elements that are involved in the reaction.

<em><u>Why is 3 wrong?</u></em>

Once again, the number of chemical formulas are meaningless.

<u><em>Why is 4 correct?</em></u>

The number of atoms are always conserved as atoms cannot be broken or created.

<em><u>Possible confusion...</u></em>

The number of atoms are always conserved but the number of molecules may not always be conserved.

Why? As 2 atoms may be a molecule on the reactant side and 4 atoms may become a molecule on the product size, causing the number of molecules to reduce by half, while the number of atoms remain constant.

Hope that makes sense!

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If the sound level at a point is 98.0 dB, what is the intensity of sound at that point (Unit=W/m^2)? please help will give brain
Firlakuza [10]

Answer: The decibel scale is a logarithmic scale where each bel or 10 decibels correspondents to a factor of ten. A power intensity of 10^(-12) watts per square meter is the standard reference for a SPL of 0 db. So an SPL of 98 db corresponds to a power intensity of 10^(9.8)*10^(-12) or 10^(9.8–12) w/m^2.

0.006309573 w/m^2.

You can also readily find the value for any given SPL using the online calculator at: http://www.sengpielaudio.com/calculator-soundlevel.htm

Explanation:

7 0
2 years ago
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You perform an experiment with a long column of air and a tuning fork. The column of air is defined by a very long vertical plas
velikii [3]

Answer:

\lambda=4L=1.33m

v=343m/s

Explanation:

We have to take into account the expressions

f=\frac{2n+1}{4}\frac{v_s}{L}\\L=(2n+1)\frac{\lambda}{4}

if we assume that 256Hz is the fundamental frequency we have

f=\frac{1}{4}\frac{v_s}{L}\\\\L=\frac{1}{4}\frac{v_s}{f}=\frac{1}{4}\frac{343\frac{m}{s}}{256Hz}=0.33m

and for wavelength

\lambda=4L=1.33m

hope this helps!!

6 0
3 years ago
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A car was at a 30km marker and began to drive. The car passed a 90km marker and the time had changed 30
dolphi86 [110]

Answer:

i think it's 2km pm

Explanation:

2km x 30 60.. start was 30, and now your at 90.. we had to determine how much time it took.. so 2 is the average.. or atleast per minute and sorry it i still didnt answer ur question lol im just trynna help

7 0
2 years ago
An insulated pipe carries steam at 300°C. The pipe is made of stainless steel (with k = 15 W/mK), has an inner diameter is 4 cm,
insens350 [35]

Answer:

The answers to the question are

(i) The rate of heat loss per-unit-length (W/m) from the pipe is 131.62 W

(ii) The temperature of the outer surface of the insulation is 49.89 °C

Explanation:

To solve the question, we note that the heat transferred is given by

Q = \frac{2\pi L(t_{hf} - t_{cf}) }{\frac{1}{h_{hf}r_1}+\frac{ln(r_2/r_1)}{k_A} + \frac{ln(r_3/r_2)}{k_B} +\frac{1}{h_{cf}r_3}}

Where

t_{hf} = Temperature at the inside of the pipe = 300 °C

t_{f} = Temperature at the outside of the pipe = 20 °C

r₁ =internal  radius of pipe = 4.0 cm

r₂ = Outer radius of pipe = 4.5 cm

r₃ = Outer radius of the insulation = r₂ + 2.5 = 7.0 cm

k_A = 15 W/m·K

k_B = 0.038 W/m·K

h_{hf} = 75 W/m²·K

h_{cf} = 10 W/m²·K

Plugging in the values in the above equation where for a unit length L = 1 m, we have

Q = 131.32 W

From which we have, for the film of air at the pipe outer boundary layer

Q = \frac{t_A-t_B}{R_T} Where R_T for the air film on the pipe outer surface is given by

R_T= \frac{1}{\alpha A}

where A =area of the outside of the pipe

= \frac{1}{10*2\pi*0.07*1 } = 0.227 K/W

Therefore

131.32 W = \frac{t_A-20}{0.227} which gives

t_A = 49.89 °C

Heat transferred by radiation = q' = ε×σ×(T₁⁴ - T₂⁴)

Where ε = 0.9, σ, = 5.67×10⁻⁸W/m²·(K⁴)

T₁ = Surface temperature of the pipe = 49.89 °C and

T₂ = Temperature of the surrounding = 20.00 °C

Plugging in the values gives, q' = 0.307 W per m²

Total heat lost per unit length = 131.32 + 0.307 =131.62 W

8 0
3 years ago
A mechanic uses a mechanical lift to raise a car. The car weighs 11,000 N. The lift raises the car 2.5 m.
KengaRu [80]

Potential energy = (weight) x (height)

After the car has been raised 2.5 meters, it has

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That's the energy that has to come from the work you do to lift it.

Since no mechanical process is ever 100% efficient, the work required
to accomplish this task is <em>at least  27,500 joules</em>.


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3 years ago
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