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Orlov [11]
2 years ago
14

A sample of ascorbic acid (vitamin C) is synthesized in the laboratory. It contains 1.50 g of carbon and 2.00 g of oxygen. Anoth

er sample of ascorbic acid isolated from citrus fruits contains 6.25 g of carbon.?
Physics
1 answer:
Karo-lina-s [1.5K]2 years ago
6 0

Answer:

8.467 gm

Explanation:

The law governing this problem is "The Law of Constant Composition "

As per this law, all compounds irrespective of their origin and source have the same composition and properties  in their purest form

It is a simple proportion and ratio related problem.

1.50 grams of carbon require 2.0 grams of oxygen

1.0 grams of carbon will require oxygen

=\frac{2}{1.5}

6.35 grams of carbon will require oxygen

\frac{2}{1.5}\times6.35= 8.4666\\= 8.467

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Answer:

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6 0
2 years ago
What's faster a slow car or a fast feather?​
krok68 [10]

Answer:

A fast feather

Explanation:

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5 0
2 years ago
Read 2 more answers
Sound level B in decibels is defined as
Lelechka [254]

Answer:

The approximate combined sound  intensity is I_{T}=1.1\times10^{-4}W/m^{2}

Explanation:

The decibel  scale intensity for busy traffic is 80 dB. so intensity will be

10log(\frac{I_{1}}{I_{0}} )=80, therefore I_{1}=1\times10^{8}I_{0}=1\times10^{8} * 1\times10^{-12}W/m^{2}=1\times10^{-4}W/m^{2}

In the same way for the loud conversation having a decibel intensity of 70 dB.

10log(\frac{I_{2}}{I_{0}} )=70, therefore I_{2}=1\times10^{7}I_{0}=1\times10^{7} * 1\times10^{-12}W/m^{2}=1\times10^{-5}W/m^{2}

Finally we add both of them I_{T}=I_{1}+I_{2}=1\times10^{-4}W/m^{2}+1\times10^{-5}W/m^{2}=1.1\times10^{-4}W/m^{2}, is the approximate combined sound  intensity.

3 0
3 years ago
A cat dozes on a stationary merry-go-round, at a radius of 4.4 m from the center of the ride. The operator turns on the ride and
monitta

Answer:

The coefficient of static friction is 0.29

Explanation:

Given that,

Radius of the merry-go-round, r = 4.4 m

The operator turns on the ride and brings it up to its proper turning rate of one complete rotation every 7.7 s.

We need to find the least coefficient of static friction between the cat and the merry-go-round that will allow the cat to stay in place, without sliding. For this the centripetal force is balanced by the frictional force.

\mu mg=\dfrac{mv^2}{r}

v is the speed of cat, v=\dfrac{2\pi r}{t}

\mu=\dfrac{4\pi^2r}{gt^2}\\\\\mu=\dfrac{4\pi^2\times 4.4}{9.8\times (7.7)^2}\\\\\mu=0.29

So, the least coefficient of static friction between the cat and the merry-go-round is 0.29.

4 0
3 years ago
Compute the work performed when 32 pounds is lifted 10 feet.
Murljashka [212]
W = force * displacement
W = 32 pounds * 10 feet
Now you need to convert it to newton and meters
W = 142 N * 3.048 m = 434 J
(I approximated the conversions- I hope it helps)
7 0
3 years ago
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