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blsea [12.9K]
3 years ago
12

A chemist working as a safety inspector finds an unmarked bottle in a lab cabinet. A note on the door of the cabinet says the ca

binet is used to store bottles of tetrahydrofuran, carbon tetrachloride, pentane, dimethyl sulfoxide, and acetone. The chemist plans to try to identify the unknown liquid by measuring the density and comparing to known densities. First, from his collection of Material Safety Data Sheets (MSDS), the chemist finds the following information: liquid density of tetrahydrofuran 0.89·gcm^−3, carbon tetrachloride 1.6·gcm^−3, pentane 0.63·gcm^−3, dimethyl sulfoxide 1.1·gcm^−3, acetone 0.79·gcm^−3. Next, the chemist measures the volume of the unknown liquid as 0.852L and the mass of the unknown liquid as 938.g . Calculate the density of the liquid. Be sure your answer has the correct number of significant digits.
Chemistry
1 answer:
lord [1]3 years ago
8 0

Answer:

DMSO

Explanation:

Volume of unknown liquid = 0.852L = 852mL or 852cm^3

Mass of unknown liquid = 938g

Density of unknown liquid = Mass of liquid / Volume of liquid

                                            = 938 / 852 = 1.100g/cm^3

From the given other data, the unknown liquid is Dimethyl sulfoxide (DMSO) as the MSDS of the mentioned liquid has been given 1.1g/cm^3

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Suppose 9 glucose molecules enter glycolysis. Calculate the number of inorganic phosphate molecules required as well as the numb
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Answer:

18 inorganic phosphates and 18 pyruvates.

Explanation:

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The net ATP production of Glycolysis involving a molecule of Glucose are 2 ATPs ,2 NADH and 2 pyruvates.

If 9 molecules of glucose enter glycolysis,then it has to be multiplied by 9 to give 18 pyruvates and 18 net ATPs

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3 years ago
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gizmo_the_mogwai [7]

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3 years ago
Three other application of Bernoulli's principle ?
Makovka662 [10]
<h2><u>Answer:</u></h2>

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7 0
3 years ago
4. DBearded waste of Co-60 must be stored until it is no longer radioactive. Cobalt-60
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<h3>Further explanation</h3>

Given

Sample waste of Co-60 = 14.5 g

26.5 years in storage

Required

Initial sample

Solution

General formulas used in decay:  

\large{\boxed{\bold{N_t=N_0(\dfrac{1}{2})^{t/t\frac{1}{2} }}}

t = duration of decay  

t 1/2 = half-life  

N₀ = the number of initial radioactive atoms  

Nt = the number of radioactive atoms left after decaying during T time  

Half-life of Co-60 = 5.3 years

Input the value :

\tt 14.5=No.\dfrac{1}{2}^{26.5/5.3}\\\\14.5=No.\dfrac{1}{2}^5\\\\No=\boxed{\bold{464~g}}

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3 years ago
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