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nekit [7.7K]
2 years ago
10

Potassium chlorate decomposes to form potassium chloride and oxygen according to the following

Chemistry
1 answer:
Tresset [83]2 years ago
8 0

Answer:

O 392 g

Explanation:

39.2 grams of oxygen (O2) can be produced.

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Answer:

C.

Explanation:

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In nematode worms, a gene that codes for an ATPase has two alternatives for exon 4 and three alternatives for exon 7. How many d
Artyom0805 [142]

Answer:

6 different forms of the protein could be made.

Explanation:

For the given nematode worm, 6 different forms of the protein could be made. This is because of the alternative splicing that will produce 6 kinds of mRNAs. We have 2 different forms for the exon 4 while we have 3 differen forms for the exon 7. Therefore, we have a total of (2*3) 6 different forms of the protein for the given nematode worm.

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3 years ago
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How many milliliters of a 0.215 molar solution are required to contain 0.0867 mol of NaBr
JulsSmile [24]

Answer: 403ml

Explanation:

M=\frac{mol}{L}

Solve for L;

L=\frac{mol}{M}\\L=\frac{0.0867mol}{0.215M}\\ L=0.403L

Convert to mililiters

0.403L(\frac{1000ml}{1L})=403ml

3 0
2 years ago
Technetium (Tc; Z = 43) is a synthetic element used as a radioactive tracer in medical studies. A Tc atom emits a beta particle
Sedaia [141]

Question in incomplete, complete question is:

Technetium (Tc; Z = 43) is a synthetic element used as a radioactive tracer in medical studies. A Tc atom emits a beta particle (electron) with a kinetic energy (Ek) of 4.71\times 10^{-15}J . What is the de Broglie wavelength of this electron (Ek = ½mv²)?

Answer:

6.762\times 10^{-12} m is the de Broglie wavelength of this electron.

Explanation:

To calculate the wavelength of a particle, we use the equation given by De-Broglie's wavelength, which is:

\lambda=\frac{h}{\sqrt{2mE_k}}

where,

= De-Broglie's wavelength = ?

h = Planck's constant = 6.624\times 10^{-34}Js

m = mass of beta particle = 9.1094\times 10^{-31} kg

E_k = kinetic energy of the particle = 4.71\times 10^{-15}J

Putting values in above equation, we get:

\lambda =\frac{6.624\times 10^{-34}Js}{\sqrt{2\times 9.1094\times 10^{-31} kg\times 4.71\times 10^{-15}J}}

\lambda = 6.762\times 10^{-12} m

6.762\times 10^{-12} m is the de Broglie wavelength of this electron.

3 0
3 years ago
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