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3 years ago

A prion is an infectious self-reproducing protein structure. true or false

Chemistry

1 answer:

Alexandra [31]3 years ago

8 0

Answer:

True is the correct answer.

Explanation:

The statement that a prion is an infectious, self-reproducing protein structure is true.

prion is an infectious particle and they do not have genetic material.

Prion present in the brain region that results in deadly neurodegenerative illnesses in humans and animals.

Prions are the self-producing proteinaceous infectious capable of transferring infection in the absence of the nucleic acids.
prions enter in the brain by an infection and they emerge from the variation in the gene that encodes the protein and once the prions present in the brain multiply by causing the benign proteins to refold into an abnormal form.

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There are two binary compounds of mercury and oxygen. heating either of them results in the decomposition of the compound, with

grandymaker [24]

$\text{Hg} \text{O}$ and $\text{Hg}_{2} \text{O}$ .

Assuming complete decomposition of both samples,

$m(\text{Hg}) = m(\text{residure})$
$m(\text{O}) = m(\text{loss})$

First compound:

$m(\text{O}) = m(\text{loss}) = 0.6498 - 0.6018 = 0.048 \; g$
$m(\text{Hg}) = m(\text{residure}) = 0.6018 \; g$

$n = m/M$ ; $0.6498 \; g$ of the first compound would contain

$n(\text{O atoms}) = 0.048 \; g / 16 \; g \cdot mol^{-1}= 0.003 \; mol$
$n(\text{Hg atoms}) = 0.6018 \; g / 200.58 \; g \cdot mol^{-1}= 0.003 \; mol$

Oxygen and mercury atoms seemingly exist in the first compound at a $1:1$ ratio; thus the empirical formula for this compound would be $\text{Hg} \text{O}$ where the subscript "1" is omitted.

Similarly, for the second compound

$m(\text{O}) = m(\text{loss}) = 0.016 \; g$
$m(\text{Hg}) = m(\text{residure}) = 0.4172 - 0.016 = 0.4012 \; g$

$n = m/M$ ; $0.4172 \; g$ of the first compound would contain

$n(\text{O atoms}) = 0.016 \; g / 16 \; g \cdot mol^{-1}= 0.001 \; mol$
$n(\text{Hg atoms}) = 0.4012 \; g / 200.58 \; g \cdot mol^{-1}= 0.002 \; mol$

$n(\text{Hg}) : n(\text{O}) \approx 2:1$ and therefore the empirical formula

$\text{Hg}_{2} \text{O}$ .

8 0

3 years ago

2.41 x 10^2 Cm to meters

Assoli18 [71]

We know that there are 100 cm in 1 m, so we can use this to convert to meters:

$(\frac{2.41x10^{2}cm}{1})*(\frac{1m}{100cm} )=2.41m$

Therefore we know that $2.41x10^{2}$ cm is equal to 2.41 m.

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n200080 [17]

The first one, my friend

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4vir4ik [10]

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Troyanec [42]

Answer: I think the answer is 1

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