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3 years ago

13

A protein is a polymer that is made of a. simple sugars. c. amino acids. b. nitrogen and carbon dioxide. d. DNA.

2 answers:

MA_775_DIABLO [31]3 years ago

7 0

Hello Lilsavage717, I believe the answer you are looking for is C. or Amino Acids. They are mainly located in living things like plants or bugs.

VladimirAG [237]3 years ago

4 0

My guess whould be a) but I'm not sure if that's right but have a good day

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What one is correct?

katen-ka-za [31]

4

Explanation:

I am not positive but it makes more sense

7 0

3 years ago

Color change is most indicative of which type of chemical reaction?

makvit [3.9K]

Answer; chemical reaction

4 0

3 years ago

the spectral lines observed for hydrogen arise from transitions from excited states back to the n=2 principle quantum level. Cal

Sunny_sXe [5.5K]

Rydberg formula is given by:

$\frac{1}{\lambda } = R_{H}\times (\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} )$

where, $R_{H}$ = Rydberg constant = $1.0973731568508 \times 10^{7} per metre$

$\lambda$ = wavelength

$n_{1}$ and $n_{2}$ are the level of transitions.

Now, for $n_{1}$ = 2 and $n_{2}$ = 6

$\frac{1}{\lambda} = 1.0973731568508 \times 10^{7} \times (\frac{1}{2^{2}}-\frac{1}{6^{2}} )$

= $1.0973731568508 \times 10^{7} \times (\frac{1}{4}-\frac{1}{36} )$

= $1.0973731568508 \times 10^{7} \times (0.25-0.0278 )$

= $1.0973731568508 \times 10^{7} \times 0.23$

= $0.2523958\times 10^{7}$

$\lambda = \frac{1}{0.2523958\times 10^{7}}$

= $3.9620\times 10^{-7} m$

= $396.20\times 10^{-9} m$

= $396.20 nm$

Now, for $n_{1}$ = 2 and $n_{2}$ = 5

$\frac{1}{\lambda} = 1.0973731568508 \times 10^{7} \times (\frac{1}{2^{2}}-\frac{1}{5^{2}} )$

= $1.0973731568508 \times 10^{7} \times (0.25-0.04 )$

= $1.0973731568508 \times 10^{7} \times (0.21 )$

= $0.230 \times 10^{7}$

$\lambda= \frac{1}{0.230 \times 10^{7}}$

= $4.3478 \times 10^{-7} m$

= $434.78\times 10^{-9} m$

= $434.78 nm$

Now, for $n_{1}$ = 2 and $n_{2}$ = 4

$\frac{1}{\lambda} = 1.0973731568508 \times 10^{7} \times (\frac{1}{2^{2}}-\frac{1}{4^{2}} )$

= $1.0973731568508 \times 10^{7} \times (0.25-0.0625 )$

= $1.0973731568508 \times 10^{7} \times (0.1875 )$

= $0.20575 \times 10^{7}$

$\lambda= \frac{1}{0.20575 \times 10^{7}}$

= $4.8602 \times 10^{-7} m$

= $486.02 \times 10^{-9} m$

= $486.02 nm$

Now, for $n_{1}$ = 2 and $n_{2}$ = 3

$\frac{1}{\lambda} = 1.0973731568508 \times 10^{7} \times (\frac{1}{2^{2}}-\frac{1}{3^{2}} )$

= $1.0973731568508 \times 10^{7} \times (0.25-0.12 )$

= $1.0973731568508 \times 10^{7} \times (0.13 )$

= $0.1426585\times 10^{7}$

$\lambda= \frac{1}{0.1426585\times 10^{7}}$

= $7.0097 \times 10^{-7} m$

= $700.97 \times 10^{-9} m$

= $700.97 nm$

5 0

3 years ago

Read 2 more answers

7. Which statement correctly describes the density of Earth's mantle compared to the density of Earth's

Mice21 [21]

Answer:

1) the mantle is less dense than the core but more dense than the crust.

5 0

3 years ago

One application of Hess's Law (which works for ΔH, ΔS, and ΔG) is calculating the overall energy of a reaction using standard en

VikaD [51]

Answer:

The standard change in free energy for the reaction = - 437.5 kj/mole

Explanation:

The standard change in free energy for the reaction:

4 KClO₃ (s) → 3 KClO₄(s) + KCl(s)

Given that ΔGf(KClO3(s)) = -290.9 kJ/mol;

ΔGf(KClO4(s)) = -300.4 kJ/mol;

ΔGf(KCl(s)) = -409 kJ/mol

According to Hess's law

ΔGr (Free energy change of reaction)= ∑(Product free energy - reactant free energy)

⇒ ΔGr⁰ = {3 x (-300.4) + (-409)} - {3 x (- 290.9)}

= - 901.2 - 409 + 872.7

= - 437.5 kj/mole

3 0

3 years ago