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Marta_Voda [28]

3 years ago

14

Use the Gizmo to estimate the age of each of the objects below. For these questions, each second in the Gizmo represents 1,000 y

ears.(8 pts)DescriptionAge (years)Egyptian papyrus with 63% of its original carbon-14 atomsAboriginal charcoal with 22% of its original carbon-14 atoms.Mayan headdress with 79% of its original carbon-14 atomsNeanderthal skull with 3% of its original carbon-14 atoms

1 answer:

ziro4ka [17]3 years ago

7 0

Answer:

Check the explanation

Explanation:

AT = A0 e(-T/H)

... where A0 is the starting activity, AT is the activity at some time T, and H is the half-life, in units of T.

Substituting what we know, we get...

0.71 = (1) e(-T/5730)

Solve for T...

loge(0.71) = -T/5730

T = -loge(0.71)(5730)

T = 1962 (conservatively rounded, T = 2000)

similarly for all

for aboriginal charcoal

0.28 = (1) e(-T/5730)

Solve for T...

loge(0.28) = -T/5730

T = -loge(0.28)(5730)

T = 7294 (conservatively rounded, T = 7000)

for mayan headdress

0.89 = (1) e(-T/5730)

Solve for T...

loge(0.89) = -T/5730

T = -loge(0.89)(5730)

T = 667 (conservatively rounded, T = 700)

for neanderthal

0.05 = (1) e(-T/5730)

Solve for T...

loge(0.05) = -T/5730

T = -loge(0.05)(5730)

T = 17165 (conservatively rounded, T = 17000)

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Hydrogen atoms in a discharge tube are in their ground state. If a beam of free electrons of energy12.9 eV are fired at the hydr

defon

Answer:

See explanation

Explanation:

-0.7ev = -13.6/n^2

n= 4

4--->1

4--->3, 3---> 2, 2--->1

E4-E1= -13.36(1/4^2 - 1/1^2) = 12.53 eV

E4-E3 = -13.36 (1/4^2 - 1/3^2) = 0.65 ev

E3-E2 = -13.36 (1/3^2 - 1/2^2) = 1.86 eV

E2- E1 = -13.6 ( 1/2^2 - 1/1^2) = 10.02 ev

a)

λ= 6.6 × 10^-34 × 3 × 10^8/12.53 × 1.6 × 10^-19

λ= 19.8 × 10^-26/20.048 ×10^-19

λ= 9.9 × 10^ -8 m

b)

λ= 6.6 × 10^-34 × 3 × 10^8/0.65 × 1.6 × 10^-19

λ= 19.8 × 10^-26/1.04 ×10^-19

λ= 19.0 × 10^ -7 m

3 0

3 years ago

Question 17 (Multiple Choice Worth 1 points)

Vesnalui [34]

17. not sure
18. not sure
19. movement of rock from one place to another
20. deposition
21. Urbanization
22. sedimentary
23. ease of traffic movement
24. absolute dating
25. fracture
26. increased farm land producing food for people

6 0

2 years ago

Are the particles in a solid motionless ?

kifflom [539]

No. Just because it is a solid doesn't mean it has no motion. The only way for particles to be motionless is for the solid to reach Absolute Zero (0* K). Which is impossible. So, therefore, it is impossible for particles in a solid to be motionless.

7 0

3 years ago

Penicillinase, also known as β‑lactamase, is a bacterial enzyme that hydrolyzes and inactivates the antibiotic penicillin. Penic

sergeinik [125]

Explanation:

The given data is as follows.

$V_{max} = 6.8 \times 10^{-10} \mu mol/min$

$K_{m} = 5.2 \times 10^{-6} M$

Now, according to Michaelis-Menten kinetics,

$V_{o} = V_{max} \times [\frac{S}{(S + Km)}]$

where, S = substrate concentration = $10.4 \times 10^{-6}$ M

Now, putting the given values into the above formula as follows.

$V_{o} = V_{max} \times [\frac{S}{(S + Km)}]$

$V_{o} = 6.8 \times 10^{-10} \mu mol/min \times [\frac{10.4 \times 10^{-6} M}{(10.4 \times 10^{-6}M + 5.2 \times 10^{-6} M)}]$

$V_{o} = 6.8 \times 10^{-10} \mu mol/min \times 0.667$

= $4.5 \times 10^{-10} \mu mol/min$

This means that $V_{o}$ would approache $V_{max}$ .

5 0

3 years ago

An oxygen atom has a mass of 2.66 x 10^-23 g and a glass of water has a mass of 0.050kg. Use this information to answer the ques

fomenos

<h2>ANSWER OF EACH PART ARE GIVEN BELOW</h2>

Explanation:

A)

We know, each mole contains $N_A=$ $6.023 \times 10^{23}$ atoms.

It is given that mass of one oxygen atom is m= $2.66\times 10^{-23}\ g$ .

Therefore, mass of one mole of oxygen, $M=m\times N_A$ .

Putting value of n and $N_A$ ,

$M=2.66\times 10^{-23}\times 6.023\times 10^{23} \ gm\\M=16.0\ gm$

B)

Given,

Mass of water in glass=0.050 kg = 50 gm.

From above part mass of one mole of oxygen atoms = 16.0 gm.

Therefore, number of mole of oxygen equivalent to 50 gm oxygen $=\dfrac{50}{16}=3.1 \ moles.$

LEARN MORE :

Avogadro's number

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3 0

3 years ago