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3 years ago

Why is it easy for archaeologists to find campsites of early explorer of over 200 years ago?

Chemistry

1 answer:

Ivenika [448]3 years ago

5 0

Here's a website with the answer and some cool reading:
http://history-world.org/archeology.htm

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A container of a mixture of 4 gases has a total pressure of 35.7 kPa. Gas A has a partial pressure of 7.8kPa. Gas B has a partia

Maurinko [17]

Answer:

partial pressure of gas D Pd = 15.5 kPa

Explanation:

As per the Dalton's law of partial pressure, in a mixture, pressure exerted by each gas when summed gives the total partial pressure exerted by mixture.

P(Total) = P1+P2+P3.....

Given P(Total) = 35.7 kPa

Partial pressure of gas A Pa = 7.8 kPa

Partial pressure of gas B Pb = 3.7 kPa

Partial pressure of gas C Pc = 8.7 kPa

There, Partial pressure of gas D Pd = P(Total) -(Pa+Pb+Pc)

Pd = 35.7-(7.8+3.7+8.7) = 35.7-20.2 kPa = 15.5 kPa

Therefore, partial pressure of gas D Pd = 15.5 kPa

4 0

3 years ago

What does scientist use to deign their experiments

joja [24]

Answer:

a chart

Explanation:

a chart a chart is the answer

8 0

2 years ago

Answer right = Brainliest!!

amid [387]

A b and e is are the answers

8 0

2 years ago

Assume an object A with a mass of 66.789 g and a volume of 10.1 mL. what is its density?

docker41 [41]

Answer: 6.61

Explanation:

5 0

3 years ago

Consider the nuclear equation below. Superscript 235 subscript 92 upper U right arrow superscript 4 subscript 2 upper H e. What

Leto [7]

Answer: The nuclide symbol of X is $^{231}_{90}\textrm{Th}$

Explanation:

The given nuclear reaction is a type of alpha decay process. In this process, the nucleus decays by releasing an alpha particle. The mass number of the nucleus is reduced by 4 units and atomic number is also decreased by 2 units. The particle released is a helium nucleus.

The general equation representing alpha decay process is:

$_{Z}^{A}\textrm{X}\rightarrow _{Z-2}^{A-4}\textrm{Y}+_2^4\textrm{He}$

For the given equation :

$^{235}_{92}\textrm{U}\rightarrow ^{A}_{Z}\textrm{X}+^4_2\textrm{He}$

As the atomic number and mass number must be equal on both sides of the nuclear equation:

$^{235}_{92}\textrm{U}\rightarrow ^{231}_{90}\textrm{Th}+^4_2\textrm{He}$

Thus the nuclide symbol of X is $^{231}_{90}\textrm{Th}$

5 0

3 years ago