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o-na [289]
3 years ago
12

A 10 wheeler truck collides with a car. Who is more likely to be hurt: the driver of the car or the driver of the truck? please

justify your answer.
THANK YOU ت︎​
Chemistry
2 answers:
yulyashka [42]3 years ago
8 0

Answer:

The driver in the car because a truck is a bigger vehicle and the impact of a truck to a car is more damaging than a car to a truck. Weight difference is the key I think.

Explanation:

Lorico [155]3 years ago
6 0
The driver in the car because the truck has more mass and sense it has more mass it has a greater acceleration than the car according to Newton’s first law. So the car would get more of an impact causing it to have a higher chance for the driver in the car to be injured!
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How many moles of co2 must enter the calvin cycle for the synthesis of one mole of glucose?
11111nata11111 [884]

Answer:3 moles

Explanation:

For every three molecules of CO2 that enters the Calvin cycle, one molecule of the three carbon glyceraldehyde 3-phosphate (G3P) is produced. Two molecules of G-3-P are required to produce one molecule of glucose. Therefore, the Calvin cycle needs to make a total of 6 turns to produce two molecules of G-3-P.

7 0
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_C4H6O3 + _H2O −−→ _C2H4O2<br><br> A) 1, 1, 2<br> B) 2, 1, 1<br> C) 2, 1, 2<br> D) 2, 2, 1
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Explanation:

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8 0
3 years ago
The atomic masses of 151eu and 153eu are 150.919860 and 152.921243 amu, respectively. The average atomic mass of europium is 151
vitfil [10]

Answer:-  The natural abundance of ^1^5^1_E_u is 0.478 or 47.8% and ^1^5^3_E_u is 0.522 or 52.2% .

Solution:- Average atomic mass of an element is calculated from the atomic masses of it's isotopes and their abundances using the formula:

Average atomic mass = mass of first isotope(abundance) + mass of second isotope(abundance)

We have been given with atomic masses for ^1^5^1_E_u and ^1^5^3_E_u as 150.919860 and 152.921243 amu, respectively.  Average atomic mass of Eu is 151.964 amu.

Sum of natural abundances of isotopes of an element is always 1. If we assume the abundance of ^1^5^1_E_u as n then the abundance of ^1^5^3_E_u would be 1-n .

Let's plug in the values in the formula:

151.964=150.919860(n)+152.921243(1-n)

151.964=150.919860n+152.921243-152.921243n

on keeping similar terms on same side:

151.964-152.921243=150.919860n-152.921243n

-0.957243=-2.001383n

negative sign is on both sides so it is canceled:

0.957243=2.001383n

n=\frac{0.957243}{2.001383}

n=0.478

The abundance of ^1^5^1_E_u is 0.478 which is 47.8%.  

The abundance of ^1^5^3_E_u is = 1-0.478

= 0.522 which is 52.2%

Hence, the natural abundance of ^1^5^1_E_u is 0.478 or 47.8% and ^1^5^3_E_u is 0.522 or 52.2% .


3 0
3 years ago
Which of the following describes what happens in a chemical reaction
Rus_ich [418]
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5 0
3 years ago
Considere un elemento "X" que posee como valencias 2, 4 y 6. En su nomenclatura tradicional, considerando la valencia cuatro, es
Ierofanga [76]

Answer:

"ite"

Explanation:

Se sabe que los elementos del grupo 16 de la tabla prioritaria muestran estados de oxidación o valencias de 2,4 y 6 respectivamente, dependiendo del compuesto formado.

Por ejemplo, el azufre forma los siguientes compuestos;

sulfato de sodio (el azufre tiene una valencia de 6)

Sulfito de sodio (el azufre tiene una valencia de 4)

Sulfuro de sodio (el azufre tiene una valencia de 2)

Por lo tanto, en compuestos en los que exhiben una valencia de 4, la terminación común es "ite"

8 0
3 years ago
Read 2 more answers
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