What particle(s) of an atom would NOT found in the nucleus?

Which represents the empirical formula for a3B9

Viktor [21]

A3B9 represents a molecular formula. The representation of the empirical formula for this compound is AB3. This is so because the empirical formula is the simplest ratio of the atoms present in the molecule. You get AB3 when you divide the subscripts of A3B9, this is 3 and 9, by the greatest common factor, which is 3. 3/3 = 1 and 9/3 = 3, so the subscripts for the empirical formula are 1 and 3, which is what AB3 represents. Answer: AB3.

3 0

3 years ago

Should existing structures build from CCA-treated wood be removed?

Arturiano [62]

There are pros and cons as to whether CCA-treated (pressure-treated) wood should be removed from existing structures, and both sides are subjective. Some of the arguments for leaving it include: *When burned, the wood can release dangerous, and sometimes, lethal fumes. *If buried in a landfill, the chemicals can soak into the ground and eventually contaminate ground water. *Removing it can expose people to arsenic *It is costly to remove an existing infrastructure that may or may not be harming people *Studies conducted within the past decade have determined structures containing CCA-treated wood pose no hazard *Studies also concluded that children who played on CCA-treated playgrounds were exposed to arsenic levels lower than those that naturally occur in drinking water Some of the arguments for removing it include: *The EPA determined that some children could face higher cancer risks from exposure to CCA-treated wood *If removed, it will need to be disposed of and, as discussed above, that creates another set of problems that could affect a community's health. A possible solution is to leave existing CCA-treated wood in place but seek viable, safe alternatives for future structures.

3 0

3 years ago

the element carbon has two common isotopes: C-12 (12 U) AND C-13 (13.003355 U). IF THE AVERAGE ATOMIC MASS OF CARBON IS 12.0107

oksian1 [2.3K]

Answer:

The percent isotopic abundance of C- 12 is 98.93 %

The percent isotopic abundance of C- 13 is 1.07 %

Explanation:

we know there are two naturally occurring isotopes of carbon, C-12 (12u) and C-13 (13.003355)

First of all we will set the fraction for both isotopes

X for the isotopes having mass 13.003355

1-x for isotopes having mass 12

The average atomic mass of carbon is 12.0107

we will use the following equation,

13.003355x + 12 (1-x) = 12.0107

13.003355x + 12 - 12x = 12.0107

13.003355x- 12x = 12.0107 -12

1.003355x = 0.0107

x= 0.0107/1.003355

x= 0.0107

0.0107 × 100 = 1.07 %

1.07 % is abundance of C-13 because we solve the fraction x.

now we will calculate the abundance of C-12.

(1-x)

1-0.0107 =0.9893

0.9893 × 100= 98.93 %

98.93 % for C-12.

7 0

3 years ago

PLEASE SOMEONE HELP ME WITH THIS!

Bad White [126]

Answer:

1) 1.202 L , 2) 1.291 dg , 3) 204.877 and 4) 1.04x $10^{3\\}$

Explanation:

You need to review about conversion factors and how to use them in the correct order. You can cancel the units and get the ones that you need if you use the appropriate conversion factors, remember is a number that you can use to multiply or divide.

For your exercise:

1) The conversion factor is: 1 L = 1000 mL

You will need to divide by 1000 mL to obtain liters L

1202.57120 mL x $\frac{1 L}{1000 mL}$ = 1.202 L

2) The conversion factor is: 1 g = 10 dg

0.1290743 g x $\frac{10 dg}{ 1 g}$ = 1.291 dg

For the next exercises, you need to follow some rules:

1. All numbers that are different from Zero (non-zero digits) are significant figures.

2.The Zeros between non-zeros digits (Imbedded zeros) always are significant, 2007.

3. If you want to be specific and want some zeros to be significant you need to add a decimal point. For example 500. or 500.0

4. Leading zeros (to the left) are not significant.

5. Trailing zeros (zeros to the right) in a whole number without decimal point are not significant.

3) 843.062 - 638.1848 = 204.8772

Now if we round to 6 significant figures we get 204.877

4)123.0 x 8.43 = 1036.89

Now we round to 3 significant figures because 8.43 has the least significant figures.

1.04x $10^{3}$

4 0

3 years ago

Provide the quantum numbers for each electron in the element gadolinium

Basile [38]

Atomic # 64
electronic configuration is *[Xe] 6s^2 4f^8
n = 4
ℓ = 3
mℓ = -3
ms = -½

7 0

3 years ago