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Why is it important for people to analyze the benefits and risks when deciding to use technology such as headphones?

solmaris [256]

Answer:

they should know things like if their hearing will be damaged or the side effects of wearing them constantly so they know the dangers before using them

4 0

3 years ago

Read 2 more answers

How do starch and ATP store and supply energy?

NeX [460]

Answer:

Explanation:

ATP is used for immediate energy and short-term storage, while starch molecules are stable and can be stored for a long time

ATP is used for immediate energy and long-term storage, while starch molecules are unstable and can be stored for a short amount of time.

8 0

3 years ago

What is the total number of oxygen atoms on the right-hand side of this chemical equation?

vivado [14]

The chemical equation is

Cu(s) +4HNO3(aq) ⇒ Cu(NO3)2(aq) + 2NO2(g) + 2H2O(g)

Answer:

12

Explanation:

In the right hand side of the equation, there are three compound which contains O2, which are;

Cu(NO3)2 , number of oxygen atoms =3*2 =6

2NO2, number of oxygen atoms = 2*2=4

2H2O, number of oxygen atoms =2*1=2

Total number of oxygen atoms on the right side of equation = 6+4+2 =12

8 0

3 years ago

[H+] for a solution is 4.59 x 10-6 M. This solution is A. acidic B. basic C. neutral

DedPeter [7]

Answer:

A. acidic

Explanation:

3 0

3 years ago

The element iridium exists in nature as two isotopes: 191Ir has a mass of 190.9606 u, and 193Ir has a mass of 192.9629 u. The av

nlexa [21]

Answer: The percentage abundance of $_{77}^{191}\textrm{Ir}$ and $_{77}^{193}\textrm{Ir}$ isotopes are 37.10% and 62.90% respectively.

Explanation:

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

$\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i$ .....(1)

Let the fractional abundance of $_{77}^{191}\textrm{Ir}$ isotope be 'x'. So, fractional abundance of $_{77}^{193}\textrm{Ir}$ isotope will be '1 - x'

For $_{77}^{191}\textrm{Ir}$ isotope:

Mass of $_{77}^{191}\textrm{Ir}$ isotope = 190.9606 amu

Fractional abundance of $_{77}^{191}\textrm{Ir}$ isotope = x

For $_{77}^{193}\textrm{Ir}$ isotope:

Mass of $_{77}^{193}\textrm{Ir}$ isotope = 192.9629 amu

Fractional abundance of $_{77}^{193}\textrm{Ir}$ isotope = 1 - x

Average atomic mass of iridium = 192.22 amu

Putting values in equation 1, we get:

$192.22=[(190.9606\times x)+(192.9629\times (1-x))]\\\\x=0.3710$

Percentage abundance of $_{77}^{191}\textrm{Ir}$ isotope = $0.3710\times 100=37.10\%$

Percentage abundance of $_{77}^{193}\textrm{Ir}$ isotope = $(1-0.3710)=0.6290\times 100=62.90\%$

Hence, the percentage abundance of $_{77}^{191}\textrm{Ir}$ and $_{77}^{193}\textrm{Ir}$ isotopes are 37.10% and 62.90% respectively.

8 0

3 years ago