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2 years ago

What systems might not work right when you have a cold

Chemistry

1 answer:

Vsevolod [243]2 years ago

5 0

Answer:

hello :3

A cold is a contagious upper respiratory infection that affects your nose, throat, sinuses and trachea (windpipe).

Explanation:

have a nice day :3

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In the balanced chemical reaction shown below, what is the molar ratio of O2 to H2O

icang [17]

Answer: 5:3

Explanation:

4 0

2 years ago

What are 2 of the characteristics of temporary physical change

Natali5045456 [20]

Answer:

Temporary in nature.

No new substance is formed.

Explanation:

Temporary in nature: Does not affect the internal structure of a substance, only the molecules are rearranged.

No new substance is formed: Most of the physical changes are reversible. We can obtain the substance back even after the change.

hope this helps

have an awesome day -TJ

8 0

1 year ago

Read 2 more answers

The two isotopes of chlorine are LaTeX: \begin{matrix}35\\17\end{matrix}Cl35 17 C l and LaTeX: \begin{matrix}37\\17\end{matrix}C

Kay [80]

Answer: The percentage abundance of $_{17}^{35}\textrm{Cl}$ and $_{17}^{37}\textrm{Cl}$ isotopes are 77.5% and 22.5% 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 $_{17}^{35}\textrm{Cl}$ isotope be 'x'. So, fractional abundance of $_{17}^{37}\textrm{Cl}$ isotope will be '1 - x'

For $_{17}^{35}\textrm{Cl}$ isotope:

Mass of $_{17}^{35}\textrm{Cl}$ isotope = 35 amu

Fractional abundance of $_{17}^{35}\textrm{Cl}$ isotope = x

For $_{17}^{37}\textrm{Cl}$ isotope:

Mass of $_{17}^{37}\textrm{Cl}$ isotope = 37 amu

Fractional abundance of $_{17}^{37}\textrm{Cl}$ isotope = 1 - x

Average atomic mass of chlorine = 35.45 amu

Putting values in equation 1, we get:

$35.45=[(35\times x)+(37\times (1-x))]\\\\x=0.775$

Percentage abundance of $_{17}^{35}\textrm{Cl}$ isotope = $0.775\times 100=77.5\%$

Percentage abundance of $_{17}^{37}\textrm{Cl}$ isotope = $(1-0.775)=0.225\times 100=22.5\%$

Hence, the percentage abundance of $_{17}^{35}\textrm{Cl}$ and $_{17}^{37}\textrm{Cl}$ isotopes are 77.5% and 22.5% respectively.

6 0

3 years ago

You see the middle layer of the sun’s atmosphere, the _____________, at the start and end of a total eclipse.

Helen [10]

Answer:

Chromosphere

Explanation:

You see the middle layer of the sun’s atmosphere, the Chromosphere, at the start and end of a total eclipse.

4 0

2 years ago

What quantity of sodium azide in grams is required to fill a 56.0 liters air bag with nitrogen gas at 1.00 atm and exactly 0 °C:

Margarita [4]

Answer:

108.6 g

Explanation:

2NaN₃(s) → 2Na(s) + 3N₂(g)

First we use the PV=nRT formula to calculate the number of nitrogen moles:

P = 1.00 atm

V = 56.0 L

n = ?

R = 0.082 atm·L·mol⁻¹·K⁻¹

T = 0 °C ⇒ 0 + 273.2 = 273.2 K

Inputting the data:

1.00 atm * 56.0 L = n * 0.082 atm·L·mol⁻¹·K⁻¹ * 273.2 K

n = 2.5 mol

Then we convert 2.5 moles of N₂ into moles of NaN₃, using the stoichiometric coefficients of the balanced reaction:

2.5 mol N₂ * $\frac{2molNaN_3}{3molN_2}$ = 1.67 mol NaN₃

Finally we convert 1.67 moles of NaN₃ into grams, using its molar mass:

1.67 mol * 65 g/mol = 108.6 g

6 0

2 years ago