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3 years ago

10 m/s to 80 m/s in 5 seconds

Chemistry

2 answers:

dusya [7]3 years ago

7 0

900 mph................

klio [65]3 years ago

7 0

Answer:

a = 0.8 m/s/s

Vav = 45m/s

Explanation:

Vav = $\frac{Vi + Vf}{2\\}$

Vav = $\frac{10 +80}{2}$

Vav = 45m/s

a = $\frac{v}{t}$

a = $\frac{4m/s}{5s}$

a= 0.8m/s/s

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Complete the following radioactive decay problem.<br>222 86 RN to 4 2 HE + ?

Andreas93 [3]

Answer:

₈₆²²²Rn → ₈₄Po²¹⁸ + H₂⁴

Explanation:

The given nuclear reaction shows alpha decay.

₈₆²²²Rn → ₈₄Po²¹⁸ + H₂⁴

Properties of alpha radiations:

Alpha radiations are emitted as a result of radioactive decay. The atom emit the alpha particles consist of two proton and two neutrons. Which is also called helium nuclei. When atom undergoes the alpha emission the original atom convert into the atom having mass number less than 4 and atomic number less than 2 as compared to parent atom the starting atom.

Alpha radiations can travel in a short distance.

These radiations can not penetrate into the skin or clothes.

These radiations can be harmful for the human if these are inhaled.

These radiations can be stopped by a piece of paper.

₉₂U²³⁸ → ₉₀Th²³⁴ + ₂He⁴ + energy

8 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]

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

<u>Explanation:</u>

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'

<u>For $_{77}^{191}\textrm{Ir}$ isotope:</u>

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

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

<u>For $_{77}^{193}\textrm{Ir}$ isotope:</u>

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

If a patient requires 30g/day of some medication and it comes in 500 mg tablets , how many tablets does this patient need in one

Anna11 [10]

Since 1000 mg=1g
500mg=?
500/1000*1g=0.5g
Since we know that 500mg is 0.5g then divide 30g by 0.5g
30/0.5=60
Therefore the patients needs to take 60 tablets a day.

4 0

4 years ago

What is the most common acid and base?<br>and why?

KatRina [158]

Answer:

Acids and bases are used in most many chemical reactions in chemistry . They are responsible for most colour changes in a chemical reaction and are used to adjust the pH of chemical solutions.

Explanation:

3 0

4 years ago

The human body needs at least 1.03 x 10-^2 mol O2 every minute. If all of this oxygen is used for the

Neko [114]

<h3>Answer:</h3>

0.3093 g of glucose are consumed each minute by the body.

<h3>Explanation:</h3>

During cellular respiration glucose is broken down in presence of oxygen to yield energy, water and carbon dioxide.
The equation for the reaction taking place during cellular respiration is;

C₆H₁₂O₆ + 6O₂ → 6H₂O + 6CO₂

We are required to calculate the amount of glucose in grams;

<h3>Step 1: Calculate the moles of glucose broken down</h3>

From the equation, the mole ratio of glucose to Oxygen is 1 : 6

Moles of Oxygen in a minute is 1.03 × 10^-2 moles

Therefore, moles of glucose will be;

= (1.03 × 10^-2)÷6

= 1.717 × 10^-3 moles

<h3>Step 2: Mass of glucose </h3>

Mass is given by multiplying the number of moles with molar mass

mass = moles × molar mass

Molar mass glucose is 180.156 g/mol

Therefore;

Mass = 1.717 × 10^-3 × 180.156 g/mol

= 0.3093 g

Hence, 0.3093 g of glucose are consumed each minute by the body.

3 0

4 years ago