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Levart [38]
2 years ago
12

The area of the circular base of a can of

Chemistry
1 answer:
zlopas [31]2 years ago
6 0

Answer:

Volume = 75 inch³

Explanation:

Given data:

Area = 12.5 inch

Height = 6 inch

Volume =?

Solution:

Volume = height × length × width

Volume =  Area × height

Now we will put the values in formula;

Volume = 12.5 inch² ×  6 inch

Volume = 75 inch³

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Số oxi hoá của nitơ trong NH3, H2SO4, NO3 lần lượt là:
NikAS [45]

Answer:

3,6,6...................................

6 0
2 years ago
How many moles of KBr are present in 500 ml of a 0.8 M KBr solution?
faltersainse [42]

Answer:

2) 0.4 mol

Explanation:

Step 1: Given data

  • Volume of the solution (V): 500 mL
  • Molar concentration of the solution (M): 0.8 M = 0.8 mol/L

Step 2: Convert "V" to L

We will use the conversion factor 1 L = 1000 mL.

500 mL × 1 L/1000 mL = 0.500 L

Step 3: Calculate the moles of KBr (solute)

The molarity is the quotient between the moles of solute (n) and the liters of solution.

M = n/V

n = M × V

n = 0.8 mol/L × 0.500 L = 0.4 mol

4 0
2 years ago
Naturally occurring silicon has an atomic mass of 28.086 and consists of three isotopes. The major isotope is 28Si, natural abun
Elden [556K]

Answer:

29Si has a natural abundance of 4.68%.

30Si has a relative atomic mass of 29.99288 and a natural abundance of 3.09%.

Explanation:

The atomic mass of silicon is given by:

Si=Si²⁸×A₁+Si²⁹×A₂+Si³⁰×A₃

Where:

Si: atomic mass of silicon (28.086)

Si²⁸: relative atomic mass of 28Si (27.97693)

A₁: natural abundance of 28Si (92.23%)

Si²⁹: relative atomic mass of 29Si (28.97649)

A₂: natural abundance of 29Si

Si³⁰: relative atomic mass of 30Si

A₃: natural abundance of 30Si

We also know that 30Si natural abundance is in the ratio of 0.6592 to that of 29Si.

We have to set up a system of three equations in three unknowns:

Si=Si²⁸×A₁+Si²⁹×A₂+Si³⁰×A₃

A₃=0.6592×A₂

A₁+A₂+A₃=1

First, we find substitute the value of A₃ in the third equation and solv teh value of A₂:

A₁+A₂+0.6592×A₂=1

A₁+1.6592×A₂=1

1.6592×A₂=1-A₁

A₂=\frac{1-A₁}{1.6592}=\frac{1-0.9223}{1.6592}=0.0468

Then, we find the value of A₃:

A₃=0.6592×A₂

A₃=0.6592×0.0468=0.0309

Finally, we find the value of Si³⁰ in the first equation:

Si=Si²⁸×A₁+Si²⁹×A₂+Si³⁰×A₃

28.086=27.97693×0.9223+28.97649×0.0468+Si³⁰×0.0309

28.086=27.15922+Si³⁰×0.0309

28.086-27.15922=Si³⁰×0.0309

\frac{0.92678}{0.0309}=Si³⁰

Si³⁰=29.99288

8 0
3 years ago
A(n) ____________________ is a negatively charged subatomic particle.
gtnhenbr [62]
An electron is a negatively charged subatomic particle, whereas a proton is positively charged, and a neutron has no charge. 
8 0
3 years ago
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