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malfutka [58]
3 years ago
11

What is the average rate of the reaction between 10 and 20 s?

Chemistry
1 answer:
kherson [118]3 years ago
8 0
The average rate of the reaction between 10 and 20 is devide by 5
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choli [55]

Answer:a series of an chemical reaction

Explanation:

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3 years ago
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The plunger on a bicycle pump with a 400 mL volume cylinder is
Kobotan [32]

Answer:

The answer to your question is V2 = 66.7 ml

Explanation:

Data

Volume 1 = V1 = 400 ml

Pressure 1 = P1 = 1 atm

Volume 2 = V2 = ?

Pressure 2 = P2 = 6 atm

Process

1.- To solve this problem use Boyle's law

                     P1V1 = P2V2

-solve for V2

                     V2 = P1V1 / P2

-Substitution

                      V2 = (1)(400) / 6

-Simplification

                      V2 = 400 / 6

-Result

                      V2 = 66.7 ml

4 0
3 years ago
4. Which of the following have mass? Select all that apply.
valkas [14]
All of these EXCEPT helium in a balloon
4 0
3 years ago
What really is the Ebola virus?
Sloan [31]
The Ebola virus belongs to a family of viruses termed Filoviridae. Filovirus particles form long sometimes branched filaments of varying shapes, as well as shorter filaments , and may measure up to 14,000 nanometers in length with diameter of 80 nanometers. Viral particles contain one molecule of single stranded RNA enveloped in a lipid membrane. New viral particle bud from the surface of their host cell. Although Ebola virus was only discovered in 1976, it is an ancient virus and is thought to have split from other viruses thousands of years ago.
7 0
3 years ago
An unknown diprotic acid (H2A) requires 44.391 mL of 0.111 M NaOH to completely neutralize a 0.58 g sample. Calculate the approx
Anna [14]

Answer:

M=235.42g/mol

Explanation:

Hello!

In this case, given this is an acid-base neutralization and we are considering a diprotic acid, we can write the following mole-mole relationship:

2n_{acid}=n_{base}

It means that the moles of acid can be computed given the volume and concentration of NaOH:

n_{acid}=\frac{M_{base}V_{base}}{2} =\frac{0.044391L*0.111mol/L}{2} \\\\n_{acid}=2.46x10^{-4}mol

It means that the approximate molar mass of the acid is:

M=\frac{m_{acid}}{n_{acid}} \\\\M=\frac{m_{acid}}{n_{acid}} =\frac{0.58g}{2.46x10^{-3}mol}\\\\M=235.42g/mol

Best regards!

5 0
3 years ago
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