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Natasha2012 [34]
3 years ago
7

How many significant digits are 6.3590x10 7 mm

Chemistry
2 answers:
Svet_ta [14]3 years ago
8 0
If you go by the decimal there should be 5 significant digits because 0 is a trailing zero and matters since it comes after a decimal. however if you do 

6.3590 x 10 ^7 = 63590000 then you would only have 4 significant digits because there is no longer a decimal to make the zero after the 9, significant. 

Hope this helps!
Tanzania [10]3 years ago
7 0

Answer:

There are 5 significant digits

Explanation:

Significant digits are essentially numbers that contribute directly towards a measurement. There are certain rules to be followed while identifying the number of significant digits in a measurement, these include:

1) Non-zero digits are significant

2) Leading zeros are insignificant

3) Trailing zeros only after a decimal point are significant

4) Zeros in between non-zero digits are significant

The given value is:

6.3590*10⁷

There are 4 non-zero digits + 1 trailing zero after the decimal.

Therefore, significant digits are 4 +  1 = 5

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A certain photon has a frequency of 634x1012 Hz. What is the photon's wavelength in a vacuum? Enter your answer in nm (x10-9 m).
Shtirlitz [24]
<h3>Answer:</h3>

4.73 × 10^4 m

<h3>Explanation:</h3>

From the question;

Frequency of the photon = 634 × 10^12 Hz

We are required to calculate the wavelength of the photon.

We need to know the relationship between wavelength and frequency of a wave.

The relationship between f and λ is given by;

c = fλ

Where c, is the speed of light, 2.998 × 10^8 m/s

Therefore, to get the wavelength we rearrange the formula such that;

λ = c ÷ f

  = 2.998 × 10^8 m/s ÷ 634 × 10^12 Hz

  = 4.73 × 10^-5 m

But we require wavelength in nm

1 M = 10^9 nm

Therefore;

Wavelength = 4.73 × 10^-5 m × 10^9 nm/m

                    = 4.73 × 10^4 m

Hence, the photon's wavelength is 4.73 × 10^4 m

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Read 2 more answers
What is the total volume of gaseous products formed when 116 liters of butane (C4H10) react completely according to the followin
Contact [7]

<u>Answer:</u> The total volume of the gaseous products is 1044.29 L

<u>Explanation:</u>

We are given:

Volume of butane = 116 L

At STP:

22.4 L of volume is occupied by 1 mole of a gas

So, 116 L of volume will be occupied by = \frac{1}{22.4}\times 116=5.18mol of butane

The chemical equation for the combustion of butane follows:

2C_4H_{10}(g)+13O_2(g)\rightarrow 8CO_2(g)+10H_2O(g)

  • <u>For carbon dioxide:</u>

By Stoichiometry of the reaction:

2 moles of butane produces 8 moles of carbon dioxide

So, 5.18 moles of butane will produce = \frac{8}{2}\times 5.18=20.72mol of carbon dioxide

Volume of carbon dioxide at STP = (20.72 × 22.4) = 464.13 L

  • <u>For water vapor:</u>

By Stoichiometry of the reaction:

2 moles of butane produces 10 moles of water vapor

So, 5.18 moles of butane will produce = \frac{10}{2}\times 5.18=25.9mol of water vapor

Volume of water vapor at STP = (25.9 × 22.4) = 580.16 L

Total volume of the gaseous products = [464.13 + 580.16] = 1044.29 L

Hence, the total volume of the gaseous products is 1044.29 L

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3 years ago
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