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Aliun [14]
3 years ago
7

The number 199,078 has how many significant digits?

Chemistry
1 answer:
nikitadnepr [17]3 years ago
8 0

I think it's 6

Explanation:

zeros between other nonzero digits are significant

ex. 50.3 has 3 significant figures. ex. 8101.01 has 6 significant figures

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An airplane flies 2100 miles in 3hours. What is it’s average speed in miles per hour?
Anestetic [448]

Answer:

700 miles per hour

Explanation:

average \: speed  \\ =  \frac{distance}{time}  \\  =  \frac{2100}{3}  \\  = 700 \: miles \: per \: hour

5 0
3 years ago
Only need help with number 1 plzzz help
just olya [345]

Answer:According to the data results and the period table, these elements/substances are similar because they are in the same group; not only that but they both discharge from metals.

Explanation: I got you

8 0
3 years ago
Which of the following is an example of a chemical change?
ad-work [718]

Answer:

burning a paper is chemical change

8 0
3 years ago
What is the lowest unoccupied molecular orbital in F2?
Juliette [100K]

Answer:

σ*2pₓ, also called 3\sigma_{u}\text{*}

Explanation:

I have drawn the MO diagram for fluorine below.

Each F atom contributes seven valence electrons, so we fill the MOs of fluorine with 14 electrons.

We have filled the \pi \text{*2p}_{y} and \pi \text{*2p}_{z} MOs.

They are the highest occupied molecular orbitals (HOMOs).

The next unfilled level (the LUMO) is the σ*2pₓ orbital. If you use the symmetry notation, it is called the 3\sigma_{u}\text{*} orbital.

This is the orbital that fluorine uses when it acts as an electron acceptor.

7 0
3 years ago
6.50 g of a certain Compound x, known to be made of carbon, hydrogen and perhaps oxygen, and to have a molecular molar mass of 1
gayaneshka [121]

Answer:

C_{10}H_8

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to find the molecular formula of the given compound by firstly calculating both moles and grams of carbon in carbon dioxide and hydrogen in water, as the only sources of these elements derived from the compound x due to its combustion:

n_C=22.35gCO_2*\frac{1molCO_2}{44.01gCO_2}*\frac{1molC}{1molCO_2}=0.51molC\\\\m_C=0.51molC*\frac{12.01gC}{1molC}   =6.10gC

n_H=3.66gH_2O*\frac{1molH_2O}{18.02gH_2O}*\frac{2molH}{1molH_2O}=0.41molH\\\\m_H=0.41molH*\frac{1.01gH}{1molH}=0.41g

Now, since the addition of carbon and hydrogen is about 6.50 grams, we infer the compound has no oxygen, that is why we now set the mole ratios in the empirical formula for both C and H as shown below:

C:\frac{0.51mol}{0.41mol}= 1.24\\\\H:\frac{0.51mol}{0.51mol}= 1\\\\C_{1.24}H

Yet it cannot be decimal, that is why we multiply by 4 to get the correct whole-numbered empirical formula:

C_5H_4

Whose molar mass is 64.09 g/mol, which makes the ratio of molar masses:

\frac{128.g/mol}{64.09g/mol} =2

Therefore, the molecular formula is twice the empirical one:

C_{10}H_8

Regards!

8 0
3 years ago
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