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3 years ago

12

A light wave traveling through an unknown medium strikes the boundary with

2 answers:

yanalaym [24]3 years ago

8 0

The unknown medium would be glass

Sever21 [200]3 years ago

5 0

Answer:

hello im breanna and i believe your answer is b.

Explanation:

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Help needed ASAP, I will mark your answer as brainliest.

beks73 [17]

Answer:

D

Explanation:

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6 0

2 years ago

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The mass of Carbon is 12 g/mol and the mass of oxygen is 16 g/mol. What is the approximate percent composition by mass of CO2 (f

nikitadnepr [17]

Answer:

the last one

27% by 73%

Explanation:

12/44*100 for C

and

16/44*100 for O

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3 years ago

Rsy-feiu-xad j0in only girlś

kirill [66]

Answer:

nahhhhhhhhhhhhhhhhhhhhhhh xD

have a good day :)

Explanation:

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3 years ago

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Consider the heating curve for water. A graph of the heating curve for water has time in minutes on the horizontal axis and Temp

Zolol [24]

Answer:

0°C.

Explanation:

Hello,

In this case, given the heating curve of water on the attached document, we can notice that at 0 °C the solid starts melting, which means that the melting point is reached. Melting point is known as a physical change whereby a solid changes to liquid by the addition of heat as it allows the molecules to separate to each other.

Best regards.

4 0

3 years ago

An unknown compound with a molar mass of 223.94 g/mol consists of 32.18% c, 4.50% h, and 63.32% cl. find the molecular formula f

Dima020 [189]

The actual number of atoms of each element present in the molecule of the compound is represented by the formula known as molecular formula.

Molar mass of the unknown compound = 223.94 g/mol (given)

Mass of each element present in the unknown compound is determined as:

Mass of carbon, $C$ :

$\frac{32.18}{100}\times 223.94 = 72.06 g$

Mass of hydrogen, $H$ :

$\frac{4.5}{100}\times 223.94 = 10.08 g$

Mass of chlorine, $Cl$ :

$\frac{63.32}{100}\times 223.94 = 141.79 g$

Now, the number of each element in the unknown compound is determined by the formula:

$number of moles = \frac{given mass}{molar mass}$

Number of moles of $C$ :

$number of moles = \frac{72.06}{12} = 6.005 mole\simeq 6 mole$

Number of moles of $H$ :

$number of moles = \frac{10.08}{1} = 10.08 mole\simeq 10 mole$

Number of moles of $Cl$

$number of moles = \frac{141.79}{35.5} = 3.99 mole\simeq 4 mole$

Dividing each mole with the smallest number of mole, to determine the empirical formula:

$C_{\frac{6}{4}}H_{\frac{10}{4}}Cl_{\frac{10}{4}}$

$C_{1.5}H_{2.5}Cl_{1}$

Multiplying with 2 to convert the numbers in formula into a whole number:

So, the empirical formula is $C_{3}H_{5}Cl_{2}$ .

Empirical mass = $12\times 3+1\times 5+2\times 35.5 = 112 g/mol$

In order to determine the molecular formula:

n = $\frac{molar mass}{empirical mass}$

n = $\frac{223.94}{112} = 1.99 \simeq 2$

So, the molecular formula is:

$2\times C_{3}H_{5}Cl_{2} = C_{6}H_{10}Cl_{4}$

6 0

3 years ago