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3 years ago

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Which of the following substances would best be measured using mole quantities?

1 answer:

galina1969 [7]3 years ago

4 0

Molecules of water in the ocean

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Which has a greater ionization energy: lithium or bromine? Which has a greater ionization energy: vanadium or sodium?

vladimir2022 [97]

1.Lithium has the greater ionization energy
2. Vanadium has a greater ionization energy

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2 years ago

The empirical formula of an organic compound is C2H4O. The molecular mass of the compound is 176g/mol.

Brrunno [24]

Answer:

The molecular formula of the compound is $C_{8}H_{16}O_{4}$ . The molecular formula is obtained by the following expression shown below

$\textrm{Molecular formula }= n\times \textrm{Empirical formula}$

Explanation:

Given molecular mass of the compound is 176 g/mol

Given empirical formula is $C_{2}H_{4}O$

Atomic mass of carbon, hydrogen and oxygen are 12 u , 1 u and 16 u respectively.

Empirical formula mass of the compound = $\left ( 2\times12+4+16 \right ) \textrm{ u} = 44 \textrm{ g/mol}$

$n = \displaystyle \frac{\textrm{Molecular formula mass}}{\textrm{Empirical formula mass}} \\n = \displaystyle \frac{176}{44} = 4$

$\textrm{Molecular formula }= n\times \textrm{Empirical formula}$

Molecular formula = 4 $\times$ $C_{2}H_{4}O$

Molecular formula is $C_{8}H_{16}O_{4}$

6 0

3 years ago

What geological processes form natural gas

Ivan

Decomposed organic matter ,mud,silt ,and sand

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3 years ago

What is the hydrogen-ion concentration of the ph is 3.7

Komok [63]

Answer:

the answer 37

Explanation:

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2 years ago

A gas has a volume of 1.75L at -23°C and 150.0 kPa. At what temperature would the gas occupy 1.30L at 210.0 kPa?

Nastasia [14]

Answer:

At -13 $^{0}\textrm{C}$ , the gas would occupy 1.30L at 210.0 kPa.

Explanation:

Let's assume the gas behaves ideally.

As amount of gas remains constant in both state therefore in accordance with combined gas law for an ideal gas-

$\frac{P_{1}V_{1}}{T_{1}}=\frac{P_{2}V_{2}}{T_{2}}$

where $P_{1}$ and $P_{2}$ are initial and final pressure respectively.

$V_{1}$ and $V_{2}$ are initial and final volume respectively.

$T_{1}$ and $T_{2}$ are initial and final temperature in kelvin scale respectively.

Here $P_{1}=150.0kPa$ , $V_{1}=1.75L$ , $T_{1}=(273-23)K=250K$ , $P_{2}=210.0kPa$ and $V_{2}=1.30L$

Hence $T_{2}=\frac{P_{2}V_{2}T_{1}}{P_{1}V_{1}}$

$\Rightarrow T_{2}=\frac{(210.0kPa)\times (1.30L)\times (250K)}{(150.0kPa)\times (1.75L)}$

$\Rightarrow T_{2}=260K$

$\Rightarrow T_{2}=(260-273)^{0}\textrm{C}=-13^{0}\textrm{C}$

So at -13 $^{0}\textrm{C}$ , the gas would occupy 1.30L at 210.0 kPa.

5 0

3 years ago