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3 years ago

How many grams of sodium chloride are required to make a 1.0 L solution with a concentration of 3.5 M?

2 answers:

8 0

The molecular weight of NaCl is 58.44; sodium =22.99; Chlorine=35.45.

A 1 molar solution is the molecular weight in grams in 1 litre of water,
So, a 3.5 molar solution would be = 58.44g * 3.5, 204.54

In short, Your Answer would be 204.54

Hope this helps!

satela [25.4K]3 years ago

5 0

3.5 M has 3.5 moles per litre
so we have one litre, so we need 3.5 moles
moles = mass/molarmass
3.5 * 23 = 80.5

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what is the proper name for the following alkyl side group where the main carbon chain is denoted with a squiggly line?

chubhunter [2.5K]

The proper name for the following alkyl side group where the main carbon chain is denoted with a squiggly line is isopropyl.

In natural chemistry, an alkyl substituent is an alkane missing one hydrogen. The term alkyl is intentionally unspecific to include many viable substitutions. An acyclic alkyl has the overall formulation of CₙH₂ₙ₊₁.

An alkyl is a purposeful institution of an organic chemical that includes only carbon and hydrogen atoms, that are organized in a chain. Examples include methyl CH3 (derived from methane) and butyl C2H5 (derived from butane). they may be now not located on their own however are discovered attached to different hydrocarbons.

what is an alkyl group? Alkyl group is shaped through putting off a hydrogen atom from the molecule of alkane. Alkanes are quite regularly represented as R-H and here R stands for alkyl group. the overall method of the alkyl organization is CₙH₂ₙ₊₁. The smallest alkyl organization is CH3 referred to as methyl.

Learn more about alkyl group here:- brainly.com/question/14272539

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4 0

1 year ago

The activation energy for a reaction is changed from 184 kJ/mol to 59.0 kJ/mol at 600. K by the introduction of a catalyst. If t

11111nata11111 [884]

Answer:

The catalyzed reaction will take 2.85 seconds to occur.

Explanation:

The activation energy of a reaction is given by:

$k = Ae^{-\frac{E_{a}}{RT}}$

For the reaction without catalyst we have:

$k_{1} = Ae^{-\frac{E_{a_{1}}}{RT}}$ (1)

And for the reaction with the catalyst:

$k_{2} = Ae^{-\frac{E_{a_{2}}}{RT}}$ (2)

Assuming that frequency factor (A) and the temperature (T) are constant, by dividing equation (1) with equation (2) we have:

$\frac{k_{1}}{k_{2}} = \frac{Ae^{-\frac{E_{a_{1}}}{RT}}}{Ae^{-\frac{E_{a_{2}}}{RT}}}$

$\frac{k_{1}}{k_{2}} = e^{\frac{E_{a_{2}} - E_{a_{1}}}{RT}$

$\frac{k_{1}}{k_{2}} = e^{\frac{59.0 \cdot 10^{3}J/mol - 184 \cdot 10^{3} J/mol}{8.314 J/Kmol*600 K} = 1.31 \cdot 10^{-11}$

Since the reaction rate is related to the time as follow:

$k = \frac{\Delta [R]}{t}$

And assuming that the initial concentrations ([R]) are the same, we have:

$\frac{k_{1}}{k_{2}} = \frac{\Delta [R]/t_{1}}{\Delta [R]/t_{2}}$

$\frac{k_{1}}{k_{2}} = \frac{t_{2}}{t_{1}}$

$t_{2} = t_{1}\frac{k_{1}}{k_{2}} = 6900 y*1.31 \cdot 10^{-11} = 9.04 \cdot 10^{-8} y*\frac{365 d}{1 y}*\frac{24 h}{1 d}*\frac{3600 s}{1 h} = 2.85 s$

Therefore, the catalyzed reaction will take 2.85 seconds to occur.

I hope it helps you!

4 0

3 years ago

How many grams of glucose are needed to form 150g of ethanol?

Svetradugi [14.3K]

1 mole glucose gives 2 moles of ethanol

moles of glucose in 2.4 kg = 2400 / 180.18 = 13.320 moles

so moles of ethanol produced = 2* 13.32 = 26.64 moles

weight of ethanol 26.64 * 46.07

=1227.30 gm or 1.23 Kg

6 0

3 years ago

What is a magnetic field what does it do??

NikAS [45]

Answer:

attract things or hold them in place

Explanation:

a magnetic field is similar to gravity. Like gravity a magnetic field can hold or pull things.

5 0

3 years ago

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