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2 years ago

What is boiling point

Chemistry

2 answers:

Aliun [14]2 years ago

8 0

Boiling point is 100°C (celcius)

klasskru [66]2 years ago

5 0

Answer:

The temperature at which the given substance starts to boil is called boiling point.

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How many moles of nitrogen are in 38 grams of nitrogen?

Lena [83]

How many grams Nitrogen in 1 mol? The answer is 14.0067. We assume you are converting between grams Nitrogen and mole. You can view more details on each measurement unit: molecular weight of Nitrogen or mol The molecular formula for Nitrogen is N. The SI base unit for amount of substance is the mole. 1 grams Nitrogen is equal to 0.071394404106606 mole. Note that rounding errors may occur, so always check the results. Use this page to learn how to convert between grams Nitrogen and mole. Type in your own numbers in the form to convert the units!

7 0

3 years ago

A histidine is involved in an interaction with a glutamic acid that stabilizes the charged form of the histidine, such that the

Greeley [361]

Answer:

pKa of the histidine = 9.67

Explanation:

The relation between standard Gibbs energy and equilibrium constant is shown below as:

$\Delta{G^0} =-RT \ln \frac{[His]}{[His+]}$

R is Gas constant having value = 0.008314 kJ / K mol

Given temperature, T = 293 K

Given, $\Delta{G^0}=15\ kJ/mol$

So, Applying in the equation as:-

$15\ kJ/mol=-0.008314\ kJ/Kmol\times 293\ K\times \ln \frac{[His]}{[His+]}$

Thus,

$15\ kJ/mol=-0.008314\ kJ/Kmol\times 293\ K\times \ln \frac{[His]}{[His+]}$

$\frac{[His]}{[His+]}=e^{\frac{15}{-0.008314\times 293}$

$\frac{[His]}{[His+]}=0.00211$

Also, considering:-

$pH=pKa+log\frac{[His]}{[His+]}$

Given that:- pH = 7.0

So, $7.0=pKa+log0.00211$

<u>pKa of the histidine = 9.67</u>

8 0

3 years ago

Read the chemical equation.

Sladkaya [172]

Answer:- D. 1.8 moles of Fe and $2.7molCO_2$ .

Solution:- The balanced equation is:

$Fe_2O_3+3CO\rightarrow 2Fe+3CO_2$

let's first figure out the limiting reactant using the given moles and mol ratio:

$1.8molFe_2O_3(\frac{3molCO}{1molFe_2O_3})$

= 5.4 mol CO

From calculations, 5.4 moles of CO are required to react completely with 1.8 moles of Iron(III)oxide but only 2.7 moles of CO are available. It means CO is limiting reactant.

Products moles depends on limiting reactant. Let's calculate the moles of each reactant formed for given 2.7 moles of CO.

$2.7molCO(\frac{2molFe}{3molCO})$