All categories

4 years ago

What kinds of substances besides water can be involved in hydrogen bonding?

1 answer:

7 0

<span>Kind of substance besides water:
The best example of hydrogen bonding excluding water is DNA. The two strands of polymers are connected by hydrogen bonds between the nucleotide bases</span>.

You might be interested in

Which element is in group 2 and period 7 of the periodic table

bekas [8.4K]

Hey
Sorry, But There Is No 7th Element in group two. only six which is Radium.
~Spades15

3 0

4 years ago

Pu is a nuclear waste byproduct with a half-life of 24,000 y. What fraction of the 239Pu present today will be present in 1000 y

olya-2409 [2.1K]

Answer:

0.9715 Fraction of Pu-239 will be remain after 1000 years.

Explanation:

$\lambda =\frac{0.693}{t_{\frac{1}{2}}}$

$A=A_o\times e^{-\lambda t}$

Where:

$\lambda$ = decay constant

$A_o$ =concentration left after time t

$t_{\frac{1}{2}}$ = Half life of the sample

Half life of Pu-239 = $t_{\frac{1}{2}}=24,000 y$ [

$\lambda =\frac{0.693}{24,000 y}=2.8875\times 10^{-5} y^{-1]$

Let us say amount present of Pu-239 today = $A_o=x$

A = ?

$A=x\times e^{-2.8875\times 10^{-5} y^{-1]\times 1000 y}$

$A=0.9715\times x$

$\frac{A}{A_0}=\frac{A}{x}=0.9715$

0.9715 Fraction of Pu-239 will be remain after 1000 years.

7 0

3 years ago

what would be the ph of an aqueous solution of sulphuric acid which is 5×10^_5 mol l^_1 in concentration

guajiro [1.7K]

Answer:

the ph of an aqueous solution of sulphuric acid which is 5*10^5 mol in concentration is basic in nature

6 0

3 years ago

onsider the following reaction: CaCN2 + 3 H2O → CaCO3 + 2 NH3 105.0 g CaCN2 and 78.0 g H2O are reacted. Assuming 100% efficiency

mestny [16]

Answer : The excess reactant is, $H_2O$

The leftover amount of excess reagent is, 7.2 grams.

Solution : Given,

Mass of $CaCN_2$ = 105.0 g

Mass of $H_2O$ = 78.0 g

Molar mass of $CaCN_2$ = 80.11 g/mole

Molar mass of $H_2O$ = 18 g/mole

Molar mass of $CaCO_3$ = 100.09 g/mole

First we have to calculate the moles of $CaCN_2$ and $H_2O$ .

$\text{ Moles of }CaCN_2=\frac{\text{ Mass of }CaCN_2}{\text{ Molar mass of }CaCN_2}=\frac{105.0g}{80.11g/mole}=1.31moles$

$\text{ Moles of }H_2O=\frac{\text{ Mass of }H_2O}{\text{ Molar mass of }H_2O}=\frac{78.0g}{18g/mole}=4.33moles$

Now we have to calculate the limiting and excess reagent.

The balanced chemical reaction is,

$CaCN_2+3H_2O\rightarrow CaCO_3+2NH_3$

From the balanced reaction we conclude that

As, 1 mole of $CaCN_2$ react with 3 mole of $H_2O$

So, 1.31 moles of $CaCN_2$ react with $1.31\times 3=3.93$ moles of $H_2O$

From this we conclude that, $H_2O$ is an excess reagent because the given moles are greater than the required moles and $CaCN_2$ is a limiting reagent and it limits the formation of product.

Left moles of excess reactant = 4.33 - 3.93 = 0.4 moles

Now we have to calculate the mass of excess reactant.

$\text{ Mass of excess reactant}=\text{ Moles of excess reactant}\times \text{ Molar mass of excess reactant}(H_2O)$

$\text{ Mass of excess reactant}=(0.4moles)\times (18g/mole)=7.2g$

Thus, the leftover amount of excess reagent is, 7.2 grams.

8 0

4 years ago

Put the following steps in order to show how a scientist follows Scientific Method

Gekata [30.6K]

1.question, 2.observe, 3.hypothesize, 4.experiment 5.conclusion, 6. record.

8 0

3 years ago