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2 years ago

11

Predict the reactants of this chemical reaction. products: KClO4 + H2O reactants: ?

1 answer:

mojhsa [17]2 years ago

3 0

Answer:

KOH + HClO4 = K(ClO4) + H2O

Explanation:

KClO4 + H2O : mix and match the letters if you need to guess

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What is the function or protons, neutrons, and electrons?<br> someone help please

lukranit [14]

Answer:

An atom is composed of them

Explanation:

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3 years ago

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Great amounts of atomic energy are released when a _______reaction occurs.

Mariulka [41]

Great amounts of atomic energy are released when a _______reaction occurs.

Great amounts of atomic energy are released when a chemical reaction occurs. The process can be an exothermic reaction or endothermic reaction depending on the substances involved in the reaction.

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3 years ago

30cm^3 of a dilute solution of Ca(OH)2 required 11 cm^3 of 0.06 mol/dm^. Hcl for complete neutralization. Calculate the concentr

Alenkasestr [34]

Answer: Thus concentration of $Ca(OH)_2$ in $mol/dm^3$ is 0.011 and in $g/dm^3$ is 0.814

Explanation:

To calculate the concentration of $Ca(OH)_2$ , we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is $HCl$

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is $Ca(OH)_2$

We are given:

$n_1=1\\M_1=0.06mol/dm^3\\V_1=11cm^3=0.011dm^3\\n_2=2\\M_2=?\\V_2=30cm^3=0.030dm^3$ $1cm^3=0.001dm^3$

Putting values in above equation, we get:

$1\times 0.06mol/dm^3\times 0.011dm^3=2\times M_2\times 0.030dm^3\\\\M_2=0.011mol/dm^3$

The concentration in $g/dm^3$ is $0.011mol/dm^3\times 74g/mol=0.814g/dm^3$

Thus concentration of $Ca(OH)_2$ is $0.011mol/dm^3$ and $0.814g/dm^3$

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3 years ago

Study of gene structure and function

gregori [183]

I believe the answer would be b

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3 years ago

What is the molecular formula of a hydrocarbon with m+ = 78?

fiasKO [112]

A molecular formula represents the exact number of atoms present for each element in the compound.

For carbon atom: $\frac{78}{12}$ = $6\frac{6}{12}$ ( as molar mass of carbon is 12 g/mol)

Now, 6 carbon atoms are present and rest are hydrogen atoms i.e. 6

Thus, formula becomes $C_{6}H_{6}$ ( $C_{x}H_{y})$

Now, check for unsaturation:

Degree of unsaturation = $\frac{2x+2-y}{2}$

Substitute the value of x and y,

Degree of unsaturation = $\frac{2(6)+2-6}{2}$

= $\frac{12-4}{2}$

= $4$ implies one ring and three double bonds.

Thus, formula comes out to be $C_{6}H_{6}$ i.e. benzene ring.

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3 years ago