Is the molecule BF3 is a resonance structure?

A chemist requires 5.00 liters of 0.420 M H2SO4 solution. How many grams of H2SO4 should the chemist dissolve in water? 129 gram

natali 33 [55]

Answer:

Approximately 206 grams.

Explanation:

How many moles of sulfuric acid $\mathrm{H_2SO_4}$ are there in this solution?

$\text{Number of moles of solute} = \text{Concentration} \times \text{Volume}$ .

The unit for concentration " $\mathrm{M}$ " is equivalent to mole per liter. In other words, $\rm 1\;M = 1\; mol\cdot L^{-1}$ . For this solution, the concentration of $\mathrm{H_2SO_4}$ is $\rm 0.420\;M = 0.420\; mol\cdot L^{-1}$ .

$\begin{aligned}n(\mathrm{H_2SO_4}) &= c(\mathrm{H_2SO_4}) \cdot V\\&= \rm 0.420\;mol\cdot L^{-1}\times 5.00\; L \\&= \rm 2.10\; mol\end{aligned}$ .

What's the mass of that $\rm 2.10\; mol$ of $\mathrm{H_2SO_4}$ ?

Start by finding the molar mass $M$ of $\mathrm{H_2SO_4}$ .

Relative atomic mass data from a modern periodic table:

H: 1.008;
S: 32.06;
O: 15.999.

$\displaystyle M(\mathrm{H_2SO_4}) = 2\times \underbrace{1.008}_{\mathrm{H}} + 1\times \underbrace{32.06}_{\mathrm{S}} + 4\times \underbrace{15.999}_{\mathrm{O}} = \rm 98.072\;g\cdot mol^{-1}$ .

$\text{Mass} = \text{Quantity in moles} \times \text{Molar Mass}$ .

$m = n \cdot M = \rm 2.10\; mol \times 98.072\;g\cdot mol^{-1} \approx 206\; g$ .

In other words, the chemist shall need approximately 206 grams of $\mathrm{H_2SO_4}$ to make this solution. As a side note, keep in mind that the 206 grams of $\mathrm{H_2SO_4}$ also take up considerable amount of space. Therefore it will take less than 5.00 L of water to make the 5.00 L solution.

6 0

4 years ago

Describe a covalent bond?

Natasha2012 [34]

A covalent bond, also called a molecular bond, is a chemical bond that involves the sharing of electron pairs between atoms. These electron pairs are known as shared pairs or bonding pairs, and the stable balance of attractive and repulsive forces between atoms, when they share electrons, is known as covalent bonding.

7 0

3 years ago

If water is present during the grignard formation, what possible byproduct may form?

Zanzabum

Important of Grignard Reagent :

Grignard reactions are main due to their capacity to form carbon-carbon bonds. Grignard reagents are powerful bases and will react with protic compounds which makes them exceptionally valuable implement for organic synthesis.

Grignard reagents react fastly with acidic hydrogen atoms in molecules such as alcohols and water. When a Grignard reagent reacts with water, a proton return back halogen, and the product is an alkane. The Grignard reagent therefore provides a trace for transforming a haloalkane to an alkane in two steps.

When reacted with water,

RMgX + H2O -------> R-H + MgOH X

R ---- Alkyl Group

X --- Halogen

To know more about Grignard Reagent here :

brainly.com/question/15876201?referrer=searchResults

#SPJ4

4 0

2 years ago

How to find the oxidation number of an element

Virty [35]

The thing you MUST do FIRST is look for any H's, O's, or F's in the equation

1)any element just by itself not in a compound, their oxidation number is 0
ex: H2's oxidation number is 0
ex: Ag: oxidation number is 0 if its just something like Ag + BLA = LALA

2) the oxidation number of H is always +1, unless its just by itself (see #1)
3) the oxidation number of O is always -2, unless its just by itself (see #1)
4) the oxidation number of F is always -1, unless its just by itself (see#1)

ok so after you have written those oxidation numbers in rules 1-4 over each H, F, or O atom in the compound, you can look at the elements that we havent talked about yet

for example::::
N2O4

the oxidation number of O is -2.

since there are 4 O's, the charge is -8. now remember that N2O4 has to be neutral so the N2 must have a charge of +8
+8 divided by 2 = +4

N has an oxidation number of +4.

more rules:
5) the sum of oxidation numbers in a compound add up to 0 (when multiplied by the subscripts!!!) (see above example)
6) the sum of oxidation numbers in a polyatomic ion is the charge (for example, PO4 has a charge of (-3) so

oxidation # of O = -2. (there are 4 O's = -8 charge on that side ) P must have an oxidation number of 5. (-8+5= -3), and -3 is the total charge of the polyatomic ion

3 0

4 years ago

If 34.5 g of Copper reacts with 70.2 g of silver nitrate, according to the following

Vlada [557]

Answer:

44,55 can be produced.

Explanation:

First, we balanced the equation

1Cu + 2AgNO3 → 1Cu(NO3)2 + Ag

Then, we find the moles of each reagent

$mol Cu = \frac{34,5g}{63,55\frac{g}{mol} } = 0,543 mol\\$