Complete and balance the following neutralization reaction, name the products, and write the net ionic equation.

What determines the ending units of a stoichiometry problem?

kicyunya [14]

Answer:

1.Write the balanced chemical equation.

2.Convert the units of the given substance (A) to moles.

3.Use the mole ratio to calculate the moles of wanted substance (B).

4.Convert moles of the wanted substance to the desired units.

Explanation:

4 0

2 years ago

How many mL of a 0.375 M solution can be made from 35 g of calcium phosphate?

andrew11 [14]

Answer:

300 mL

Explanation:

the unit formula of calcium phosphate is Ca3(PO4)2

molar mass of Ca3(PO4)2 = (3×40 + 2×31 + 8×16) g/mol = 310 g/mol

n = m/M = 35 g/(310 g/mol)

c = n/V

V = n/c = [35 g/(310 g/mol)]/0.375 mol/L

V = 0.30 L = 300 mL

3 0

2 years ago

Plz help I’m pretty confused on this question

sergeinik [125]

Answer:

Bohr model A

Explanation:

It has more valance electrons therefore has more interaction between the atoms and has more electronegativity.

7 0

2 years ago

How many moles of chromium III nitrate are produced When chromium reacts with 0.85 moles of lead for nitrate to produce chromium

Pepsi [2]

0.85 moles formula units of lead nitrate will produce 0.57 moles formula units of chromium (III) nitrate.

<h3>Explanation</h3>

Typically, the oxidation state of Pb in lead nitrate tend to be +2. In other words, Pb in lead nitrate tends to exist as $\text{Pb}^{2+}$ ions. The formula for a nitrate ion is ${\text{NO}_3}^{-}$ . The charge on each of the nitrate ion is -1. The charge on the two ions should balance. As a result, each $\text{Pb}^{2+}$ ion in lead nitrate would pair up with two ${\text{NO}_3}^{-}$ ions. The formula for lead nitrate will be $\text{Pb}({\text{NO}_3})_2$ . Each formula unit of lead nitrate will contain one $\text{Pb}^{2+}$ ion and two ${\text{NO}_3}^{-}$ ions.

The "III" in the name "chromium (III) nitrate" is a Roman Numeral. It indicates that the oxidation state of Cr in chromium (III) nitrate is +3. The Cr in that compound will exist as $\text{Cr}^{3+}$ . Similarly, each $\text{Cr}^{3+}$ will pair up with three ${\text{NO}_3}^{-}$ ions. The formula for chromium (III) nitrate will be $\text{Cr}(\text{NO}_3})_3$ . Each formula unit of chromium (III) nitrate will contain one ${\text{NO}_3}^{-}$ ion and three ${\text{NO}_3}^{-}$ ions.

0.85 moles formula units of lead nitrate will contain 0.85 × 2 = 1.7 moles of ${\text{NO}_3}^{-}$ ions. Those nitrate ions will end up in 1.7 / 3 = 0.57 moles formula units of chromium (III) nitrate. As a result, the reaction will produce 0.57 moles formula units of chromium (III) nitrate.

7 0

3 years ago

How many atoms are in 1.75 mol CHCL3

vaieri [72.5K]

1 mol = 6.022 x 10²³ atoms

In order to find how many atoms, dimly multiply the amount of moles you have by 6.022 x 10²³ or Avogadro's number.

So you have 1.75 mol CHC1₃ x (6.022x10²³) = 1.05385 x 10²⁴ atoms of CHCl₃

But now you have to round because of the rules of significant figures so you get 1.05 x 10²⁴ atoms of CHCl₃

6 0

3 years ago

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