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3 years ago

8

Using the activity series provided. Which reactants will form products? Na > Mg > Al > Mn > Zn > Cr > Fe >

Cd > Co > Ni > Sn > Pb > H > Sb > Bi > Cu > Ag Ag +NaNO3 Right arrow. Fe + Al2O3 Right arrow. Ni + NaCl Right arrow. Fe + Cu(NO3)2 Right arrow.

1 answer:

Elodia [21]3 years ago

3 0

Answer:

Fe + Cu(NO3)2 →

Explanation:

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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 do you balance this chemical equation?

garik1379 [7]

Answer:

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Explanation:

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2 years ago

The acid-dissociation constant for benzoic acid (C6H5COOH) is 6.3×10−5. Calculate the equilibrium concentration of H3O+ in the s

raketka [301]

Answer : The equilibrium concentration of $H_3O^+$ in the solution is, $2.1\times 10^{-3}M$

Explanation :

The dissociation of acid reaction is:

$C_6H_5COOH+H_2O\rightarrow H_3O^++C_6H_5COO^-$

Initial conc. c 0 0

At eqm. c-x x x

Given:

c = $7.0\times 10^{-2}M$

$K_a=6.3\times 10^{-5}$

The expression of dissociation constant of acid is:

$K_a=\frac{[H_3O^+][C_6H_5COO^-]}{[C_6H_5COOH]}$

$K_a=\frac{(x)\times (x)}{(c-x)}$

Now put all the given values in this expression, we get:

$6.3\times 10^{-5}=\frac{(x)\times (x)}{[(7.0\times 10^{-2})-x]}$

$x=2.1\times 10^{-3}M$

Thus, the equilibrium concentration of $H_3O^+$ in the solution is, $2.1\times 10^{-3}M$

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2 years ago

How much 6.0 m hno3 is needed to neutralize 39ml of 2 m koh

Sever21 [200]

Answer:

13mL

Explanation:

Step 1:

The balanced equation for the reaction. This is given below:

HNO3 + KOH —> KNO3 + H2O

From the balanced equation above, we obtained the following data:

Mole ratio of the acid (nA) = 1

Mole ratio of the base (nB) = 1

Step 2:

Data obtained from the question.

This includes the following:

Molarity of the acid (Ma) = 6M

Volume of the acid (Va) =?

Volume of the base (Vb) = 39mL

Molarity of the base (Mb) = 2M

Step 3:

Determination of the volume of the acid.

Using the equation:

MaVa/MbVb = nA/nB, the volume of the acid can be obtained as follow:

MaVa/MbVb = nA/nB

6 x Va / 2 x 39 = 1/1

Cross multiply to express in linear form

6 x Va = 2 x 39

Divide both side by 6

Va = (2 x 39)/6

Va = 13mL

Therefore, the volume of the acid (HNO3) needed for the reaction is 13mL

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3 years ago

Which of the following is the best example of an open system? (2 points)

Dmitriy789 [7]

1) A Car engine.............. Hope it helps, Have a nice day :)

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2 years ago

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