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Xelga [282]
3 years ago
7

Acids

Chemistry
1 answer:
Marizza181 [45]3 years ago
6 0

-Acids release hydrogen ions

-Acids taste sharp

-Some Acids can cause serious burns

The answer is "All of the above"

<em />

<em>Hope this helps! Please mark brainliest!!!</em>

You might be interested in
For the following balanced equation: 3 Cu(s) + 8 HNO3(aq) → 3 Cu(NO3)2(aq) + 2 NO(g) + 4 H2O(l) a) How many moles of HNO3 will r
s344n2d4d5 [400]

Answer:

a) <u>26.67 moles HNO3 </u>

b) <u>0.33 moles NO</u>

c) <u>0.40 moles NO is produced</u>

d)<u>.157 moles Cu</u>

e) <u>0.105 moles NO</u>

f) <u>26.4 grams HNO3</u>

g) <u>Cu is in excess</u>

h) <u>2.41 grams Cu remain</u>

i) <u>2.37 grams NO</u>

Explanation:

Step 1: Data given

Molar mass of Cu = 63.55 g/mol

Molar mass of HNO3 = 63.01 g/mol

Molar mass of Cu(NO3)2 = 187.56 g/mol

Molar mass of NO = 30.01 g/mol

Molar mass of H2O = 18.02 g/mol

Step 2: The balanced equation

3 Cu(s) + 8 HNO3(aq) → 3 Cu(NO3)2(aq) + 2 NO(g) + 4 H2O(l)

a) How many moles of HNO3 will react with 10 moles of Cu?

For 3 moles Cu we need 8 moles HNO3 to produce 3 moles Cu(NO3)2, 2 moles NO and 4 moles H2O

For 10 moles Cu we need 8/3 *10 = <u>26.67 moles HNO3 </u>

b) How many moles of NO will form if 0.50 moles of Cu reacts?

For 3 moles Cu we need 8 moles HNO3 to produce 3 moles Cu(NO3)2, 2 moles NO and 4 moles H2O

For 0.50 moles Cu we'll have 2/3 *0.50 = <u>0.33 moles NO</u>

c) If 0.80 moles of H2O forms, how much NO must also form?

For 3 moles Cu we need 8 moles HNO3 to produce 3 moles Cu(NO3)2, 2 moles NO and 4 moles H2O

If 0.80 moles H2O is produced, 0.80/2 = <u>0.40 moles NO is produced</u>

d) How many moles of Cu are in 10.0 grams of Cu?

Moles Cu = 10.0 grams / 63.55 g/mol = 0.157 moles

In 10.0 grams Cu we have <u>0.157 moles Cu</u>

e) If 10.0 g of Cu reacts, how many moles of NO will form?

10.0 grams Cu = 0.157 moles

For 3 moles Cu we need 8 moles HNO3 to produce 3 moles Cu(NO3)2, 2 moles NO and 4 moles H2O

For 0.157 moles Cu we'll have 2/3 * 0.157 = <u>0.105 moles NO</u>

f) If 10.0 g of Cu reacts, how many grams of HNO3 are required?

10.0 grams Cu = 0.157 moles

For 3 moles Cu we need 8 moles HNO3 to produce 3 moles Cu(NO3)2, 2 moles NO and 4 moles H2O

For 0.157 moles Cu we'll need 0.419 moles HNO3

This is 0.419 moles * 63.01 g/mol = <u>26.4 grams HNO3</u>

g) If 10.0 g of Cu and 20.0 g of HNO3 are put together in a reaction vessel, which one will be in excess?

Moles Cu = 0.157 moles

Moles HNO3 = 20.0 grams / 63.01 g/mo = 0.317 moles

For 3 moles Cu we need 8 moles HNO3 to produce 3 moles Cu(NO3)2, 2 moles NO and 4 moles H2O

The limiting reactant is HNO3. It will completely be consumed (0.317 moles). <u>Cu is in excess.</u> There will react 3/8 * 0.317 = 0.119 moles Cu

There will remain 0.157 - 0.119 = 0.038 moles

h) How many grams of the excess substance will be left over?

There will react 3/8 * 0.317 = 0.119 moles Cu

There will remain 0.157 - 0.119 = 0.038 moles

This is 0.038 moles * 63.55 g/mol = 2.41 grams

i) How many grams of NO will form in the reaction described in part g?

For 3 moles Cu we need 8 moles HNO3 to produce 3 moles Cu(NO3)2, 2 moles NO and 4 moles H2O

For 0.317 moles HNO3 we'll have 0.317/4 = 0.0793 moles NO

This is 0.079 mol * 30.01 g/mol =<u> 2.37 grams NO</u>

3 0
3 years ago
Observational data is easier to analyze and interpret when it is organized into tables, charts, or graphs.
chubhunter [2.5K]

Than Answer is: True

8 0
4 years ago
Read 2 more answers
BRAINLIESTTT ASAP!! PLEASE HELP ME :)
horsena [70]

Answer: Hard working,satisfied. (i think)

~Please mark as brainiest~

3 0
3 years ago
Read 2 more answers
The following reaction: HF(aq) &lt;--&gt; H+(aq) + F-(aq) has an equilbirum constant (K) value of 7.2 x 10-4. This means that th
andriy [413]

Given an equilibrium constant value of 7.2 x 10-4 it is false to say that the reaction proceeds essentially to completion.

<h3>What is the equilibrium constant?</h3>

In a reaction, we can judge using the value of the equilibrium constant weather or not the reaction moves on to completion. If the reaction moves up to completion, it the follows that the value of the equilibrium constant ought to be large.

On the other hand, when we have a case that the equilibrium constant is small and is not so large, then the reaction does not proceed essentially to completion.

Given an equilibrium constant value of 7.2 x 10-4 it is false to say that the reaction proceeds essentially to completion.

Learn more about equilibrium constant:brainly.com/question/10038290

#SPJ1

6 0
2 years ago
How do you do problem #15?
SVETLANKA909090 [29]
You would convert the cg to g. 1.66 cg is 0.0166 g. then you add. 0.398 + 0.0166 = 0.4146. the answer is 0.4146 grams.
7 0
3 years ago
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