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Orlov [11]
2 years ago
5

What does the symbol Cp

Chemistry
1 answer:
xenn [34]2 years ago
8 0

Answer:

From what i've learned so far, the correct answer is "Heat at a constant Pressure" or "Specific Heat"

Explanation:

Hope this helps!

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What does the oxidizing agent do in a redox reaction apex?
densk [106]
Same as balancing a regular chemical reaction! Please see the related question to the bottom of this answer for how to balance a normal chemical reaction. This is for oxidation-reduction, or redox reactions ONLY! These instructions are for how to balance a reduction-oxidation, or redox reaction in aqueous solution, for both acidic and basic solution. Just follow these steps! I will illustrate each step with an example. The example will be the dissolution of copper(II) sulfide in aqueous nitric acid, shown in the following unbalanced reaction: CuS (s) + NO 3 - (aq) ---> Cu 2+ (aq) + SO 4 2- (aq) + NO (g) Step 1: Write two unbalanced half-reactions, one for the species that is being oxidized and its product, and one for the species that is reduced and its product. Here is the unbalanced half-reaction involving CuS: CuS (s) ---> Cu 2+ (aq) + SO 4 2- (aq) And the unbalanced half-reaction for NO 3 - is: NO 3 - (aq) --> NO (g) Step 2: Insert coefficients to make the numbers of atoms of all elements except oxygen and hydrogen equal on the two sides of each half-reaction. In this case, copper, sulfur, and nitrogen are already balanced in the two half-reaction, so this step is already done here. Step 3: Balance oxygen by adding H 2 O to one side of each half-reaction. CuS + 4 H 2 O ---> Cu 2+ + SO 4 2- NO 3 - --> NO + 2 H 2 O Step 4: Balance hydrogen atoms. This is done differently for acidic versus basic solutions. . For acidic solutions: Add H 3 O + to each side of each half-reaction that is "deficient" in hydrogen (the side that has fewer H's) and add an equal amount of H 2 O to the other side. For basic solutions: add H 2 O to the side of the half-reaction that is "deficient" in hydrogen and add an equal amount of OH - to the other side. Note that this step does not disrupt the oxygen balance from Step 3. In the example here, it is in acidic solution, and so we have: CuS + 12 H 2 O ---> Cu 2+ + SO 4 2- + 8 H 3 O + . NO 3 - + 4 H 3 O + --> NO + 6 H 2 O Step 5: Balance charge by inserting e - (electrons) as a reactant or product in each half-reaction. Oxidation: CuS + 12 H 2 O ---> Cu 2+ + SO 4 2- + 8 H 3 O + + 8 e - . Reduction: NO 3 - + 4 H 3 O + + 3 e - --> NO + 6 H 2 O . Step 6: Multiply the two half-reactions by numbers chosen to make the number of electrons given off by the oxidation step equal to the number taken up by the reduction step. Then add the two half-reactions. If done correctly, the electrons should cancel out (equal numbers on the reactant and product sides of the overall reaction). If H 3 O + , H 2 O, or OH - appears on both sides of the final equation, cancel out the duplication also. Here the oxidation half-reaction must be multiplied by 3 (so that 24 electrons are produced) and the reduction half-reaction must by multiplied by 8 (so that the same 24 electrons are consumed). 3 CuS + 36 H 2 O ---> 3 Cu 2+ + 3 SO 4 2- + 24 H 3 O + + 24 e - 8 NO 3 - + 32 H 3 O + + 24 e - ---> 8 NO + 48 H 2 O Adding these two together gives the following equation: 3 CuS + 36 H 2 O + 8 NO 3 - + 8 H 3 O + ---> 3 Cu 2+ + 3 SO 4 2- + 8 NO + 48 H 2 O Step 7: Finally balancing both sides for excess of H 2 O (On each side -36) This gives you the following overall balanced equation at last: 3 CuS (s) + 8 NO 3 - (aq) + 8 H 3 O + (aq) ---> 3 Cu 2+ (aq) + 3 SO 4 2- (aq) + 8 NO (g) + 12 H 2 O (l)


6 0
3 years ago
Read 2 more answers
What causes waves to occur?
Flura [38]

Answer:

definitely the wind lol

5 0
2 years ago
Radio waves travel at the speed of light which is 3.00 x 10^8. How many minutes does it take for a radio message to reach saturn
ryzh [129]

Answer:

43.89 min

Explanation:

Given that:-

The speed of light = 3.00\times 10^8\ m/s

The distance = 7.9\times 10^8\ km

The conversion of distance in km to distance into m is shown below as:-

1 km = 1000 m

So,

Distance = 7.9\times 10^8\times 1000\ m=7.9\times 10^{11}\ m

The relation between speed distance and time is shown below as:-

Speed=\frac{Distance}{Time}

Thus,

3.00\times 10^8=\frac{7.9\times 10^{11}}{Time}

300000000\times time=10^{11}\times \:7.9\ s

Time = 2633.33 seconds

Also, 1 s = 1/60 min

So,

Time=\frac{2633.33}{60}\ min=43.89\ min

3 0
3 years ago
If you mix water with your urine drug test will you pass
masya89 [10]
<span>A "dilute specimen" is a urine sample that has a higher than average water ... even if the drug is detected, it will not be marked as a positive result because ... Powdered urine can be purchased online in packets, and then mixed with water to form urine. ... so they would report your test as negative andyou'd get hired anyway.</span><span>
</span>
6 0
3 years ago
if three oxygen particles are needed to form ozone, how many units of ozone could be formed from 6 oxygen particles? from 9? fro
tensa zangetsu [6.8K]
<h3>Answers:</h3>

             1) 2 Units of Ozone

             2)  3 Units of Ozone

              3)  9 Units of Ozone

<h3>Solution:</h3>

1)  From 6 Oxygen Particles;

As given,

                          3 Oxygen Particles form  =  1 Unit of Ozone

So,

                      6 Oxygen Particles will form  =  X Units of Ozone

Solving for X,

                      X =  (6 O Particles × 1 Unit of Ozone) ÷ 3 O Particles

                      X =  2 Units of Ozone

2) From 9 Oxygen Particles;

As given,

                           3 Oxygen Particles form  =  1 Unit of Ozone

So,

                      9 Oxygen Particles will form  =  X Units of Ozone

Solving for X,

                      X =  (9 O Particles × 1 Unit of Ozone) ÷ 3 O Particles

                      X =  3 Units of Ozone

3)  From 27 Oxygen Particles;

As given,

                             3 Oxygen Particles form  =  1 Unit of Ozone

So,

                      27 Oxygen Particles will form  =  X Units of Ozone

Solving for X,

                      X =  (27 O Particles × 1 Unit of Ozone) ÷ 3 O Particles

                      X =  9 Units of Ozone

3 0
3 years ago
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