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irga5000 [103]
3 years ago
7

How many atoms of Carbon are found on the PRODUCTS side?

Chemistry
1 answer:
Semenov [28]3 years ago
5 0
Secrets undiscovered is correct, but this chemical formula is not balanced. Double check your question to make sure
You might be interested in
What is a reaction rate?
Kay [80]

Answer:

Reaction rate, in chemistry, the speed at which a chemical reaction proceeds. It is often expressed in terms of either the concentration (amount per unit volume)

Explanation:

8 0
2 years ago
How many molecules of O2 will be required to produce 28.8 g of water?
Mrac [35]
I think 1.67 x 10^20
5 0
3 years ago
PLEASE ANSWER FAST AND YOU WILL GET BRAINLIESTTTTTTT!!!!!!! And POINTSSSSSS!! PLEASE ANSWER FAST AND YOU WILL GET points and BRA
stiks02 [169]

Answer:

B on both questions

Explanation:

6 0
2 years ago
Read 2 more answers
Any help would be appreciated. Confused.
masya89 [10]

Answer:

q(problem 1) = 25,050 joules;  q(problem 2) = 4.52 x 10⁶ joules

Explanation:

To understand these type problems one needs to go through a simple set of calculations relating to the 'HEATING CURVE OF WATER'. That is, consider the following problem ...

=> Calculate the total amount of heat needed to convert 10g ice at -10°C to steam at 110°C. Given are the following constants:

Heat of fusion (ΔHₓ) = 80 cal/gram

Heat of vaporization (ΔHv) = 540 cal/gram

specific heat of ice [c(i)] = 0.50 cal/gram·°C

specific heat of water [c(w)] = 1.00 cal/gram·°C

specific heat of steam [c(s)] = 0.48 cal/gram·°C

Now, the problem calculates the heat flow in each of five (5) phase transition regions based on the heating curve of water (see attached graph below this post) ...   Note two types of regions (1) regions of increasing slopes use q = mcΔT and (2) regions of zero slopes use q = m·ΔH.

q(warming ice) =  m·c(i)·ΔT = (10g)(0.50 cal/g°C)(10°C) = 50 cal

q(melting) = m·ΔHₓ = (10g)(80cal/g) 800 cal

q(warming water) = m·c(w)·ΔT = (10g)(1.00 cal/g°C)(100°C) = 1000 cal

q(evaporation of water) =  m·ΔHv = (10g)(540cal/g) = 5400 cal

q(heating steam) = m·c(s)·ΔT = (10g)(0.48 cal/g°C)(10°C) = 48 cal

Q(total) = ∑q = (50 + 800 + 1000 + 5400 + 48) = 7298 cals. => to convert to joules, multiply by 4.184 j/cal => q = 7298 cals x 4.184 j/cal = 30,534 joules = 30.5 Kj.

Now, for the problems in your post ... they represent fragments of the above problem. All you need to do is decide if the problem contains a temperature change (use q = m·c·ΔT) or does NOT contain a temperature change (use q = m·ΔH).    

Problem 1: Given Heat of Fusion of Water = 334 j/g, determine heat needed to melt 75g ice.

Since this is a phase transition (melting), NO temperature change occurs; use q = m·ΔHₓ = (75g)(334 j/g) = 25,050 joules.

Problem 2: Given Heat of Vaporization = 2260 j/g; determine the amount of heat needed to boil to vapor 2 Liters water ( = 2000 grams water ).

Since this is a phase transition (boiling = evaporation), NO temperature change occurs; use q = m·ΔHf = (2000g)(2260 j/g) = 4,520,000 joules = 4.52 x 10⁶ joules.

Problems containing a temperature change:

NOTE: A specific temperature change will be evident in the context of problems containing temperature change => use q = m·c·ΔT. Such is associated with the increasing slope regions of the heating curve.  Good luck on your efforts. Doc :-)

5 0
3 years ago
100 POINTS! Final Honor Activity Question
castortr0y [4]

The change in temperature had the greatest effect at changing the volume of the balloon.

<h3>What are the gas laws?</h3>

The gas laws are used to describe the parameters that has to do with gases.

Given that;

P1 = 98.5 kPa

T1 = 18oC or 291 K

V1 =  74.0 dm3

P2 =  7.0 kPa

V2 = ?

T2 = 18oC or 291 K

P1V1/T1 = P2V2/T2

P1V1T2 =P2V2T1

V2= P1V1T2/P2T1

V2 =  98.5 kPa *  74.0 dm3 * 291 K/ 7.0 kPa * 291 K

V2 = 1041.3 dm3

When;

V1 = 1041.3 dm3

T1 = 291 K

V2 = ?

T2 = 80oC or 353 K

V1/T1 = V2/T2

V1T2 = V2T1

V2 = V1T2/T1

V2 = 1041.3 dm3 * 353 K/291 K

V2 = 1263 dm3

The change in temperature had the greatest effect at changing the volume of the balloon.

Given that

V1 =  100 cm^3

T1 = 273 K

P1 = 1.01 * 10^5 Pa

V2 = ?

P2 =  3.00 x 10^-4 Pa

T2 = -180oC or 255 K

V2= P1V1T2/P2T1

V2 =  1.01 * 10^5 Pa * 100 cm^3 * 255 K / 3.00 x 10^-4 Pa * 273 K

V2 = 3.14 * 10^10 cm^3

Learn more about gas laws:brainly.com/question/12669509

#SPJ1

7 0
2 years ago
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