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WARRIOR [948]
3 years ago
14

What is 0.00055030 in Scientific form

Chemistry
1 answer:
kipiarov [429]3 years ago
8 0

5.503 * 10^-4 Is the answer.

Gl

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NEED HELP ASAPPP<br><br> What is the number of beryllium atoms in 78.0g of Be?
polet [3.4K]

Answer: number of atoms is 5.21 · 10^24

Explanation: Atomic mass of Be is 9.012 g/mol.

Number of moles n = m/M = 78.0 g / 9.012 g/mol =

Multiply this with Avogadro number Na = 6.022*10^23 1/mol

4 0
3 years ago
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
The mass of an electron is about 9.11 × 10 -9 kg., write this whole number​
Pepsi [2]

Answer:0.00000009

Explanation:

9.11 × (1/1000000000)=0.00000009

8 0
3 years ago
Which of the following is a true
Katarina [22]

❤️Hello!❤️ The answer is A. Energy can change from one form to another. Hope this helps! ↪️ Autumn ↩️

5 0
3 years ago
Blackmail threat of informational disclosure is an example of which threat category?
Rashid [163]

Answer:

Information Extortion.

Explanation:

Computer oriented crime also known as cyber crime that intentionally harm the victims. The complete data of the company or net banking information can be hacked easily by the cyber crime.

Different category of the threats are included in the cyber crime. The information extortion threat is the crime in which the hackers hack complete information and control the data of the victim. The criminal an black mail the victim regarding the hacked data.

Thus, the correct answer is information extortion.

6 0
3 years ago
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