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Taya2010 [7]
3 years ago
10

Jojojoojojoojoojoj , sdincicnjojo

Chemistry
2 answers:
motikmotik3 years ago
8 0
Whta the haha hahaha
zvonat [6]3 years ago
5 0

Answer:

?

Explanation:

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1.34 milligrams is the same as _______kg and ______g
Maurinko [17]
There are 1000 mg in 1 g
and there are 1000 g in 1 kg

Start by converting 1.34 mg to grams by dividing 1.34 mg by 1000 g = 0.00134 g

Then convert 0.00134 g to kg by dividing 0.00134 g by 1000 kg = 1.34×10^-6 kg OR 0.00000134 kg
3 0
4 years ago
Read 2 more answers
Suppose that there is an air sample and a water sample, each of the same mass and each with an initial temperature, Tinitial, of
klasskru [66]

Answer:

The air will reach a higher final temperature because its specific heat is lower.

Explanation:

6 0
3 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
Which of the following is the SI unit used to measure mass? a. kilogram b. liter c. meter d. Kelvin Please select the best answe
Ket [755]

Answer: d

Explanation: because it’s right

5 0
4 years ago
Isotopes (such as hydrogen-1, hydrogen-2, and hydrogen-3) are atoms of the same element that differ in:
maria [59]
They have a different amount of neutrons.
8 0
3 years ago
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