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Lapatulllka [165]
2 years ago
15

Two students performed the same experiment separately and each one of them recorded two readings of mass which are given below.

Correct reading of mass is 3.0 g. On the basis of given data, mark the correct option out of the following statements. Student Readings (i) (ii) A 3.01 2.99 B 3.05 2.95 (i) Results of both the students are neither accurate nor precise. (ii) Results of student A are both precise and accurate. (iii) Results of student B are neither precise nor accurate. (iv) Results of student B are both precise and accurate.
answer it clearly ​
Chemistry
2 answers:
enyata [817]2 years ago
6 0

Answer:

The answer is option ii ,hope it helps

Lapatulllka [165]2 years ago
5 0

Answer:

Hey weirdo what's up?

So you got a question huh?

Lemme answer

As said the correct reading is 3.0grams

And the option A has 3.01 and 2.99 which are very mush precise and accurate to 3.0 gram don't you think?

So the answer is

Option ii the result of students A is both precise and accurate

Loye ya

Peace out

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3 years ago
12. A beginner's bowling ball has a mass of 4.9 kg and a volume of 5.4 liters. Will it float in water,
Greeley [361]

Taking into account the definition of density and Archimedes' principle, the beginner bowling ball will float on the water.

But first it is neccesary to know that density is a quantity referred to the amount of mass in a certain volume of a substance or a solid object.

In other words, the density is the relationship between the weight (mass) of a substance and the volume that the same substance occupies.

The expression for the calculation of density is the quotient between the mass of a body and the volume it occupies:

density=\frac{mass}{volume}

In this case, a beginner's bowling ball has a mass of 4.9 kg and a volume of 5.4 liters. This is:

  • mass= 4.9 kg= 4900 g (being 1 kg= 1000 kg)
  • volume= 5.4 L= 5400 mL (being 1L=1000 mL)

Replacing in the definition of density:

density=\frac{4900 g}{5400 mL}

Solving:

<u><em>density=0.907 </em></u>\frac{g}{mL}<u><em /></u>

On the other hand, Archimedes' principle says that an object immersed in a liquid experiences an upward vertical force equal to the weight of the volume of the dislodged liquid.

The sinking or floating of an object is determined by its density with respect to that of the liquid in which it is submerged.

Considering water as the liquid where the object is submerged in this case, an object with a higher density than water will sink. In contrast, an object with a lower density than water will float.

In this case, considering that water has a density of 1 \frac{g}{mL}, the bowling ball for beginners has a lower density. This indicates that, having a lower density than water, the object will float.

In summary, the beginner bowling ball will float on the water.

Learn more about density:

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5 0
2 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
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