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LUCKY_DIMON [66]
3 years ago
15

If an isotope of an element has 33 neutrons and a mass number of 64, how many electrons must it have?

Chemistry
2 answers:
larisa [96]3 years ago
7 0
The mass number is the sum of the protons and neutrons
64 = #protons + #neutrons
Find the number of protons given the number of neutrons
64 = #protons + 33
#protons = 31
In a neutral element (such as this), the number of electrons equals the number of protons
#electrons = #protons
#electrons = 31
andrew-mc [135]3 years ago
6 0

Answer:

The isotope must have 31 electrons

Explanation:

  • Mass number = (number of neutrons)+(number of protons)
  • Here, mass number of element is 64 and number of neutrons is 33
  • So, number of protons i the element = (mass number)-(number of neutrons) = (64)-(33) = 31
  • Isotope of an element remain present in neutral form.
  • So, number of protons should be equal to number of electrons in that isotope. Because proton has +1 charge and electron has -1 charge
  • So, the isotope must have 31 electrons.
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5 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
An iron block of mass 18 kg is heated from 285 K to 318 K. If 267.3 kJ is required, what is the specific heat of iron? A. 450.00
valkas [14]

Answer:

  • <u>Option A. 450.00</u>

Explanation:

<u>1) Data:</u>

a) m = 18 kg

b) T₁ = 285 K

c) T₂ = 318 K

d) Q = 267.3 kJ

e) S = ?

<u>2) Principles and equations</u>

The specific heat of a substance is the amount of heat energy absorbed to increase the temperature of certain amount (gram, kg, or moles, depending on the definition or units) of the substance in 1 ° C or 1 K.

The mathematical relation between the specific heat and the heat energy absorbed is:

  • Q = m × S × ΔT

Where,

  • Q is the heat absorbed,
  • S is the specific heat, and
  • ΔT is the temperature increase (T₂ - T₁)

<u>3) Solution:</u>

<u>a) Substitute the data into the equation:</u>

  • 267.3 kJ = 18 kg × S × (318 K - 285 K)

<u>b) Solve for S and compute:</u>

  • S = 267.3 kJ / (18 kg × 33 K) = 0.45 kJ / (Kg . K)

The options have not units, but I notice that the first answer is 1,000 times the answer I obtained, so I will make a conversion of units.

<u>c) Convert to J /( kg . k):</u>

  • 0.45 kJ / (Kg . K) × 1,000 J / kJ = 450 J / (kg . K)

Now we can see that the option A is is the answer, assuming the units.

6 0
3 years ago
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