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vekshin1
3 years ago
7

Which of the following is NOT true of white dwarfs?

Chemistry
2 answers:
ella [17]3 years ago
8 0
The answer is they are cold because white dwarfs are hot when they form.
liq [111]3 years ago
6 0

Answer:

A white dwarf is what stars like the Sun become after they have exhausted their nuclear fuel. Near the end of its nuclear burning stage, this type of star expels most of its outer material, creating a planetary nebula. Only the hot core of the star remains. ... That means a white dwarf is 200,000 times as dense.

Explanation:

they are cold

they are about the size of Earth

these both are not true

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The measure of the attraction that an atom has for electrons involved in chemical bonds is known as: ? electron affinity / radio
swat32
The answer is the last option. Electronegativity is the measure of the attraction that an atom has for electrons involved in chemical bonds.  It relates to covalent bonds where electrons are shared. The Pauling scale is the most used.
5 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
The smallest part of a molecule is called
yarga [219]

Answer:

D atom

Explanation:

The smallest particle of a substance that retains the chemical and physical properties of the substance and is composed of two or more atoms

8 0
2 years ago
Read 2 more answers
At the beginning of an experiment, a scientist has 176 grams of radioactive goo. After 165 minutes, her sample has decayed to 5.
Paul [167]
The half-life equation is written as:

An = Aoe^-kt

We use this equation for the solution. We do as follows:

5.5 = 176e^-k(165)
k = 0.02
<span>What is the half-life of the goo in minutes? 
</span>
0.5 = e^-0.02t
t = 34.66 minutes <----HALF-LIFE


Find a formula for G(t) , the amount of goo remaining at time t.G(t)=? 

G(t) = 176e^-0.02t

How many grams of goo will remain after 50 minutes? 

G(t) = 176e^-0.02(50) = 64.75 g
6 0
3 years ago
If steel (a metal) is hard and granite (a nonmetal) is hard, why don’t we make automobile engines out of granite?
Charra [1.4K]
Because. just because.
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