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liraira [26]
3 years ago
7

Object A at 40ºC and object B at 80ºC are placed in contact with each other. Which statement describes the heat flow between the

objects?
A)Heat flows from object A to object B.B)Heat flows from object B to object A.C)Heat flows in both directions between the objects.D)No heat flow occurs between the objects.
Chemistry
2 answers:
Arada [10]3 years ago
6 0
By the second law of thermodynamics:
Heat can not spontaneously flow from cold regions to hot regions without external work being performed on a system.
Heat transfer is the passage of thermal energy from a hot ( t B = 80° C ) to a colder body ( t A = 40° C ). 
Answer:  B ) Heat flows from object B to object A.
ozzi3 years ago
5 0
Option B is right that is Heat flows from object <span>B </span>to object because if 2 bodies have different temperature so heat always transfer from higher to lower ,from hot to cold .
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At a certain temperature, the solubility of n2 gas in water at 3.08 atm is 72.5 mg of n2 gas/100 g water . calculate the solubil
8090 [49]

According to Henry's law, solubility of solution is directly proportional to partial pressure thus,

\frac{S_{1}}{P_{1}}=\frac{S_{2}}{P_{2}}

Solubility at pressure 3.08 atm is 72.5/100, solubility at pressure 8 atm should be calculated.

Putting the values in equation:

\frac{0.725}{3.08}=\frac{S_{2}}{8}

On rearranging,

S_{2}=\frac{0.725\times 8}{3.08}=1.88

Therefore, solubility will be 1.88 mg of N_{2} gas in 1 g of water or, 188 mg of tex]N_{2}[/tex] gas in 100 g of water.

3 0
3 years ago
A) 30.22 g NaCl x 1 mol NaC l =      58.4430 Molar mass (g) NaCl=0.5171 mol NaCl
jekas [21]
The third question requires you to solve for the weight of sodium (Na) and weight of Chloride (Cl) from the calculated moles of each element Na, and Cl.

So, you need to multiply the calculated moles of Na with its molar mass (23 g/ mol) to get the answer for Na. And multiply the calculated moles of Cl with its molar mass (35.45 g/mol) to get the answer for Cl. 
6 0
3 years ago
Is there are a total of 7.3 x 10^29 atoms in a sample of glucose, C6H12O6(s),what amount in miles of glucose is in the sample?
Vadim26 [7]

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5.1 x 10^4 mol C6H12O6

4 0
3 years ago
Iron(II) sulfide has a primitive cubic unit cell with sulfide ions at the lattice points.
masya89 [10]

We have that the  the density of FeS  is mathematically given as

  •  \phi=2.56h/cm^3

From the question we are told

Iron(II) sulfide has a primitive <em>cubic</em> unit cell with <em>sulfide</em> ions at the <em>lattice points.</em>

The ionic radii of iron(II) ions and sulfide ions are 88 pm and 184 pm, respectively.

What is the density of FeS (in g/cm3)?

<h3>Density</h3>

Generally the equation for the Velocity  is mathematically given as

V_c=a^3=(2\pi Fe^{2+}+2\pi S^{2-})^3\\\\Therefore\\\\V=a^3(2\pi*0.088+2\pi 0.184)^3\\\\V=16.98*10^{-23}\\\\Therefore\\\\\phi=n\frac{PFeion+PSion}{VNa}\\\\\phi=3*\frac{55.85+32}{16.9*10^{-23}*6.023*10^{23}}

  • \phi=2.56h/cm^3
  • \phi=2.56h/cm^3

For more information on ionic radii visit

brainly.com/question/13981855

4 0
3 years ago
Which has the highest boiling point .33 m NH3 or .10 m Na2SO4​
Aleksandr-060686 [28]

Answer:

0.33 mol/kg NH₃

Explanation:

Data:

     b(NH₃) = 0.33 mol/kg

b(Na₂SO₄) = 0.10 mol/ kg

Calculations:

The formula for the boiling point elevation ΔTb is

\Delta T_{b} = iK_{b}b

i is the van’t Hoff factor — the number of moles of particles you get from a solute.

(a) For NH₃,

The ammonia is a weak electrolyte, so it exists almost entirely as molecules in solution.  

1 mol NH₃ ⟶  1 mol particles

i ≈ 1, and ib = 1 × 0.33 = 0.33 mol particles per kilogram of water

(b) For Na₂SO₄,

Na₂SO₄(aq) ⟶ 2Na⁺(aq) + 2SO₄²⁻(aq)

1 mol Na₂SO₄ ⟶ 3 mol particles

i = 1 and ib = 3 × 0.10 = 0.30 mol particles per kilogram of water

The NH₃ has more moles of particles, so it has the higher boiling point.

5 0
3 years ago
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