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Marrrta [24]
4 years ago
6

Heat will flow from a hot object to a cold object until the objects reach A. stability. B. boiling point. C. collapse. D. equili

brium.
Chemistry
2 answers:
Aleonysh [2.5K]4 years ago
6 0
The thermal energy will flow from the hotter object to the cooler object until their temperatures are equal i.e. until they reach equilibrium.

Answer:
D. equilibrium
Luba_88 [7]4 years ago
3 0
HEY THERE;))

The answer to your question is D. Equilibrium.

Hope this helps:)

~ TRUE BOSS
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What is the density of lead (in g/cm^3 3 ) if a rectangular bar measuring 0.500 cm in height, 1.55 cm in width, and 25.00 cm in
Viktor [21]

Answer:

Density = 11.4 g/cm³

Explanation:

Given data:

Density of lead = ?

Height of lead bar = 0.500 cm

Width of lead bar = 1.55 cm

Length of lead bar = 25.00 cm

Mass of lead bar = 220.9 g

Solution:

Density = mass/ volume

Volume of bar = length × width × height

Volume of bar = 25.00 cm × 1.55 cm × 0.500 cm

Volume of bar = 19.4 cm³

Density of bar:

Density = 220.9 g/ 19.4 cm³

Density = 11.4 g/cm³

3 0
3 years ago
Express 749 000 000 in scientific notation,
Dima020 [189]

Answer:

7.49 × 108

Explanation:

Scientific notation is a way to express numbers in a form that makes numbers that are too small or too large more convenient to write. It is commonly used in mathematics, engineering, and science, as it can help simplify arithmetic operations. In scientific notation, numbers are written as a base, b, referred to as the significant, multiplied by 10 raised to an integer exponent, n, which is referred to as the order of magnitude:

8 0
3 years ago
When there are more electrons than protons an object has an overall<br> charge.
azamat

Answer:it would have a negative charge because  electrons are negative

Explanation:

7 0
3 years ago
The standard enthalpy of formation for glucose [c6h12o6(s)] is −1273.3 kj/mol. what is the correct formation equation correspond
balu736 [363]
The standard formation equation for glucose C6H12O6(s) that corresponds to the standard enthalpy of formation or enthalpy change ΔH°f = -1273.3 kJ/mol is 
     C(s) + H2(g) + O2(g) → C6H12O6(s)
and the balanced chemical equation is 
     6C(s) + 6H2(g) + 3O2(g) → C6H12O6(s)

Using the equation for the standard enthalpy change of formation 
     ΔHoreaction = ∑ΔHof(products)−∑ΔHof(Reactants)
     ΔHoreaction = ΔHfo[C6H12O6(s)] - {ΔHfo[C(s, graphite) + ΔHfo[H2(g)] + ΔHfo[O2(g)]}

C(s), H2(g), and O2(g) each have a standard enthalpy of formation equal to 0 since they are in their most stable forms:
     ΔHoreaction = [1*-1273.3] - [(6*0) + (6*0) + (3*0)]
                           = -1273.3 - (0 + 0 + 0)
                           = -1273.3
8 0
3 years ago
How do the reaction rates change as the system approaches equilibrium?
Leni [432]
It becomes more stable
delta G gets closer to zero
7 0
4 years ago
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