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3 years ago

Sentence using Dune fence

Chemistry

1 answer:

tino4ka555 [31]3 years ago

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Which has more energy, 500g of ice or 500g of steam

liraira [26]

Answer:

Steam because steam pushes out and ice stays dormant.

Explanation:

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3 years ago

A sample may contain any or all of the following ions: Hg22+, Ba2+, Mn2+. No precipitate formed when an aqueous solution of NaCl

Bess [88]

Answer:

Only Mn⁺² is present.

Explanation:

When an aqueous solution of NaCl is added, the Cl⁻ species is introduced. In the presence of Cl⁻, Hg₂²⁺ would precipitate as Hg₂Cl₂.

When an aqueous solution of Na₂SO₄ is added, the SO₄⁻² species is introduced. In the presence of SO₄⁻², Ba⁺² would precipitate as BaSO₄.

No precipitate formed when either of these solutions were added, thus the sample does not contain Hg₂⁺² nor Ba⁺².

Under basic conditions Mn⁺² would precipitate as Mn(OH)₂. A precipitate formed once the solution was made basic, so the sample contains Mn⁺².

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3 years ago

Which of the following is the correct way to express 0.0000321 in scientific notation?

scZoUnD [109]

The answer is 3.21*10^-5

7 0

3 years ago

A mixture whoose parts are the same throughout

Anit [1.1K]

A homogeneous mixture is a mixture whose parts are the same throughout.

4 0

4 years ago

Read 2 more answers

A sample of tin (Cp = 0. 227 J/gâ€˘Â°C) is placed in a freezer. Its temperature decreases from 15. 0Â°C to â’10. 0Â°C as it rele

pshichka [43]

Taking into account the definition of calorimetry, the mass of the sample is 478.41 grams.

Calorimetry is the measurement and calculation of the amounts of heat exchanged by a body or a system.

Sensible heat is defined as the amount of heat that a body absorbs or releases without any changes in its physical state (phase change).

So, the equation that allows to calculate heat exchanges is:

Q = c× m× ΔT

where Q is the heat exchanged by a body of mass m, made up of a specific heat substance c and where ΔT is the temperature variation.

In this case, you know:

Q= -543 J because the energy is releases
c= 0.227 $\frac{J}{gC}$
m= ?
ΔT= Tfinal - Tinitial= 10 C - 15 C= -5 C

Replacing:

-543 J= 0.227 $\frac{J}{gC}$ × m× (-5 C)

Solving:

$m=\frac{-543 J}{0.227 \frac{J}{gC}x(-5 C)}$

m=478.41 grams

In summary, the mass of the sample is 478.41 grams.

Learn more about calorimetry:

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3 years ago