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3 years ago

HELP NOW!!!!!! When you put water into a hot metal pan of equal mass, why does the pan cool down more than the water heats up?

Chemistry

1 answer:

Salsk061 [2.6K]3 years ago

8 0

Answer:

Because the specific heat of the metal is less than the specific heat of water.

Explanation:

Hello, happy to help you today!

In this case, we need to analyze a property called "specific heat" which accounts for how much energy is required to increase or decrease the temperature of 1 g of the substance by 1 °C.

In this case, since the specific heat of water is about 4.184 J/g°C and the specific heat of metals in general is greater than zero, of course, but less than one, we can infer that for the same amount of energy, when they are in contact, more grams of metal will be cooled down to those of water heated up, because the specific heat of the metal is less than the specific heat of water.

Best regards!.

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The _______ of an atom are negatively charged. a. protons b. negatrons c. neutrons d. electrons

PIT_PIT [208]

Protons = +1
negatrons = -1
neutrons = 0
electrons = -1

negatrons are like positrons and they are subatomic particles and are not present in atoms

so electrons are negatively charged and are present in atoms

hope that helps

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3 years ago

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Answer:

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3 years ago

How can you convert diamond into graphite ?

Setler79 [48]

Answer:

➢There are different ways to trigger the conversion of diamond to graphite, for instance by simply heating the diamond under exclusion of oxygen or even with an aimed mechanical stroke.

Explanation:

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3 years ago

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3.15.8 cm3 of H2SO4 completely neutralised 20 cm3 of NaOH. The NaOH was 1.5 mol/dm3.

Furkat [3]

Answer:

Molarity = 0.95 mol/dm³

Explanation:

Given data:

Volume of H₂SO₄ = 15.8 cm³

Volume of NaOH = 20 cm³

Concentration of NaOH = 1.5 mol/dm³

Concentration of H₂SO₄ = ?

Solution:

Chemical equation:

NaOH + H₂SO₄ → Na₂SO₄ + H₂O

First of all we will calculate the number of moles of NaOH and for that we will convert the units first,

Volume = 20 cm³/1000 = 0.02 L

Concentration of NaOH = 1.5 mol/dm³

1 mol/dm³ = 1 mol/L

Concentration of NaOH = 1.5 mol/L

Number of moles of NaOH:

Molarity = number of moles / volume in L

1.5 M = number of moles / 0.02 L

Number of moles = 1.5 M ×0.02 L

Number of moles = 0.03 mol

Now we will compare the moles of NaOH and H₂SO₄

NaOH : H₂SO₄

2 : 1

0.03 : 1/2×0.03 = 0.015 mol

Concentration of H₂SO₄:

Volume of H₂SO₄:

15.8 cm³/1000 = 0.0158 L

Molarity = number of moles / volume in L

Molarity = 0.015 mol / 0.0158 L

Molarity = 0.95 mol/L

1 mol/L = 1 mol/dm³

Molarity = 0.95 mol/dm³

6 0

3 years ago

stbank, Question 075 Get help answering Molecular Drawing questions. Compound A, C6H12 reacts with HBr/ROOR to give compound B,

Law Incorporation [45]

Answer:

Explanation:

In this case we want to know the structures of A (C6H12), B (C6H13Br) and C (C6H14).

A and C reacts with two differents reagents and conditions, however both of them gives the same product.

Let's analyze each reaction.

First, C6H12 has the general formula of an alkene or cycloalkane. However, when we look at the reagents, which are HBr in ROOR, and the final product, we can see that this is an adition reaction where the H and Br were added to a molecule, therefore we can conclude that the initial reactant is an alkene. Now, what happens next? A is reacting with HBr. In general terms when we have an adition of a molecule to a reactant like HBr (Adding electrophyle and nucleophyle) this kind of reactions follows the markonikov's rule that states that the hydrogen will go to the carbon with more hydrogens, and the nucleophyle will go to the carbon with less hydrogen (Atom that can be stabilized with charge). But in this case, we have something else and is the use of the ROOR, this is a peroxide so, instead of follow the markonikov rule, it will do the opposite, the hydrogen to the more substituted carbon and the bromine to the carbon with more hydrogens. This is called the antimarkonikov rule. Picture attached show the possible structure for A. The alkene would have to be the 1-hexene.

Now in the second case we have C, reacting with bromine in light to give also B. C has the formula C6H14 which is the formula for an alkane and once again we are having an adition reaction. In this case, conditions are given to do an adition reaction in an alkane. bromine in presence of light promoves the adition of the bromine to the molecule of alkane. In this case it can go to the carbon with more hydrogen or less hydrogens, but it will prefer the carbon with more hydrogens. In this case would be the terminal hydrogens of the molecules. In this case, it will form product B again. the alkane here would be the hexane. See picture for structures.

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4 years ago