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3 years ago

In an endothermic reaction which bonds are stronger? The reactants or products? Explain

Chemistry

1 answer:

Marina86 [1]3 years ago

6 0

reactants because it had other chemicals, when more are mixed it will be more stronger.

It is the correct answer

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Evaluate ( a=2 b=3)...3a - 2b =

skad [1K]

Hey!

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Steps To Solve:

~Substitute

3(2) - 2(3)

~Subtract

6 - 6

~Simplify

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Answer:

$\Large\boxed{\mathsf{0}}$

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Hope This Helped! Good Luck!

4 0

3 years ago

Read 2 more answers

if 1.386 g of mg ribbon combusts to form 2.309 g of oxide product, calculate the experimental mass percent of oxygen from this d

Pavlova-9 [17]

1.386 g of Mg ribbon combusts to form 2.309 g of oxide product. The mass percent of oxygen in the oxide is 40.0 %.

Let's consider the reaction for the combustion of Mg.

Mg + 1/2 O₂ ⇒ MgO

1.386 g of Mg combusts to form 2.309 g of MgO. We want to determine the mass of oxygen in MgO. According to Lavoisier's law of conservation of mass, matter is not created nor destroyed over the course of a chemical reaction. Then, the mass of Mg in the reactants is equal to the mass of Mg in MgO. The mass of the magnesium oxide is the sum of the masses of magnesium and oxygen. The <u>mass of oxygen in the oxide</u> is:

$mMgO = mMg + mO\\mO = mMgO - mMg = 2.309 g - 1.386 g = 0.923 g$

We can calculate the mass percent of O in MgO using the following expression.

$\% O = \frac{mO}{mMgO} \times 100\% = \frac{0.923 g}{2.309g} \times 100\% = 40.0 \%$

You can learn more about mass percent here: brainly.com/question/14990953

3 0

2 years ago

An unknown compound is processed using elemental analysis and found to contain 117.4g of platinum 28.91 carbon and 33.71g nitrog

dlinn [17]

Answer:

1 mole of platinum

Explanation:

To obtain the number of mole(s) of platinum present, we need to determine the empirical formula for the compound.

The empirical formula for the compound can be obtained as follow:

Platinum (Pt) = 117.4 g

Carbon (C) = 28.91 g

Nitrogen (N) = 33.71 g

Divide by their molar mass

Pt = 117.4 / 195 = 0.602

C = 28.91 / 12 = 2.409

N = 33.71 / 14 = 2.408

Divide by the smallest

Pt = 0.602 / 0.602 = 1

C = 2.409 / 0.602 = 4

N = 2.408 / 0.602 = 4

The empirical formula for the compound is PtC₄N₄ => Pt(CN)₄

From the formula of the compound (i.e Pt(CN)₄), we can see clearly that the compound contains 1 mole of platinum.

8 0

3 years ago

Use your periodic table and calculator as needed for the following question.

sveticcg [70]

Hope you find this answer I need points

7 0

3 years ago

Can someone help me with this I've been stuck on it for a few days

Fiesta28 [93]

So I haven’t got time to answer all of it for you but the id you look at the picture of the periodic table I’ve added the top number in the red boxes are the groups and the period is how many elements down from the top it is (remember that the hydrogen and helium make up period ONE) so remember to include them when counting the elements as you go down the table

5 0

2 years ago