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3 years ago

When you burn a log in your fireplace you are converting ________?

1 answer:

3 0

When you burn a log in your fireplace you are converting cellulose into carbon dioxide and water.

Wood is made from cellulose. When wood burns, cellulose reacts with oxygen from the air and it forms carbon dioxide and water vapor.

The reaction of the combustion of glucose is as follows:

C₆H₁₂O₆ + 6O₂ = 6 CO₂ + 6H₂O
When wood burns, we notice that it slowly disappears because it passes into the mentioned gases. The ash that is created is part of the wood that did not burn.

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A student is doing an experiment that stoichiometry says should produce 34.6 grams of product. The student actually makes 25.2 g

attashe74 [19]

25.2 / 34.6 x 100
which would get you 73%

if you found my answer helpful pls give brainliest

6 0

3 years ago

Given the data calculated in Parts A, B, C, and D, determine the initial rate for a reaction that starts with 0.85 M of reagent

elixir [45]

Answer : The initial rate for a reaction will be $3.8\times 10^{-4}Ms^{-1}$

Explanation :

Rate law : It is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

The chemical equation will be:

$A+B+C\rightarrow P$

Rate law expression for the reaction:

$\text{Rate}=k[A]^a[B]^b[C]^c$

where,

a = order with respect to A

b = order with respect to B

c = order with respect to C

Expression for rate law for first observation:

$6.1\times 10^{-5}=k(0.2)^a(0.2)^b(0.2)^c$ ....(1)

Expression for rate law for second observation:

$1.8\times 10^{-4}=k(0.2)^a(0.2)^b(0.6)^c$ ....(2)

Expression for rate law for third observation:

$2.4\times 10^{-4}=k(0.4)^a(0.2)^b(0.2)^c$ ....(3)

Expression for rate law for fourth observation:

$2.4\times 10^{-4}=k(0.4)^a(0.4)^b(0.2)^c$ ....(4)

Dividing 1 from 2, we get:

$\frac{1.8\times 10^{-4}}{6.1\times 10^{-5}}=\frac{k(0.2)^a(0.2)^b(0.6)^c}{k(0.2)^a(0.2)^b(0.2)^c}\\\\3=3^c\\c=1$

Dividing 1 from 3, we get:

$\frac{2.4\times 10^{-4}}{6.1\times 10^{-5}}=\frac{k(0.4)^a(0.2)^b(0.2)^c}{k(0.2)^a(0.2)^b(0.2)^c}\\\\4=2^a\\a=2$

Dividing 3 from 4, we get:

$\frac{2.4\times 10^{-4}}{2.4\times 10^{-4}}=\frac{k(0.4)^a(0.4)^b(0.2)^c}{k(0.4)^a(0.2)^b(0.2)^c}\\\\1=2^b\\b=0$

Thus, the rate law becomes:

$\text{Rate}=k[A]^2[B]^0[C]^1$

Now, calculating the value of 'k' by using any expression.

Putting values in equation 1, we get:

$6.1\times 10^{-5}=k(0.2)^2(0.2)^0(0.2)^1$

$k=7.6\times 10^{-3}M^{-2}s^{-1}$

Now we have to calculate the initial rate for a reaction that starts with 0.85 M of reagent A and 0.70 M of reagents B and C.

$\text{Rate}=k[A]^2[B]^0[C]^1$

$\text{Rate}=(7.6\times 10^{-3})\times (0.85)^2(0.70)^0(0.70)^1$

$\text{Rate}=3.8\times 10^{-3}Ms^{-1}$

Therefore, the initial rate for a reaction will be $3.8\times 10^{-3}Ms^{-1}$

6 0

3 years ago

Calcium is a nontoxic metal that is used in supplements to maintain strong bones. Under high pressure conditions, over about 320

Lera25 [3.4K]

Explanation:

It is known that in a simple cubic unit cell the atoms are only present at the corner of the unit cell. This means that there are in total 8 atoms present in a simple cubic unit cell.

Therefore, in one simple cubic unit cell sharing of one atom is only $\frac{1}{8}$ .

Hence, the total number of atoms in a unit cell will be as follows.

$8 \times \frac{1}{8}$ = 1

Thus, we can conclude that there is 1 calcium atom present in each unit cell.

4 0

3 years ago

Which of the following diagrams represents a single replacement reaction?

kari74 [83]

Answer:

It's a, trust me I'm very good with chemistry

3 0

2 years ago

How does beta decay affect the atomic number?

MA_775_DIABLO [31]

The atomic mass number does not change because a beta particle has a much smaller mass than the atom. The atomic number goes up because a neutron has turned into an extra proton. Beta decay is fundamentally different from alpha decay. An alpha particle is made of two protons and two neutrons.

6 0

3 years ago