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3 years ago

8. A log is burning in a fireplace. If the amount of oxygen reaching the log is decreased, which of the

2 answers:

8 0

Answer: the reaction rate decreases.

Fire is the result of a combustion reaction, which is the reactionof the organic matter with oxygen. The rate is proportional to amountof oxygen. The, when the amount of oxygen that reaches the organic matter decreases, the rate of combustion decreases too.

EastWind [94]3 years ago

3 0

Answer is: The reaction rate decreases.

Balanced chemical reaction of burning log:

C(s) + O₂(g) → CO₂(g).

The reaction rate is the speed at which reactants are converted into products.

Reaction rate depends on concentration (amount) of oxygen.

If amount of oxygen decreases, reaction rate also decreases and vice versa.

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How to solve 2=-log(x)

ch4aika [34]

First, you want to extract the negative from -log(x).

So now you have log(x) = -2

Now you have to use the property loga(x)=b is the same as x=a^2

So now, it is x = 10^-2. Remember that when there is just a log, it is implied that it is ‘a’ is 10.

Then you evaluate the negative square to 1/10^2

Answer is 1/100

8 0

3 years ago

The mass of an atom of element x is equivalent to the total mass of 7 hydrogen atoms.

Agata [3.3K]

Answer:

Nitrogen

Explanation:

7 0

3 years ago

Calculate the number of grams of solute in 500.0 mL of 0.189 M KOH.

KIM [24]

Answer : The number of grams of solute in 500.0 mL of 0.189 M KOH is, 5.292 grams

Solution : Given,

Volume of solution = 500 ml

Molarity of KOH solution = 0.189 M

Molar mass of KOH = 56 g/mole

Formula used :

$Molarity=\frac{\text{Mass of KOH}\times 1000}{\text{Molar mass of KOH}\times \text{Volume of solution in ml}}$

Now put all the given values in this formula, we get the mass of solute KOH.

$0.189M=\frac{\text{Mass of KOH}\times 1000}{(56g/mole)\times (500ml)}$

$\text{Mass of KOH}=5.292g$

Therefore, the number of grams of solute in 500.0 mL of 0.189 M KOH is, 5.292 grams

7 0

3 years ago

Read 2 more answers

a compound has a molar mass of 129 g/mol if its empirical formula is C2H5N then what is the molecular formula

Elena-2011 [213]

Given :

A compound has a molar mass of 129 g/mol .

Empirical formula of compound is C₂H₅N .

To Find :

The molecular formula of the compound.

Solution :

Empirical mass of compound :

$M_e = ( 2 \times 12 ) + ( 5 \times 1 ) + ( 1 \times 14 )\\\\M_e = 43\ gram/mol$

Now, n-factor is :

$n = \dfrac{M}{M_e}\\\\n = \dfrac{129}{43}\\\\n = 3$

Multiplying each atom in the formula by 3 , we get :

Molecular Formula, C₆H₁₅N₃

3 0

3 years ago

When equal volumes of 0.5 M HCl and 0.5 M Ca(OH)2 are mixed, the resulting solution is

Mademuasel [1]

The concentration of mixed solution = 0.5 M

<h3> Further explanation </h3>

Given

0.5 M HCl

0.5 M Ca(OH)₂

Required

The concentration

Solution

Molarity from 2 solutions :

Vm Mm = V₁. M₁ + V₂. M₂

m = mixed solution

V = volume

M = molarity

V = mixed volume

1 = solution 1

2 = solution 2

Vm = V₁+V₂

Equal volumes⇒V₁=V₂, and Vm = 2V, then equation becomes :

2V.Mm = V(M₁+M₂)

2V.Mm = V(0.5+0.5)

Mm=0.5 M

8 0

3 years ago