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2 years ago

What is 0.151 g in scientific notation

Chemistry

2 answers:

8090 [49]2 years ago

5 0

1.51x10^-1 i did this a while ago

Ahat [919]2 years ago

4 0

Answer: Scientific notation- $1.51 x 10^{-1}$

Explanation: I got this by moving the decimal, so that there is one non-zero digit to the left of that decimal point. Tip- The number of decimal places you move will be the exponent on the 10.

Moved right- Negative

Moved left- Positive

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A period is a horizontal row on the periodic table. a. True b. False

Kay [80]

It is True because in the periodic table of elements, there are seven horizontal rows of elements called periods. The vertical columns of elements are called groups, or families.

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2 years ago

How many moles of iron can be produced from 4.20 moles of iron (III) oxide?

Kay [80]

Hope this helps, have a nice day ahead!

6 0

3 years ago

Which of the following you is true for a limiting reactant

skad [1K]

Answer:

C) The limiting reactant has the lowest ratio of moles available / coefficient in the balanced equation.

Explanation:

Please, find attached a complete question to determine which of the statements is or are true for a limiting reactant in a chemical equation.

First, remember that the limiting reactant is the substance that is consumed completely while the excess reactant is the substance that does not react completely.

The limiting reactant is found comparing the stoichiometry ratio and the actual ratio between the reactants.

The stoichiometry ratio is found using the coefficientes of the chemical equation.

For illustration, assume the general chemical equation:

$aA+bB\rightarrow cC+dD$

The stoichiometric ratio of the reactants is:

$a\text{ }moles\text{ }of\text{ }A/b\text{ }moles\text{ }of\text{ }B$

If the ratio of the available moles of substance A to the available moles of substance B is greater than the stoichiometric ratio, it means that there are more moles of the substance A than what is needed to react with the available moles of substance B, then A will be in excess and B will B the limiting reactant.

If, on the contrary, the ratio of the available moles of substance A to the available moles of substance B is is less than the stoichiometric ratio, then substance A is less than the necessary to make the all the moles of substance B react, meaning that the substance A will limit the reaction (it will be consumed completely), while the substance B will be in excess.

As for the options:

A) The limiting reactant is has the lowest coefficient in a balanced equation.

This is false, since it is not the magnitude of the coefficiente what determines the limiting reactant, but the comparison of the ratios.

B) The limiting reactant is the reactant for which you have the fewest number of moles.

This is false because it is not the number of moles what determines the limiting reactant , but the comparison of the ratios.

C) The limiting reactant has the lowest ratio of moles available / coefficient in the balanced equation.

This is true as proved below.

The stoichiometric ratio of the reactants is:

$a\text{ }moles\text{ }of\text{ }A/b\text{ }moles\text{ }of\text{ }B$

The actual ratio is:

$available\text{ }moles\text{ }of\text{ }A/available\text{ }moles\text{ }of\text{ }B$

Assume the first ratio is less than the second (which describes when the substance A is in excess and the limiting reactant is the substance B).

$a\text{ }moles\text{ }of\text{ }A/b\text{ }moles\text{ }of\text{ }B$

Change the relation to show the ratios of moles available of each substance to the cofficient in the chemical equation:

$available\text{ }moles\text{ }of\text{ }B/b\text{ }moles\text{ }of\text{ }B$

Then, in the scenary that the limiting reactant is the substance B, the ratio of the left is lower than the ratio of the right, which is the same that limiting reactant has the lowest ratio of moles available / coefficient in the balanced equation.

D) The limiting reactant has the lowest ratio of coefficients in the balanced eqution/moles available.

This ratio is the inverse of the ratio of the previous statement, thus the relation is inverse, and, since the previous statement was true, this statement is false.

3 0

2 years ago

Calculate the pH for the following weak acid. A solution of HCOOH has 0.12M HCOOH at equilibrium. The Ka for HCOOH is 1.8×10−4.

Shtirlitz [24]

Answer:

the pH of HCOOH solution is 2.33

Explanation:

The ionization equation for the given acid is written as:

$HCOOH\leftrightarrow H^++HCOO^-$

Let's say the initial concentration of the acid is c and the change in concentration x.

Then, equilibrium concentration of acid = (c-x)

and the equilibrium concentration for each of the product would be x

Equilibrium expression for the above equation would be:

$\Ka= \frac{[H^+][HCOO^-]}{[HCOOH]}$

$1.8*10^-^4=\frac{x^2}{c-x}$

From given info, equilibrium concentration of the acid is 0.12

So, (c-x) = 0.12

hence,

$1.8*10^-^4=\frac{x^2}{0.12}$

Let's solve this for x. Multiply both sides by 0.12

$2.16*10^-^5=x^2$

taking square root to both sides:

$x=0.00465$

Now, we have got the concentration of $[H^+] .$

$[H^+] = 0.00465 M$

We know that, $pH=-log[H^+]$

pH = -log(0.00465)

pH = 2.33

Hence, the pH of HCOOH solution is 2.33.

7 0

2 years ago

Read 2 more answers

Androstenedione, which contains only carbon, hydrogen, and oxygen, is a steroid hormone produced in the adrenal glands and the g

Gennadij [26K]

Answer: The molecular formula for androstenedione is, $C_{19}H_{26}O_2$

Explanation:

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of $CO_2=5.527g$

Mass of $H_2O=1.548g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 5.527 g of carbon dioxide, $\frac{12}{44}\times 5.527=1.507g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18 g of water, 2 g of hydrogen is contained.

So, in 1.548 g of water, $\frac{2}{18}\times 1.548=0.172g$ of hydrogen will be contained.

For calculating the mass of oxygen:

Mass of oxygen in the compound = $(1.893g)-[(1.507g)+(0.172g)]=0.214g$

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{1.507g}{12g/mole}=0.126moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.172g}{1g/mole}=0.172moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.214g}{16g/mole}=0.0133moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is $0.0133$ moles.

For Carbon = $\frac{0.126}{0.0133}=9.5$

For Hydrogen = $\frac{0.172}{0.0133}=12.9\approx 13$

For Oxygen = $\frac{0.0133}{0.0133}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 9.5 : 13 : 1

To make in a whole number we are multiplying the ratio by 2, we get:

The ratio of C : H : O = 19 : 26 : 2

Thus, the empirical formula for the given compound is $C_{19}H_{26}O_2$

The empirical formula weight of $C_{19}H_{26}O_2$ = 19(12) + 26(1) + 2(16) = 286 gram/eq

Now we have to calculate the molecular formula of the compound.

Formula used :

$n=\frac{\text{Molecular formula}}{\text{Empirical formula weight}}$

$n=\frac{286.4}{286}=1$

Molecular formula = $C_{19}H_{26}O_2$

Therefore, the molecular formula for androstenedione is, $C_{19}H_{26}O_2$

3 0

2 years ago