Calculate the molarity of a MgSosolution

The mass of solute per 100 mL of solution is abbreviated as (m/v). Mass is not technically the same thing as weight, but the abb

Nimfa-mama [501]

Answer:

$\boxed{\text{254 g}}$

Explanation:

$\begin{array}{rcl}\text{\% m/V} & = & \dfrac{\text{Mass of sucrose}}{\text{Volume of solution}}\\\\\text{Let m}& = &\text{mass of sucrose}\\\dfrac{\text{35.0 g}}{\text{100 mL}}& = & \dfrac{m}{\text{725 mL}}\\\\m & = &\dfrac{\text{35.0 g}\times 725}{100}\\\\ & = &\textbf{254 g}\\\end{array}\\\text{You need $\boxed{\textbf{254 g}}$ of sucrose}$

3 0

3 years ago

Determine the name of the compound Li2O

eduard

Lithium Oxide
I just search it up to be honest

7 0

3 years ago

Please and thank you!

Valentin [98]

Answer:

7.5 moles of O₂.

Explanation:

We'll begin by writing the balanced equation for the reaction. This is illustrated below:

2KClO₃ —> 2KCl + 3O₂

From the balanced equation above,

2 moles of KClO₃ decomposed to produce 3 moles of O₂.

Finally, we shall determine the number of mole of O₂ produced by the decomposition of 5 moles of KClO₃. This can be obtained as follow:

From the balanced equation above,

2 moles of KClO₃ decomposed to produce 3 moles of O₂.

Therefore, 5 moles of KClO₃ will decompose to produce = (5 × 3)/ 2 = 7.5 moles of O₂.

Thus, 7.5 moles of O₂ were obtained from the reaction.

3 0

3 years ago

Describe the changes in properties (from metals to nonmetals or from nonmetals to metals) as we move (a) down a periodic group a

defon

Answer:

Explanation:

The changes in properties from metals to non-metals on a periodic table can be measured and determined by the metallicity or electropositivity of elements.

Metallicity is a measure of the tendency of atoms of an element to lose electrons.

a.

Down a periodic group, metallicity increases.

b.

Across a period from left to right electropositivity or metallicity decreases.

Metals are found in the left part of periodic table and the most reactive metal sits in the lower left corner. Non-metals are towards the right side of the table.

7 0

3 years ago

Chlorine has two naturally occurring isotopes, 35C1 Gsotopic mass 34.9689 amu) and 370 (isotopic mass 36.9659 amu). If chlorine

Lilit [14]

Answer: The percentage abundance of $_{17}^{35}\textrm{Cl}$ and $_{17}^{37}\textrm{Cl}$ isotopes are 75.77% and 24.23% respectively.

Explanation:

Average atomic mass of an element is defined as the sum of masses of each isotope each multiplied by their natural fractional abundance.

Formula used to calculate average atomic mass follows:

$\text{Average atomic mass }=\sum_{i=1}^n\text{(Atomic mass of an isotopes)}_i\times \text{(Fractional abundance})_i$ .....(1)

Let the fractional abundance of $_{17}^{35}\textrm{Cl}$ isotope be 'x'. So, fractional abundance of $_{17}^{37}\textrm{Cl}$ isotope will be '1 - x'

For $_{17}^{35}\textrm{Cl}$ isotope:

Mass of $_{17}^{35}\textrm{Cl}$ isotope = 34.9689 amu

Fractional abundance of $_{17}^{35}\textrm{Cl}$ isotope = x

For $_{17}^{37}\textrm{Cl}$ isotope:

Mass of $_{17}^{37}\textrm{Cl}$ isotope = 36.9659 amu

Fractional abundance of $_{17}^{37}\textrm{Cl}$ isotope = 1 - x

Average atomic mass of chlorine = 35.4527 amu

Putting values in equation 1, we get:

$35.4527=[(34.9689\times x)+(36.9659\times (1-x))]\\\\x=0.7577$

Percentage abundance of $_{17}^{35}\textrm{Cl}$ isotope = $0.7577\times 100=75.77\%$

Percentage abundance of $_{17}^{37}\textrm{Cl}$ isotope = $(1-0.7577)=0.2423\times 100=24.23\%$

Hence, the percentage abundance of $_{17}^{35}\textrm{Cl}$ and $_{17}^{37}\textrm{Cl}$ isotopes are 75.77% and 24.23% respectively.

6 0

3 years ago