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1 year ago

What is the formula of manganese (Il) phosphate? chemistry related

Chemistry

1 answer:

xeze [42]1 year ago

4 0

The chemical formula of manganese (Il) phosphate is Mn3(PO4)2, which is option 2.

<h3>What is chemical formula?</h3>

The chemical formula of a compound is the notation indicating the number of atoms of each element present in a compound.

The chemical formula of a compound tells us the type of elements in the compound, the number atoms of each element in the compound and in what proportion these elements are present.

According to this question, manganese II phosphate has the chemical formula of Mn3(PO4)2.

This chemical formula shows that it posseses the following:

3 manganese atoms
2 phosphorus atoms
8 oxygen atoms

Learn more about chemical formula at: brainly.com/question/11995171

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anyanavicka [17]

Answer:

by watching and writing down the information that is provided

Explanation:

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3 years ago

The cylinder of aluminum in class was approximately 13mm in circumference and 42mm in length. What would be it's approximate mas

Roman55 [17]

Answer:

Mass, m = 1.51 grams

Explanation:

It is given that,

The circumference of Aluminium cylinder, C = 13 mm = 1.3 cm

Length of the cylinder, h = 4.2 cm

We know that the density of the Aluminium is 2.7 g/cm³

Circumference, C = 2πr

$r=\dfrac{C}{2\pi}\\\\r=\dfrac{1.3}{2\pi}\\\\r=0.206\ cm$

Density is equal to mass per unit volume.

$d=\dfrac{m}{V}$

m is mass of the cylinder

V is the volume of the cylinder

$V=\pi r^2h\\\\V=\dfrac{22}{7}\times0.206^2\times 4.2\\\\V=0.5601\ cm^3$

So,

$m=d\times V\\\\m=2.7\times 0.5601\\\\m=1.51\ g$

So, the mass of the cylinder is 1.51 grams.

5 0

3 years ago

44.8% of a 250. mL acid solution is used in an experiment, what volume of acid (in mL) was used?

postnew [5]

Answer:

112 mL

Explanation:

The formula for percent by volume is

$\text{Percent by volume} = \dfrac{\text{Volume of solute}}{\text{Volume of solution}}\times 100 \, \%$

If you have 250 mL of a solution that is 44.8 % v/v,

$\begin{array}{rcl}44.8\, \% & = & \dfrac{\text{Volume of solute}}{\text{250 mL}}\times 100 \, \%\\\\44.8 \times \text{ 250 mL} & = & \text{Volume of solute} \times 100\\\text{Volume of solute} & = & \dfrac{44.8 \times 250\text{ mL}}{100}\\\\ & = & \textbf{112 mL}\\\end{array}$