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Elena L [17]
3 years ago
11

When an alkaline earth metal, a, reacts with a halogen, x, the formula of the covalent compund formed should be a2x?

Chemistry
1 answer:
Ierofanga [76]3 years ago
3 0

Alkaline earth metal are the elements present in II group in the periodic table and are known as 'Metals' and have a charge of +2.

Alkaline earth metals - Be , Mg Ca, Sr , Ba, Ra

Halogens are present in VII A group in the periodic table and are 'Non-metals' and have a charge of -1.

Halogens - F, Cl, Br, I, At

When Alkaline earth metal (metals) combine with Halogens (non-metals) the compound formed will be ionic compound and the formula of the compound will be based on the charges of the element.

When we write the formula of the ionic compound the charges of the elements get criss crossed.

For example - Mg (Alkaline earth metal) have a charge of +2 and Cl (Halogen) have a charge of -1 and when they combine to form the formula their charges get criss crossed and we will get Mg_{1}Cl_{2} or MgCl_{2}

When an alkaline earth metal, A, reacts with a halogen, X, the formula of the Ionic compound formed should be AX_{2}


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The concentration of a biomolecule inside a rod‑shaped prokaryotic cell is 0.0035 M . Calculate the number of molecules inside t
timama [110]

Answer: This rod-shape prokaryotic cell has 3.740,734725‬ molecules

Explanation:

<u>Step 1 :</u> given data

Molarity of the prokaryotic cell = 0.0035 M

Length of the cell = 4.2 μm = 4.2 * 10^-6 m

diameter of the cell = 1.3 μm = 1.3 * 10^-6 m

<u>Step 2: </u>calculate volume

To calculate volume of a rod, weneef to know the radius.

V = r ² × l

The radius = half of the diameter : r = d/2 ⇒ (1.3 * 10^-6 m)/2 = 0.65 * 10^-6 m

V= (0.65 * 10^-6 m)² * 4.2 * 10^-6 m = 1.7745 * 10 ^-18 L

<u>Step 3:</u> Calculating number of moles

Number of moles = Concentration * Volume

moles = 0.0035 M * 1.7745 * 10 ^-18 L = 6.21075 * 10^-21 moles‬

<u>Step 4:</u> calculating number of molecules

1 mole contains 6.023 * 10 ^-23 molecules

6.21075 * 10^-21 moles contain : 6.21075 * 10^-21 * 6.023 * 10 ^-23 molecules = 3.740,734725‬ molecules

This rod-shape prokaryotic cell has 3.740,734725‬ molecules

4 0
3 years ago
What season does the sourtther hempishere experience when it is away from the sun?
kakasveta [241]
I believe Winter is <span>your answer.</span>
4 0
3 years ago
Morphine has the formula c17h19no3. it is a base and accepts one proton per molecule. it is isolated from opium. a 0.685 −g samp
kotykmax [81]
2 C₁₇H₁₉NO₃ + H₂SO₄ → Product
Moles of H₂SO₄ = M x V(liters) = 0.0116 x 8.91/1000 = 1.033 x 10⁻⁴ mole 
moles of morphine = 2 x moles of H₂SO₄ = 2.066 x 10⁻⁴
Mass of morphine = moles x molar mass of morphine = 2.066 x 10⁻⁴ x 285.34 
                             = 0.059 g
percent morphine = \frac{mass of morphine}{mass of opium} x 100
                             = \frac{0.059}{0.685} x 100 = 8.6 %   
7 0
3 years ago
Find the volume of 20g of benzene
sergey [27]

(d) Density of Benzene: 0.8786 g/cm cubed

(m) Mass: 20.00g

Formula to solve (v) volume of benzene: V=m/d

V=20g / 0.8786g/cm cubed

Answer: Volume of benzene is: 22.8 cm cubed  


Explanation: Well, density is mass divided volume. In this, it is what the volume is. All you need is mass divided by density. Hope this helps!

6 0
3 years ago
A sample of CaCO3 (molar mass 100. g) was reported as being 30. percent Ca. Assuming no calcium was present in any impurities, c
natka813 [3]

Answer:

Approximately 75%.

Explanation:

Look up the relative atomic mass of Ca on a modern periodic table:

  • Ca: 40.078.

There are one mole of Ca atoms in each mole of CaCO₃ formula unit.

  • The mass of one mole of CaCO₃ is the same as the molar mass of this compound: \rm 100\; g.
  • The mass of one mole of Ca atoms is (numerically) the same as the relative atomic mass of this element: \rm 40.078\; g.

Calculate the mass ratio of Ca in a pure sample of CaCO₃:

\displaystyle \frac{m(\mathrm{Ca})}{m\left(\mathrm{CaCO_3}\right)} = \frac{40.078}{100} \approx \frac{2}{5}.

Let the mass of the sample be 100 g. This sample of CaCO₃ contains 30% Ca by mass. In that 100 grams of this sample, there would be \rm 30 \% \times 100\; g = 30\; g of Ca atoms. Assuming that the impurity does not contain any Ca. In other words, all these Ca atoms belong to CaCO₃. Apply the ratio \displaystyle \frac{m(\mathrm{Ca})}{m\left(\mathrm{CaCO_3}\right)} \approx \frac{2}{5}:

\begin{aligned} m\left(\mathrm{CaCO_3}\right) &= m(\mathrm{Ca})\left/\frac{m(\mathrm{Ca})}{m\left(\mathrm{CaCO_3}\right)}\right. \cr &\approx 30\; \rm g \left/ \frac{2}{5}\right. \cr &= 75\; \rm g \end{aligned}.

In other words, by these assumptions, 100 grams of this sample would contain 75 grams of CaCO₃. The percentage mass of CaCO₃ in this sample would thus be equal to:

\displaystyle 100\%\times \frac{m\left(\mathrm{CaCO_3}\right)}{m(\text{sample})} = \frac{75}{100} = 75\%.

3 0
3 years ago
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