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kondaur [170]
3 years ago
12

Two atoms that share one electron each between have a ?

Chemistry
1 answer:
evablogger [386]3 years ago
8 0

Answer:

Covalent bond

Explanation:

A covalent bond is defined as the sharing of election pairs between atoms

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ohaa [14]
6 ( tell me if you got it right)
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3 years ago
A student measures the mass of a 6.0 cm3 block of brown sugar to be 10.0 g. What is the density of the brown sugar?
Alex17521 [72]

Answer:

1.67g/cm3

Explanation:

The formula for density is d=\frac{m}{v} . The m variable stands for mass and the v variable stands for volume.

The mass of the brown sugar is 10.0g and the volume is 6.0cm3, so we can plug those values into the equation.

d=\frac{10g}{6cm^{3} }

d=1.67\frac{g}{cm^{3} }

Rounded to 3 significant figures, the density of the block of brown sugar is 1.67 g/cm3. If the mass is in grams and the volume is in cm3, the unit for the final answer is \frac{g}{cm^{3} } (grams per centimetres cubed).

8 0
3 years ago
Read 2 more answers
What is the concentration of H+ in a 2.5 M HCl solution?
Pavel [41]

Answer:

The concentration of H⁺ in a 2.5 M HCl solution is 2.5 M

Explanation:

As HCl is a strong acid and hence a strong electrolyte, it will dissociate as

HCl ⟶ H⁺ + Cl⁻

So, The concentration of H⁺ will be 2.5 M (same as HCl)

Thus, The concentration of H⁺ in a 2.5 M HCl solution is 2.5 M

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6 0
3 years ago
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The sun radiates different type of Electromagnetic energy (TRUE or FALSE)
Paha777 [63]
That statment is true
7 0
3 years ago
What is the molality of an aqueous solution containing FeCl3 (MM = 162.2 g/mol) with a mole fraction of FeCl3 of 0.15?
Natalka [10]

Answer:

10 m

Explanation:

The mole fraction of FeCl₃ of 0.15, that is, per mole of solution, there are 0.15 moles of FeCl₃ and 1 - 0.15 = 0.85 moles of water.

The molar mass of water is 18.02 g/mol. The mass corresponding to 0.85 moles is:

0.85 mol × 18.02 g/mol = 15 g = 0.015 kg

The molality of FeCl₃ is:

m = moles of solute / kilogram of solvent

m = 0.15 mol / 0.015 kg

m = 10 m

3 0
3 years ago
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