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Brut [27]
3 years ago
11

Hen one atom contributes both bonding electrons in a single covalent bond, the bond is called a(n) ____.

Chemistry
1 answer:
Kaylis [27]3 years ago
4 0
Hello  <span>Squirrel9177 
</span><span>

Answer: Wh</span>en one atom contributes both bonding electrons in a single covalent bond, the bond is called a coordinate covalent bond


Hope This Helps!
-Chris
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Answer the type of cell

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3 years ago
A gas that exerts a pressure of
irina [24]

Answer:

V1 = 3.09 L

Explanation:

Initial Pressure, P1 = 15.6 psi

Initial Volume, V1 = ?

Final Pressure, P2 = 25.43 psi

Final Volume, V2 = 1.895 L

The relationship between these quantities is given by boyles law;

V1P1 = V2P2

V1 = V2P2 / P1 = 1.895 * 25.43 / 15.6

V1 = 3.09 L

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3 years ago
sarah measures out 151 grams of SO2. how many moles is this? Express your answer to three significant figures. ​
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150 g SO₂ = 2.36 mol SO₂


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4 0
3 years ago
Calculate the freezing temperature of the following solution of 0.50 M glucose (a covalent compound). Assume that the molality o
kirza4 [7]

Answer:

-0.93 °C

Explanation:

Hello,

The freezing-point depression is given by:

T_f-T_f^*=-iK_{solvent}m_{solute}

Whereas T_f is the freezing temperature of the solution, T_f^* is the freezing temperature of the pure solvent (0 °C since it is water), i the Van't Hoff factor (1 since the solute is covalent), K_{f,solvent} the solvent's freezing point depression point constant (in this case 1.86 C\frac{kg}{mol}) and m_{solute} the molality of the glucose.

As long as the unknown is T_f, solving for it:

T_f=T_f^*-iK_fm\\T_f=0C-1*1.86C\frac{kg}{mol}*0.5\frac{mol}{kg}  \\T_f=-0.93C

Best regards.

4 0
3 years ago
A 25.0 mL sample of a solution of a monoprotic acid is titrated with a 0.115 M NaOH solution. The end point was obtained at abou
kati45 [8]

Answer:

The concentration of the acid is about 0.114 M (option E)

Explanation:

Step 1: Data given

Volume of the monoprotic acid = 25.0 mL = 0.025 L

Molarity of the monoprotic acid = ?

Molarity of the NaOH solution = 0.115 M

Volume NaOH = 24.8 mL = 0.0248 L

Step 2: Calculate the concentration

a*Cb * Vb = b * Ca * Va

⇒ a = the coeficient of NaOH = 1

⇒ Cb = the molarity of the acid = TO BE DETERMINED

⇒ Vb = the volume of the acid = 0.025 L

⇒ b = the coefficient of the acid = monoprotic = 1

⇒ Ca = the moalrity of NaOH = 0.115 M

⇒ Va = the volume of NaOH = 0.0248 L

1 * Cb * 0.025 = 1 * 0.115 * 0.0248

0.025 Cb = 0.002852

Cb = 0.11408 M

The concentration of the acid is about 0.114 M

5 0
4 years ago
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