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katen-ka-za [31]
4 years ago
10

In the reaction between Li and O, there is a transfer of electrons making an ionic bond. In the bond, lithium would be a (n) bec

ause it an electron, and oxygen would be an) because it electrons. O A anion, loses, cation, gains B cation, gains, anion, loses Oc cation, loses, anion, gains O D anion, gains, cation, loses INTL 5​
Chemistry
1 answer:
galben [10]4 years ago
6 0

Answer:

C) cation, loses, anion, gains

Explanation:

Lithium is a metal from Group 1, so it has 1 valence electron. Thus, it loses 1 electron to complete its octet and form the cation Li⁺.

Oxygen is a nonmetal from Group 16, so it has 6 valence electrons. Thus, it gains 2 electrons to complete its octet and form the anion O²⁻.

In the reaction between Li and O, there is a transfer of electrons making an ionic bond. In the bond, lithium would be a cation because it loses an electron, and oxygen would be an anion because it gains 2 electrons.

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Which of the following statements is true?
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I believe the answer would be D.
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A closed container has 5.60 x 1023 molecules of oxygen gas (O2). How many moles is that?
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3 years ago
Perform the following for Part C of this lab:
kaheart [24]

Answer:

a. 0.0110 L

b. 0.0020 L

c. 0.011 mol

d. 5.5 M

e. 0.66 g

f. 33%

Explanation:

There is some info missing. I will use some values to show you the procedure and then you can replace them with your values.

<em>Titrant (NaOH) concentration: 1.0 M</em>

<em>Vinegar volume: 2.0 mL</em>

<em>Initial buret reading (initial NaOH volume): 0.1 mL</em>

<em>Final buret reading (final NaOH volume): 11.1 mL</em>

<em>a. Calculate the volume of NaOH that was added to the vinegar. Convert this volume to liters. Show your work.</em>

The volume of NaOH is the difference between the final and the initial buret reading.

11.1 mL - 0.1 mL = 11.0 mL × (1 L/1000 mL) = 0.0110 L

<em>b. Convert the measured volume of vinegar to liters. Show your work.</em>

2.0 mL × (1 L/1000 mL) = 0.0020 L

<em>c. Calculate the moles of NaOH using the volume and molarity of NaOH. Show your work. moles = molarity x volume</em>

moles = molarity × volume

moles = (1.0 mol/L) × 0.0110 L = 0.011 mol

<em>d. Since the reaction ratio is 1:1, the moles of acetic acid in the vinegar is equal to the moles of NaOH reacted during the titration. Calculate the molarity of the acetic acid in the vinegar. Show your work. molarity = moles / volume</em>

molarity = moles / volume

molarity = 0.011 mol/0.0020 L = 5.5 M

<em>e. Calculate the grams of acetic acid in the vinegar. Show your work. mass = moles x molar mass (g/mol)</em>

mass = moles × molar mass

mass = 0.011 mol × 60.05 g/mol = 0.66 g

<em>f. Assuming that the density of vinegar is very close to 1.0 g/mL, the 2.0 mL sample of vinegar used in the titration should weigh 2.0  g. Use this to calculate the mass % of acetic acid in the vinegar sample. mass % = (mass acetic acid / mass vinegar) * 100%</em>

mass % = (mass acetic acid / mass vinegar) * 100%

mass % = (0.66 g /2.0 g) * 100% = 33%

6 0
3 years ago
John dissolves .5g of a white powder in 25g of benzene (FP 5oC) (kf benzene is 5.1) and finds the solution freezes at 3.7oC. Det
navik [9.2K]

Answer:

The compound has a molar mass of 78.4 g/mol

Explanation:

Step 1: data given

Mass of a sample = 0.5 grams

Mass of benzene = 25 grams

Freezing poing = 5 °C

Kf of benzene = 5.1 °C/m

Freezing point solution = 3.7 °C

Step 2: Calculate molality

ΔT = i*Kf*m

⇒with ΔT = the freezing point depression = 5.0 - 3.7 = 1.3 °C

⇒with i = the can't hoff factor = 1

⇒with Kf = the freezing point depression constant of benzene = 5.1 °C/m

⇒with m = the molality

1.3 = 5.1 * m

m = 1.3 / 5.1

m = 0.255 moles /kg

Step 3: Calculate moles

Molality = moles / mass benzene

0.255 molal = moles / 0.025 kg

Moles = 0.255 molal * 0.025 kg

Moles = 0.006375 moles

Step 4: Calculate molar mass of the compound

Molar mass compund = mass / moles

Molar mass compound = 0.5 grams / 0.006375 moles

Molar mass compound = 78.4 g/mol

The compound has a molar mass of 78.4 g/mol

7 0
3 years ago
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