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ArbitrLikvidat [17]
1 year ago
11

Calculate the number of moles in 8 g of water.

Chemistry
1 answer:
polet [3.4K]1 year ago
4 0

Answer:

refer the above attachment

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Which intermolecular forces increase rapidly with increasing molecular size?
blondinia [14]
LONDON DISPERSION FORCES
8 0
3 years ago
Is water made of plant cells or animal cells<br> •plant<br> •animals<br> •niether
Pepsi [2]

Answer:

i think niether

Explanation:

5 0
2 years ago
Which of the following liquids would have the highest viscosity at room temperature?
Liono4ka [1.6K]

Answer:

5)HOCH2CH2OH

Explanation:

This is also known as ethylene glycol. An increase in hydrogen bonds of a compound means an increase in the viscosity. Hydrogen bonds occur as a result of bonding with electronegative elements such as Oxygen, Nitrogen etc.

The compounds with the highest amount of Hydrogen bond represents the one with the highest viscosity which is B) HOCH2CH2OH

3 0
3 years ago
Bob measured out 1.60 grams of sodium. He calculates that 1.60 g of
saul85 [17]

Answer:

84.8%

Explanation:

Step 1: Given data

Bob measured out 1.60 g of Na. He forms NaCl according to the following equation.

Na + 1/2 Cl₂ ⇒ NaCl

According to this equation, he calculates that 1.60 g of sodium should produce 4.07 g of NaCl, which is the theoretical yield. However, he carries out the experiment and only makes 3.45 g of NaCl, which is the real yield.

Step 2: Calculate the percent yield.

We will use the following expression.

%yield = real yield / theoretical yield × 100%

%yield = 3.45 g / 4.07 g × 100% = 84.8%

6 0
3 years ago
Someone answer the limited regent...
OleMash [197]

Answer:

20 g Ag

General Formulas and Concepts:

<u>Chemistry - Stoichiometry</u>

  • Using Dimensional Analysis

<u>Chemistry - Atomic Structure</u>

  • Reading a Periodic Table

Explanation:

<u>Step 1: Define</u>

[RxN]   Cu (s) + AgNO₃ (aq) → CuNO₃ (aq) + Ag (s)

[Given]   10 g Cu

<u>Step 2: Identify Conversions</u>

[RxN]   1 mol Cu = 1 mol Ag

Molar Mass of Cu - 63.55 g/mol

Molar Mass of Ag - 197.87 g/mol

<u>Step 3: Stoichiometry</u>

<u />10 \ g \ Cu(\frac{1 \ mol \ Cu}{63.55 \ g \ Cu})(\frac{1 \ mol \ Ag}{1 \ mol \ Cu} )(\frac{197.87 \ g \ Ag}{1 \ mol \ Ag} ) = 16.974 g Ag

<u>Step 4: Check</u>

<em>We are given 1 sig fig. Follow sig fig rules and round.</em>

16.974 g Ag ≈ 20 g Ag

6 0
3 years ago
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