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Irina18 [472]
3 years ago
7

NH3 is nitrogen trihydride (ammonia) and BF3 is boron trifluoride. What is the name for AlCl3 and why?

Chemistry
1 answer:
Korvikt [17]3 years ago
5 0
The answer is A. Aluminum Chloride it does not have the prefix "tri-" since it is ionic.
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10. Lead nitrate solution mixed with sodium sulfate solution forms lead sulfate as a
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Answer:

88%.

Explanation:

The percentage yield of lead sulfate in this experiment is 88% if 2.53 is divided by 2.85 and multiply by 100. The percentage yield can be calculated when the experimental yield is divided by theoretical yield and then multiply by 100. The percentage yield tells us about the actual yield that is gained in the end of experiment which is lower than theoretical yield.

8 0
3 years ago
How many independent variables can be tested in a standard scientific experiment
Shalnov [3]

One independent variable

6 0
3 years ago
There are 0.3 moles of xenon gas in a 0.5-liter container at 30 degrees C. What is the pressure exerted by the xenon gas
Kaylis [27]
PV = nRT
P = (nRT)/V
P = (0.3 mol × 0.08206 atm-l/(mol-K) × (273.15 + 30) K)/(0.5 l)
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3 0
2 years ago
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Use the balanced equation given below to solve the following problem; Calculate the volume in liters of CO produced by the react
Elan Coil [88]

Answer: 40.3 L

Explanation:

To calculate the moles :

\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}

\text{Moles of} Sb_2O_3=\frac{175g}{291.5g/mol}=0.600moles

Sb_2O_3+3C\rightarrow 2Sb+3CO  

According to stoichiometry :

1 moles of Sb_2O_3 produces = 3 moles of CO

Thus 0.600 moles of Sb_2O_3 will produce=\frac{3}{1}\times 0.600=1.80moles  of CO

Volume of CO=moles\times {\text {Molar volume}}=1.80moles\times 22.4L/mol=40.3L

Thus 40.3 L of CO is produced.

6 0
2 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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