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1 year ago

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Chemistry

2 answers:

gtnhenbr [62]1 year ago

5 0

Answer:

1001 g and 120.0 mL

Explanation:

Zeros in-between natural numbers are significant

If your number has a decimal place, and you have zeroes after a natural number, then all the zeros are sig figs.

Ex: 10000.0 = 6 sig fig

Not: 0.00001 = 1 sig fig

Not: 0012.0 = 3 sig fig

inysia [295]1 year ago

3 0

Answer:

1001

Explanation:

Zeroes between non zero digits are significant

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How many grams of CO2 are produced from 6.7 L of O2 gas at STP?

Tasya [4]

<h3>Answer:</h3>

13 g CO₂

<h3>General Formulas and Concepts:</h3>

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<u>Atomic Structure</u>

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STP (Standard Conditions for Temperature and Pressure) = 22.4 L per mole at 1 atm, 273 K

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<h3>Explanation:</h3>

<u>Step 1: Define</u>

<em>Identify variables</em>

[Given] 6.7 L O₂

[Solve] g O₂

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

[STP] 22.4 L = 1 mol

[PT] Molar Mass of O: 16.00 g/mol

[PT] Molar Mass of C: 12.01 g/mol

Molar Mass of CO₂: 12.01 + 2(16.00) = 44.01 g/mol

<u>Step 3: Convert</u>

[DA] Set up: $\displaystyle 6.7 \ L \ O_2(\frac{1 \ mol \ O_2}{22.4 \ L \ O_2})(\frac{44.01 \ g \ O_2}{1 \ mol \ O_2})$
[DA] Divide/Multiply [Cancel out units]: $\displaystyle 13.1637 \ g \ O_2$

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 2 sig figs.</em>

13.1637 g CO₂ ≈ 13 g CO₂

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3 years ago

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Julli [10]

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3 years ago

7. What is the molarity of the nitrate ion that is found in a solution made by dissolving 6.25g

tekilochka [14]

Answer:

0.271 M NO₃⁻

Explanation:

To find the molarity of the nitrate ion (NO₃⁻), you need to (1) convert grams to moles (via molar mass), then (2) convert moles Al(NO₃)₃ to moles NO₃⁻, then (3) convert mL to L, and then (4) calculate the molarity. When (Al(NO₃)₃) dissolves in water, it dissociates into 3 nitrate ions. The final answer should have 3 sig figs.

(Steps 1 + 2)

Molar Mass (Al(NO₃)₃): 26.982 g/mol + 3(14.007 g/mol) + 9(15.998 g/mol)

Molar Mass (Al(NO₃)₃): 212.985 g/mol

1 Al(NO₃)₃ = 1 Al³⁺ and 3 NO₃⁻

6.25 g Al(NO₃)₃ 1 mole 3 moles NO₃⁻
------------------------- x ----------------- x ----------------------- = 0.0880 moles NO₃⁻
212.985 g 1 mole Al(NO₃)₃

(Steps 3 + 4)

325.0 mL / 1,000 = 0.3250 L

Molarity = moles / volume

Molarity = 0.0880 moles / 0.3250 L

Molarity = 0.271 M

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1 year ago