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2 years ago

Using g= 10 m/s?, find the weight of a 3 kg mass.

Chemistry

1 answer:

Wewaii [24]2 years ago

3 0

Answer:

30 N

Explanation:

weight = mass * g

w = mg

w 3*10

w = 30

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When sugar is dissolved in hot tea, physical changes occur. Write two physical properties of sugar

kompoz [17]

Answer:

When you mix the sugar into the tea and stir, it dissolves so you can't see it. Also when you stir the sugar into the tea the taste changes and it turns sweeter.

6 0

2 years ago

CH4 + 202 → CO2 + 2H2O<br> How many grams of O2 needed to produce 36 grams of H2O?

Kipish [7]

<h3>Answer:</h3>

64 g O₂

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

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

<u>Chemistry</u>

<u>Atomic Structure</u>

Reading a Periodic Table

<u>Stoichiometry</u>

Using Dimensional Analysis

<h3>Explanation:</h3>

<u>Step 1: Define</u>

[RxN - Balanced] CH₄ + 2O₂ → CO₂ + 2H₂O

[Given] 36 g H₂O

[Solve] x g O₂

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

[RxN] 2 mol O₂ → 2 mol H₂O

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

[PT] Molar Mas of H - 1.01 g/mol

Molar Mass of O₂ - 2(16.00) = 32.00 g/mol

Molar Mass of H₂O - 2(1.01) + 16.00 = 18.02 g/mol

<u>Step 3: Stoichiometry</u>

Set up conversion: $\displaystyle 36 \ g \ H_2O(\frac{1 \ mol \ H_2O}{18.02 \ g \ H_2O})(\frac{2 \ mol \ O_2}{2 \ mol \ H_2O})(\frac{32.00 \ g \ O_2}{1 \ mol \ O_2})$
Divide/Multiply [Cancel Units]: $\displaystyle 63.929 \ g \ O_2$

<u>Step 4: Check</u>

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

63.929 g O₂ ≈ 64 g O₂

8 0

3 years ago

Read 2 more answers

How did you get the 3.2?

Helga [31]

I don't understand what you mean??

8 0

2 years ago

During the daytime, wind blows off of the __________________ because that material takes ___________ time to get warm, creating

Butoxors [25]

Answer: sea, more

I hope this helps

5 0

2 years ago

explain why it is a common laboratory procedure to heat analytical reagents and store them in a dessicated atmosphere (a sealed

Readme [11.4K]

Explanation:

Most reagent forms are going to absorb water from the air; they're called "hygroscopic". Water presence can have a drastic impact on the experiment being performed For fact, it increases the reagent's molecular weight, meaning that anything involving a very specific molarity (the amount of molecules in the final solution) will not function properly.

Heating will help to eliminate water, although some chemicals don't react well to heat, so it shouldn't be used for all. A dessicated environment is simply a means to "dry." That allows the reagent with little water in the air to attach with.

6 0

2 years ago