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katrin2010 [14]
3 years ago
14

Can someone help me with this question ASAP I’ll give 30 point

Chemistry
1 answer:
Brut [27]3 years ago
8 0

Answer:

B

Explanation:

Because B is stable square both sides are equal

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About how many harvests of bamboo can be collected during the time it takes to fully grow one pine tree?
Tomtit [17]

Answer:

bamboo can grow 910 mm (36 in) within a 24-hour period,at a rate of almost 40 mm (1 1⁄2 in) an hour (a growth around 1 mm every 90 seconds, or 1 inch {2.54 centimeters} every 40 minutes).

Explanation:

6 0
3 years ago
Why does a carbonated soft drink lose carbonation when the container is left open?
DENIUS [597]

Pressure decreases when you open the top. Pressure decreases, which favors the formation of H2CO3.

I better get your number.

5 0
3 years ago
How many moles of methane are in 7.31*10^25 molecules?
GrogVix [38]
<h3>Answer:</h3>

121 mol CH₄

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

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

  1. Brackets
  2. Parenthesis
  3. Exponents
  4. Multiplication
  5. Division
  6. Addition
  7. Subtraction
  • Left to Right

<u>Chemistry</u>

<u>Organic</u>

  • Writing chemical compounds
  • Writing organic structures
  • Prefixes
  • Alkanes, Alkenes, Alkynes

<u>Atomic Structure</u>

  • Using Dimensional Analysis
  • Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.
<h3>Explanation:</h3>

<u>Step 1: Define</u>

7.31 × 10²⁵ molecules CH₄

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

Avogadro's Number

<u>Step 3: Convert</u>

<u />\displaystyle 7.31 \cdot 10^{25} \ molecules \ CH_4(\frac{1 \ mol \ CH_4}{6.022 \cdot 10^{23} \ molecules \ CH_4} ) = 121.388 \ mol \ CH_4<u />

<u />

<u>Step 4: Check</u>

<em>Follow sig fig rules and round. We are given 3 sig figs.</em>

121.388 mol CH₄ ≈ 121 mol CH₄

7 0
3 years ago
Which substance is not a solid at 20˚C and one atmosphere of pressure?
levacccp [35]
Choose A, Kr is noble gas
3 0
2 years ago
How many kilojoules of energy would be required to heat a 225g block of aluminum from 23.0 C to 73.5 C?
gulaghasi [49]

Answer:

\boxed {\boxed {\sf 10.2 \ kJ}}

Explanation:

We are asked to find how many kilojoules of energy would be required to heat a block of aluminum.

We will use the following formula to calculate heat energy.

q=mc \Delta T

The mass (m) of the aluminum block is 225 grams and the specific heat (c) is 0.897 Joules per gram degree Celsius. The change in temperature (ΔT) is the difference between the final temperature and the initial temperature.

  • ΔT = final temperature - inital temperature

The aluminum block was heated from 23.0 °C to 73.5 °C.

  • ΔT= 73.5 °C - 23.0 °C = 50.5 °C

Now we know all three variables and can substitute them into the formula.

  • m= 225 g
  • c= 0.897 J/g° C
  • ΔT= 50.5 °C

q= (225 \ g )(0.897 \ J/g \textdegree C)(50.5 \textdegree C)

Multiply the first two numbers. The units of grams cancel.

q= (225 \ g  * 0.897 \ J/g \textdegree C)(50.5 \textdegree C)

q= (225   * 0.897 \ J / \textdegree C)(50.5 \textdegree C)

q= (201.825\ J / \textdegree C)(50.5 \textdegree C)

Multiply again. This time, the units of degrees Celsius cancel.

q= 201.825 \ J * 50.5

q= 10192.1625 \ J

The answer asks for the energy in kilojoules, so we must convert our answer. Remember that 1 kilojoule contains 1000 joules.

\frac { 1  \ kJ}{ 1000 \ J}

Multiply by the answer we found in Joules.

10192.1625 \ J * \frac{ 1 \ kJ}{ 1000 \ J}

10192.1625  * \frac{ 1 \ kJ}{ 1000 }

\frac {10192. 1625}{1000} \ kJ

10.1921625 \ kJ

The original values of mass, temperature, and specific heat all have 3 significant figures, so our answer must have the same. For the number we found, that is the tneths place. The 9 in the hundredth place tells us to round the 1 up to a 2.

10.2 \ kJ

Approximately <u>10.2 kilojoules</u> of energy would be required.

3 0
3 years ago
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