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Juliette [100K]

3 years ago

12

Hen the space shuttle blasts into orbit, a hot gas forms when hydrogen and oxygen join to make water. this helps the space shutt

le to blast into orbit. the formation of this gas is an example of

1 answer:

sukhopar [10]3 years ago

3 0

Answer: compound

Explanation:

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Calculate the mass in grams in nine molecules of CH3COOH? Please show how you got your answer.

WINSTONCH [101]

M CH₃COOH: 12u×2 + 1u×4 + 16u×2 =<u> 60u</u>

m 9CH₃COOH: 60u×9 = <u>540u</u>

<em>(1u ≈ 1,66·10⁻²⁴g)</em>
-----------------------------
1u ------- <span>1,66·10⁻²⁴g
540u ---- X
X = 540</span>×<span>1,66·10⁻²⁴g
<u>X = 896,4</u></span><span><u>·10⁻²⁴g

</u></span>

7 0

3 years ago

Choose the number of significant figures indicated. 90

kvv77 [185]

There are 2 significant figures. All numbers in a whole number are significant.

3 0

3 years ago

Read 2 more answers

20cm of 0.09M solution of H2SO4. requires 30cm of NaOH for complete neutralization. Calculate the

kirill115 [55]

Answer:

Choice A: approximately $0.12\; \rm M$ .

Explanation:

Note that the unit of concentration, $\rm M$ , typically refers to moles per liter (that is: $1\; \rm M = 1\; \rm mol\cdot L^{-1}$ .)

On the other hand, the volume of the two solutions in this question are apparently given in $\rm cm^3$ , which is the same as $\rm mL$ (that is: $1\; \rm cm^{3} = 1\; \rm mL$ .) Convert the unit of volume to liters:

$V(\mathrm{H_2SO_4}) = 20\; \rm cm^{3} = 20 \times 10^{-3}\; \rm L = 0.02\; \rm L$ .
$V(\mathrm{NaOH}) = 30\; \rm cm^{3} = 30 \times 10^{-3}\; \rm L = 0.03\; \rm L$ .

Calculate the number of moles of $\rm H_2SO_4$ formula units in that $0.02\; \rm L$ of the $0.09\; \rm M$ solution:

$\begin{aligned}n(\mathrm{H_2SO_4}) &= c(\mathrm{H_2SO_4}) \cdot V(\mathrm{H_2SO_4})\\ &= 0.02 \; \rm L \times 0.09 \; \rm mol\cdot L^{-1} = 0.0018\; \rm mol \end{aligned}$ .

Note that $\rm H_2SO_4$ (sulfuric acid) is a diprotic acid. When one mole of $\rm H_2SO_4$ completely dissolves in water, two moles of $\rm H^{+}$ ions will be released.

On the other hand, $\rm NaOH$ (sodium hydroxide) is a monoprotic base. When one mole of $\rm NaOH$ formula units completely dissolve in water, only one mole of $\rm OH^{-}$ ions will be released.

$\rm H^{+}$ ions and $\rm OH^{-}$ ions neutralize each other at a one-to-one ratio. Therefore, when one mole of the diprotic acid $\rm H_2SO_4$ dissolves in water completely, it will take two moles of $\rm OH^{-}$ to neutralize that two moles of $\rm H^{+}$ produced. On the other hand, two moles formula units of the monoprotic base $\rm NaOH$ will be required to produce that two moles of $\rm OH^{-}$ . Therefore, $\rm NaOH$ and $\rm H_2SO_4$ formula units would neutralize each other at a two-to-one ratio.

$\rm H_2SO_4 + 2\; NaOH \to Na_2SO_4 + 2\; H_2O$ .

$\displaystyle \frac{n(\mathrm{NaOH})}{n(\mathrm{H_2SO_4})} = \frac{2}{1} = 2$ .

Previous calculations show that $0.0018\; \rm mol$ of $\rm H_2SO_4$ was produced. Calculate the number of moles of $\rm NaOH$ formula units required to neutralize that

$\begin{aligned}n(\mathrm{NaOH}) &= \frac{n(\mathrm{NaOH})}{n(\mathrm{H_2SO_4})}\cdot n(\mathrm{H_2SO_4}) \\&= 2 \times 0.0018\; \rm mol = 0.0036\; \rm mol\end{aligned}$ .

Calculate the concentration of a $0.03\; \rm L$ solution that contains exactly $0.0036\; \rm mol$ of $\rm NaOH$ formula units:

$\begin{aligned}c(\mathrm{NaOH}) &= \frac{n(\mathrm{NaOH})}{V(\mathrm{NaOH})} = \frac{0.0036\; \rm mol}{0.03\; \rm L} = 0.12\; \rm mol \cdot L^{-1}\end{aligned}$ .

3 0

3 years ago

C6H12O6 + 6 O2 --> 6 CO2 + 6 H2O + energy In the reaction listed above, 1 molecule of glucose reacts with 6 molecules of oxyg

kirill115 [55]

Answer:

Third choice:<em> energy present in the glucose and oxygen that is not needed for the formation of carbon dioxide and water is released to form energy/ATP.</em>

Explanation:

<u>1) Chemical equation (given):</u>

C₆H₁₂O₆ + 6 O₂ --> 6 CO₂ + 6 H₂O + energy

<u>2) Chemical potential energy:</u>

Each compound stores chemical potential energy. This energy is stored in the chemical bonds.

Due to every substance has its own unique chemical potential energy, when a chemical reaction takes plase, yielding to the change of some substances, some energy is absorbed (when bonds are formed) and some energy is released (when bonds are broken).

<u>3) Conservation of energy:</u>

Then, if the sum of the bond energies of the final products is less than the sum of the bond energies of the reactants, the<em> law of conservation of energy</em> rules that the difference between the total energies of the products and reactants must be released to the surroundings.

That is what is happening in the given reaction:

C₆H₁₂O₆ + 6 O₂ --> 6 CO₂ + 6 H₂O + energy

The term energy in the product side means that energy is conserved because it is being released due to the the glucose and oxygen (reactant side) have more energy stored in their bonds than the energy needed for the formation of carbon dioxide and water, so that excess of energy is released to form energy/ATP.

<u>Summarizing:</u>

The energy on the product side added to the energy of carbon dioxide and water equals the energy of the glucose and oxygen and the final balance is:

∑ Energy of the reactants = ∑energy of the products + released energy, supporting the law of conservation of energy.

5 0

4 years ago

If carbon-15 has a half-life of 2.5 seconds, % of the sample would still be

kramer

Answer:

After 5 second 25% C-15 will remain.

Explanation:

Given data:

Half life of C-15 = 2.5 sec

Original amount = 100%

Sample remain after 5 sec = ?

Solution:

Number of half lives = T elapsed / half life

Number of half lives = 5 sec / 2.5 sec

Number of half lives = 2

At time zero = 100%

At first half life = 100%/2 = 50%

At second half life = 50%/2 = 25%

Thus after 5 second 25% C-15 will remain.

8 0

4 years ago