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never [62]
3 years ago
7

Is Lilium lancifolium Heterotroph, Autotroph, or Both

Chemistry
1 answer:
Sphinxa [80]3 years ago
4 0
Lilium lancifolium is an autotroph since it is a plant and makes its own food through photosynthesis.
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Thirty grams of lead oxide and fifteen grams of ammonia react completely to produce solid lead, nitrogen gas, and liquid water.
Kitty [74]

45 g  Thirty grams of lead oxide and fifteen grams of ammonia react completely to produce solid lead, nitrogen gas, and liquid water.

8 0
3 years ago
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Two solutions namely, 500 ml of 0.50 m hcl and 500 ml of 0.50 m naoh at the same temperature of 21.6 are mixed in a constant-pre
weeeeeb [17]

24.6 ℃

<h3>Explanation</h3>

Hydrochloric acid and sodium hydroxide reacts by the following equation:

\text{HCl} \; (aq) + \text{NaOH} \; (aq) \to \text{NaCl} \; (aq) + \text{H}_2\text{O} \; (aq)

which is equivalent to

\text{H}^{+} \; (aq) + \text{OH}^{-} \; (aq) \to \text{H}_2\text{O}\; (l)

The question states that the second equation has an enthalpy, or "heat", of neutralization of -56.2 \; \text{kJ}. Thus the combination of every mole of hydrogen ions and hydroxide ions in solution would produce 56.2 \; \text{kJ} or 56.2 \times 10^{3}\; \text{J} of energy.

500 milliliter of a 0.50 mol per liter "M" solution contains 0.25 moles of the solute. There are thus 0.25 moles of hydrogen ions and hydroxide ions in the two 0.500 milliliter solutions, respectively. They would combine to release 0.25 \times 56.2 \times 10^{3} = 1.405 \times 10^{4} \; \text{J} of energy.

Both the solution and the calorimeter absorb energy released in this neutralization reaction. Their temperature change is dependent on the heat capacity <em>C</em> of the two objects, combined.

The question has given the heat capacity of the calorimeter directly.

The heat capacity (the one without mass in the unit) of water is to be calculated from its mass and <em>specific</em> heat.

The calorimeter contains 1.00 liters or 1.00 \times 10^{3} \; \text{ml} of the 1.0 gram per milliliter solution. Accordingly, it would have a mass of 1.00 \times 10^{3} \; \text{g}.

The solution has a specific heat of 4.184 \; \text{J} \cdot \text{g}^{-1} \cdot \text{K}^{-1}. The solution thus have a heat capacity of 4.184 \times 1.00 \times 10^{3} = 4.184 \times 10^{3} \; \text{J} \cdot\text{K}^{-1}. Note that one degree Kelvins K is equivalent to one degree celsius ℃ in temperature change measurements.

The calorimeter-solution system thus has a heat capacity of 4.634 \times 10^{3} \; \text{J} \cdot \text{K}^{-1}, meaning that its temperature would rise by 1 degree celsius on the absorption of 4.634 × 10³ joules of energy. 1.405 \times 10^{4} \; \text{J} are available from the reaction. Thus, the temperature of the system shall have risen by 3.03 degrees celsius to 24.6 degrees celsius by the end of the reaction.

4 0
3 years ago
1 x 10^-3 m is the same as which of the following? Select all that apply.
Yuri [45]

Answer: b and d

Explanation:

1×10^-3=1×1/10³=1×1/1000=1/1000 so b is correct

1×10^-3=1/1000=0.001

so d is correct

8 0
3 years ago
What mass of lithium phosphate would you mass to make 2.5 liter of 1.06 M lithium
alex41 [277]

Answer:

Approximately 3.06 \times 10^{2}\; \rm g (approximately 306\; \rm g.)

Explanation:

Calculate the quantity n of lithium phosphate in V = 2.5\; \rm L of thisc = 1.06\; \rm M = 1.06\; \rm mol \cdot L^{-1} lithium phosphate solution.

\begin{aligned}n &= c \cdot V\\ &= 2.5\; \rm L \times 1.06\; mol \cdot L^{-1}\\ &= 2.65\; \rm mol\end{aligned}.

Empirical formula of lithium phosphate: \rm Li_3PO_4.

Look up the relative atomic mass of \rm Li, \rm P,and \rm O on a modern periodic table:

  • \rm Li: 6.94.
  • \rm P: 30.974.
  • \rm O: 15.999.

Calculate the formula mass of \rm Li_3PO_4:

M(\rm Li_3PO_4) = 3 \times 6.94 + 30.974 + 4 \times 15.999 = 115.79\; \rm g \cdot mol^{-1}.

Calculate the mass of that n = 2.65\; \rm mol of \rm Li_3PO_4 formula units:

\begin{aligned}m &= n \cdot M \\ &= 2.65\; \rm mol \times 115.79\; \rm g\cdot mol^{-1} \\ &\approx 3.06 \times 10^{2}\; \rm g \end{aligned}.

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3 years ago
Please help this chemistry question!<br> ty if u answer
Strike441 [17]
H20) Is the answer Right

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