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3 years ago

When sugar is burned, water vapor and carbon dioxide are produced.

Chemistry

1 answer:

Natalija [7]3 years ago

4 0

Answer:

12 carbon, 22 hydrogen, 11 oxygen

Explanation:

Our equation is: Sugar + $12O_2$ ⇒ $11H_2O$ + $12CO_2$

This is already balanced which means that the number of atoms for each type of element on each side should be the same.

On the right side, there are 11 * 2 = 22 hydrogen atoms, 11 * 1 + 12 * 2 = 11 + 24 = 35 oxygen atoms, and 12 * 1 = 12 carbon atoms.

Currently on the left, there are 12 * 2 = 24 oxygen atoms. That means that sugar must have 11 oxygen atoms to balance it out. In addition, since the left side currently has no hydrogen or carbon atoms, those must belong to the sugar molecule.

Thus, the sugar molecule has 12 carbon atoms, 22 hydrogen atoms, and 11 oxygen atoms.

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Which of the following steps will slow down the reaction between coal and oxygen?

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During laparoscopic surgery , carbon dioxide gas is used to expand the abdomen to help create a larger working space. If 4.80 L

Studentka2010 [4]

Answer:

5.37 L

Explanation:

To solve this problem we need to use the PV=nRT equation.

First we calculate the amount of CO₂, using the initial given conditions for P, V and T:

P = 785 mmHg ⇒ 785/760 = 1.03 atm

V = 4.80 L

T = 18 °C ⇒ 18 + 273.16 = 291.16 K

1.03 atm * 4.80 L = n * 0.082 atm·L·mol⁻¹·K⁻¹ * 291.16 K

We solve for n:

n = 0.207 mol

Then we use that value of n for another PV=nRT equation, where T=37 °C (310.16K) and P = 745 mmHg (0.98 atm).

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And we solve for V:

V = 5.37 L

7 0

3 years ago

Which solution has the same boiling point as 0.25 mol CaCl2 dissolved in 1000 g water?

Rufina [12.5K]

The boiling point of water at 1 atm is 100 degrees celsius. However, when water is added with another substance the boiling point of it rises than when it is still a pure solvent. This called boiling point elevation, a colligative property. The equation for the boiling point elevation is expressed as the product of the ebullioscopic constant (0.52 degrees celsius / m) for water), the vant hoff factor and the concentration of solute (in terms of molality).

ΔT(CaCl2) = i x K x m = 3 x 0.52 x 0.25 = 0.39 °C
 ΔT(Sucrose) = 1 x 0.52 x 0.75 = 0.39 °C
 ΔT(Ethylene glycol) = 1 x 0.52 x 1 = 0.52 °C
 ΔT(CaCl2) = 3 x 0.52 x 0.50 = 0.78 °C
 ΔT(NaCl) = 2 x 0.52 x 0.25 = 0.26 °C

Thus, from the calculated values, we see that 0.75 mol sucrose dissolved on 1 kg water has the same boiling point with 0.25 mol CaCl2 dissolved in 1 kg water.

5 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 this $c = 1.06\; \rm M = 1.06\; \rm mol \cdot L^{-1}$ 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