Ask question

Ask question

All categories

3 years ago

11

Carbon burns in the presence of oxygen to give carbon dioxide. Which chemical equation describes this reaction? carbon + oxygen

+ carbon dioxide carbon + oxygen → carbon dioxide carbon dioxide → carbon + oxygen carbon dioxide + carbon → oxygen NextReset

2 answers:

zmey [24]3 years ago

8 0

Answer: CO2

Explanation:

frutty [35]3 years ago

6 0

C (solid) + O2 (gas) -> C02 (gas)

You might be interested in

A 50/50 blend of engine coolant and water (by volume) is usually used in an automobile's engine cooling system. If a car's cooli

Diano4ka-milaya [45]

Answer:

$\large \boxed{109.17 \, ^{\circ}\text{C}}$

Explanation:

Data:

50/50 ethylene glycol (EG):water

V = 4.70 gal

ρ(EG) = 1.11 g/mL

ρ(water) = 0.988 g/mL

Calculations:

The formula for the boiling point elevation ΔTb is

$\Delta T_{b} = iK_{b}b$

i is the van’t Hoff factor — the number of moles of particles you get from 1 mol of solute. For EG, i = 1.

1. Moles of EG

$\rm n = 0.50 \times \text{4.70 gal} \times \dfrac{\text{3.785 L}}{\text{1 gal}} \times \dfrac{\text{1000 mL}}{\text{1 L}} \times \dfrac{\text{1.11 g}}{\text{1 mL}} \times \dfrac{\text{1 mol}}{\text{62.07 g}} = \text{159 mol}$

2. Kilograms of water

$m = 0.50 \times \text{4.70 gal} \times \dfrac{\text{3.785 L}}{\text{1 gal}} \times \dfrac{\text{998 g}}{\text{1 L}} \times \dfrac{\text{1 kg}}{\text{1000 g}} = \text{8.88 kg}$

3. Molal concentration of EG

$b = \dfrac{\text{159 mol}}{\text{8.88 kg}} = \text{17.9 mol/kg}$

4. Increase in boiling point

$\rm \Delta T_{b} = iK_{b}b = 1 \times 0.512 \, \, ^{\circ}\text{C} \cdot kg \cdot mol^{-1} \, \times 17.9 \cdot mol \cdot kg^{-1} = 9.17 \, ^{\circ}\text{C}$

5. Boiling point

$\rm T_{b} = T_{b}^{\circ} + \Delta T_{b} = 100.00 \, ^{\circ}\text{C} + 9.17 \, ^{\circ}\text{C} = \mathbf{109.17 \, ^{\circ}C}\\\rm \text{The boiling point of the solution is $\large \boxed{\mathbf{109.17 \, ^{\circ}C}}$}$

7 0

3 years ago

For the reaction 2hbr(g)⇌h2(g)+br2(g), k= 2.00×10−19 at 298 k what can be said about this reaction at this temperature? hints fo

Olin [163]

The equilibrium constant k is actually the ratio of the concentration of the products over the concentration of reactants at equilibrium. So if the concentration of products < concentration of reactants, therefore the constant k will be small. But if the concentration of products > concentration of reactants, the constant k will be large. In this case the value is too small (x10^-19), therefore we can say that the reaction favors the reactant side:

the equilibrium lies far to the left

7 0

3 years ago

Plz 50 points and brainliest but give me the true answer with steps

Ilya [14]

Answer:

i think A...

Explanation:

4 0

2 years ago

Which conclusion could be made from Ernest Rutherford’s gold foil experiment?

Tamiku [17]

The nucleus while large has a small ratio of mass when dealing with atoms, they are very spread out and are mostly solid with a blocky shape.

8 0

3 years ago

Read 2 more answers

If poking a skunk will always result in the emission (release) of toxic fumes

Dahasolnce [82]

Control: skunk
Independent: poking (the skunk)
dependent: emission of toxic fumes

3 0

3 years ago