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vovikov84 [41]
3 years ago
12

_______________ is when a compound containing carbon and hydrogen (and sometimes oxygen) combines with oxygen gas to produce car

bon dioxide and water. *
Fill in the blank:
synthesis
decomposition
combustion
single replacement
double replacement
Chemistry
1 answer:
kotykmax [81]3 years ago
4 0

Answer:

C. Combustion

Explanation:

Combustion- an organic compound containing carbon, hydrogen and sometimes oxygen reacts  with oxygen gas to form carbon dioxide and water.

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Why do the gases on the periodic table tend to form negative ions?
boyakko [2]
Because the valence shell of gases wants to become full
8 0
3 years ago
Hydrogen gas reacts with chlorine gas to form hydrogen chloride as shown in the following reaction: H2 (9) + Cl2 (g) + 2HCl (9)
vivado [14]

Answer:

467

Explanation:

ncl2 = 454.4x1/(71.0 g/mol) = 6.40 mols cl2

6.40 mols cl2 x 2molsHCL/1moleCL2 x 36.5g/1moleHCL = <u>467 g HCL</u>

8 0
3 years ago
Read 2 more answers
Engineers are starting on a new project to develop technology. Which of these
KATRIN_1 [288]

Answer:

D.

Explanation:

Deciding whether the best product has been designed,should be the last step.

3 0
3 years ago
What volume of a 0.155 M potassium hydroxide solution is required to neutralize 25.7 mL of a 0.388 M hydrobromic acid solution
vekshin1

Answer: Therefore, the volume of a 0.155 M potassium hydroxide solution  is 56.0 ml

Explanation:

Molarity of a solution is defined as the number of moles of solute dissolved per Liter of the solution.

According to the neutralization law,

n_1M_1V_1=n_2M_2V_2

where,

M_1 = molarity of HBr solution = 0.338 M

V_1 = volume of HBr solution = 25.7 ml

M_2 = molarity of KOH solution = 0.155 M

V_2 = volume of KOH solution = ?

n_1 = valency of HBr = 1

n_2 = valency of KOH = 1

1\times 0.338\times 25.7=1\times 0.155\times V_2

V_2=56.0ml

Therefore, the volume of a 0.155 M potassium hydroxide solution  is 56.0 ml

8 0
3 years ago
I NEED HELP PLEASE, THANKS! :)
EleoNora [17]

Answer:

\large \boxed{\text{2.20 g Pb}}

Explanation:

They gave us the masses of two reactants and asked us to determine the mass of the product.

This looks like a limiting reactant problem.

1. Assemble the information

We will need a chemical equation with masses and molar masses, so, let's gather all the information in one place.

Mᵣ:       239.27   32.00        207.2

            2PbS   +   3O₂   ⟶  2Pb   +   2SO₃

m/g:      2.54        1.88

2. Calculate the moles of each reactant

\text{Moles of PbS} = \text{2.54 g PbS } \times \dfrac{\text{1 mol PbS}}{\text{239.27 g PbS}} = \text{0.010 62 mol PbS}\\\\\text{Moles of O}_{2} = \text{1.88 g O}_{2} \times \dfrac{\text{1 mol O}_{2}}{\text{32.00 g O}_{2}} = \text{0.058 75 mol O}_{2}

3. Calculate the moles of Pb from each reactant

\textbf{From PbS:}\\\text{Moles of Pb} =  \text{0.010 62 mol PbS} \times \dfrac{\text{2 mol Pb}}{\text{2 mol PbS}} = \text{0.010 62 mol Pb}\\\\\textbf{From O}_{2}:\\\text{Moles of Pb} =\text{0.058 75 mol O}_{2} \times \dfrac{\text{2 mol Pb}}{\text{3 mol O}_{2}}= \text{0.039 17 mol  Pb}\\\\\text{PbS is the $\textbf{limiting reactant}$ because it gives fewer moles of Pb}

4. Calculate the mass of Pb

\text{ Mass of Pb} = \text{0.010 62 mol Pb} \times \dfrac{\text{207.2 g Pb}}{\text{1 mol Pb}} = \textbf{2.20 g Pb}\\\\\text{The reaction produces $\large \boxed{\textbf{2.20 g Pb}}$}

4 0
3 years ago
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