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Leona [35]
3 years ago
6

What coefficients are needed to balance the equation for the complete combustion of methane? Enter the coefficients in the order

ch4
Chemistry
2 answers:
jeka943 years ago
8 0

Answer:

CH4 + 202 ---> 2H20 + CO2

Vaselesa [24]3 years ago
4 0

Answer:

\huge \boxed{\mathrm{CH_4+ 2O_2 \Rightarrow CO_2 +2 H_2O}}

\rule[225]{225}{2}

Explanation:

\sf CH_4+ O_2 \Rightarrow CO_2 + H_2O

Balancing the Hydrogen atoms on the right side,

\sf CH_4+ O_2 \Rightarrow CO_2 +2 H_2O

Balancing the Oxygen atoms on the left side,

\sf CH_4+ 2O_2 \Rightarrow CO_2 +2 H_2O

\rule[225]{225}{2}

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NaOH + H2CO3 → Na2CO3 + H2O<br> Balance Chemical Eguation
IrinaVladis [17]

Answer:

2NaOH + H2CO3 —> Na2CO3 + 2H2O

Explanation:

5 0
4 years ago
Group VIIA onmetals are called hologen?why​
denis-greek [22]

Answer:

<em>The elements of Group VIIA (new Group 17 – fluorine, chlorine, bromine, iodine, and astatine) are called the halogens (tan column). The term “halogen” means “salt-former” because these elements will readily react with alkali metal and alkaline earth metals to form halide salts</em>

4 0
3 years ago
A 1.00 L volume of HCl reacted completely with 2.00 L of 1.50 M Ca(OH)2 according to the balanced chemical equation below. 2HCl
wolverine [178]

Answer:

\boxed{\text{6.00 mol/L}}

Explanation:

(a) Balanced equation

2HCl + Ca(OH)₂ ⟶ CaCl₂ + 2H₂O  

(b) Moles of Ca(OH)₂

\text{Moles of base} = \text{2.00L} \times \dfrac{\text{1.50 mol}}{\text{1 L}} = \text{3.000 mol base}

(c) Moles of HCl

\text{Moles of HCl} = \text{3.000 mol base} \times \dfrac{ \text{2 mol HCl}}{\text{1 mol base}} = \text{6.000 mol HCl}

(d) Molar concentration of HCl

\text{Molar concentration} = \dfrac{\text{moles of solute}}{\text{litres of solution}}\\\\c = \dfrac{ n }{ V}\\\\c= \dfrac{ \text{6.000 mol}}{ \text{1.000 L}} = \text{6.00 mol/L}

The molar concentration of the HCl was \boxed{\textbf{6.00 mol/L}}

6 0
4 years ago
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Natasha_Volkova [10]

Answer:

They both have a cell wall

Explanation:

7 0
3 years ago
What is the density of an object with 525 grams and a volume of 15cm3?
Hunter-Best [27]

Answer: 35 g/cm

Explanation:

Density equals mass over volume. 525 divided by 15 is 35

7 0
3 years ago
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