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tamaranim1 [39]
3 years ago
7

An arrhenius base is a compound consisting of hydroxide ions and _____.

Chemistry
1 answer:
Bad White [126]3 years ago
4 0
Acids are hydrogen-containing compound that ionize to yield hydrogen ions (H⁺) in an aqueous solution
-bases are compound that ionize to yield hydroxide ions (OH⁻) in an aqueous solution
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Which of the following would most likely result in an increase in reaction rate? placing the reactants on a hotplate
puteri [66]

Answer:

placing the reactants on a hot plate

Explanation:

If the temperature goes up, the reaction rate will increase. Because the particle will move faster and makes the kinetic energy larger.

4 0
3 years ago
In the alkane series of hydrocarbons, as the number of carbon atoms decreases, the normal boiling point ofthe compounds
noname [10]
<span>The relationship between the number of carbon atoms and boiling point is inversely proportional. In the alkane series of hydrocarbons, as the number of carbon atoms decreases, the normal boiling point of the compounds decreases. The reason behind this is that longer chains of molecules require more energy to separate the bonds while shorter chains or molecules with lower number of carbon atoms require less energy to break away from each other. Thus, low carbon molecules have lower boiling point.</span>
3 0
3 years ago
In a titration experiment 12.5 ml of 0.500 m h2so4 neutralized 50.0 ml of naoh. the concentration of the naoh solution is ____.
s344n2d4d5 [400]

0.250 mol/L

<em>Step 1</em>. Write the chemical equation

H2SO4 + 2NaOH → Na2SO4 + 2H2O

<em>Step 2</em>. Calculate the moles of H2SO4

Moles of H2SO4 = 12.5 mL H2SO4 × (0.500 mmol H2SO4/1 mL H2SO4)

= 6.25 mmol H2SO4

<em>Step 3</em>. Calculate the moles of NaOH

Moles of NaOH = 6.25 mmol H2SO4 × (2 mmol NaOH/(1 mmol H2SO4)

= 12.5 mmol NaOH

<em>Step 4</em>. Calculate the concentration of the NaOH

[NaOH] = moles/litres = 12.5 mmol/50.0 mL = 0.250 mol/L

4 0
3 years ago
How many moles are in 7.46 x 1025 particles of iron
sladkih [1.3K]

Answer:

1 mole of iron =6.023×10^23 particles

1 particles of iron=1/6.023×10^23 mole

7.46×10^25 particles =1/6.023×10^23×7.46×10^25

=1.238×10^48 mole is a required answer.

4 0
2 years ago
CO(g)+2H2(g)⇌CH3OH(g)CO(g)+2H2(g)⇌CH3OH(g) This reaction is carried out at a different temperature with initial concentrations o
Katarina [22]

Answer:

9.4

Explanation:

The equation for the reaction can be represented as:

CO_{(g)}    +      2H_2O_{(g)}   ⇄  CH_3OH_{(g)}

The ICE table can be represented as:

                                  CO_{(g)}    +      2H_2_{(g)}   ⇄  CH_3OH_{(g)}

Initial                          0.27             0.49              0.0

Change                      -x                  -2x                 x

Equilibrium               0.27 - x         0.49 -2x          x

We can now say that the concentration of  CH_3OH_{(g)} at equilibrium is x;

Let's not forget that at equilibrium  CH_3OH_{(g)} = 0.11 M

So:

x =  [CH_3OH_{(g)}] = 0.11 M

[CO_{(g)}] = 0.27 - x

[CO_{(g)}] = 0.27 - 0.11

[CO_{(g)}] = 0.16 M

[2H_2_{(g)}] = (0.49 - 2x)

[2H_2_{(g)}] = (0.49 - 2(0.11))

[2H_2_{(g)}] = 0.49 - 0.22

[2H_2_{(g)}] = 0.27 M

K_C = \frac{[CH_3OH]}{[CO][H_2]^2}

K_C = \frac{(0.11)}{(0.16)[(0.27)^2}

K_C = 9.4307

K_C = 9.4

∴ The equilibrium constant at that temperature = 9.4

8 0
2 years ago
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