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4 years ago

What happened to the rate of the reaction when the concentration was increased?

Chemistry

1 answer:

pickupchik [31]4 years ago

4 0

Answer:

If you increase the concentration of a reactant, there will be more of the chemical present. More reactant particles moving together allow more collisions to happen and so the reaction rate is increased. The higher the concentration of reactants, the faster the rate of a reaction will be.

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Which element must be present in an organic compound?(1) hydrogen (3) carbon(2) oxygen (4) nitrogen

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CARBON is the element that must be present in an organic compound.

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3 years ago

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Calculate the mass of water produced when 7.26 g of butane reacts with excess oxygen

MaRussiya [10]

Answer:

11.3 g.

Explanation:

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In this case, since the combustion of butane is:

$C_4H_{10}+\frac{13}{2} O_2\rightarrow 4CO_2+5H_2O$

Thus, since there is a 1:5 mole ratio between butane and water, we obtain the following mass of water:

$m_{H_2O}=7.26gC_4H_{10}*\frac{1molC_4H_{10}}{58.14gC_4H_{10}}*\frac{5molH_2O}{1molC_4H_{10}} *\frac{18.02gH_2O}{1molH_2O}$

Therefore, the resulting mass of water is:

$m_{H_2O}=11.3gH_2O$

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3 years ago

What is the mass of a neutron? 1/2,000 amu 1 amu 2,000 amu 1/200 amu

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3 years ago

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Calculate the pH of a 1.00 L buffer of 0.97 M CH3COONa / 1.02 M CH3COOH before and after the addition of the following species.

ivanzaharov [21]

Answer:

a) 4.73

b) 4.78

c) 4.66 (further addition) or 4.60 (starting from the original buffer solution)

Explanation:

Step 1: Data given

volume of the buffer = 1.00 L

Buffer = 0.97 M CH3COONa / 1.02 M CH3COOH

pKa CH3COOH = 4.75

Step 2: pH = pKa + log [CH3COONa]/[CH3COOH]

pH = 4.75 + log (0.97/1.02)

pH = 4.73

(b) pH after addition of 0.065 mol NaOH

Adding 0.065 mol NaOH will reduce the acid by that amount leaving 1.02 - 0.065 = 0.955 moles HA in 1 L so [HA] = 0.955; the neutralized acid produces A- in the same amount, increasing [A-] to 0.97 +0.065 = 1.035

pH = pKa + log[CH3COONa]/[CH3COOH]

pH = 4.75 + log(1.035/0.955)

pH = 4.78

c) pH after further addition of 0.144 mol HCl

The reverse will happen after the addition of HCl:

[HA] = 0.955 + 0.144 = 1.099

[A-] = 1.035 - 0.144 = 0.891

pH = 4.75 + log(0.891/1.099)

pH = 4.66

If we add 0.144 mol of HCl to the original buffer we will get:

[HA] = 1.02 + 0.144 = 1.164

[A-] = 0.97 - 0.144 = 0.826