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aleksley [76]
3 years ago
5

If a buffer solution is 0.130 M in a weak acid (K_a = 1.7 x 10^-5) and 0.590 M in its conjugate base, what is the pH?

Chemistry
1 answer:
Serggg [28]3 years ago
8 0
Use the Henderson-Hasselbach equation:
pH = pKa + log[base]/[acid]
pH = -log(1.7 x 10^-5) + log(0.590/0.130) = 5.43
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What do the superscripts and subscripts in the notation ^40 K represent?
raketka [301]

Explanation:

subscript is K

superscript is ^

subscript K means a unit of temperature like F or C

superscript ^ means to the power of

So all together it means to the power of 40 K

3 0
4 years ago
A student wanted to investigate the effect of light on the growth of
timofeeve [1]

Answer:

D.phototropism

Explanation:

Phototropism is a type of tropism in which a plant or plant part responds to light. According to this question, a student wanted to investigate the effect of light on the growth of cress seedlings. The student used three different pots for the experiment.

Pot 1 was placed with light from above. Pot 2 was placed in a cupboard with no light. Pot 3 was placed in a window with light from one direction only. However, the image attached to this question shows that the plants in the different pots face different directions in response to light, which depicts phototropism

5 0
3 years ago
Whats the answer?? Please help
Katarina [22]

Answer:

.

Explanation:

4 0
3 years ago
How many moles of aspartame are present in 1.00 mg of aspartame?
Diano4ka-milaya [45]

Answer:- There are 3.40*10^-^6 moles.

Solution:- It is a unit conversion problem where we are asked to convert mg of aspartame to moles. Aspartame is C_1_4H_1_8N_2O_5 and it's molar mass is 294.31 grams per mole.

mg are converted to grams and then the grams are converted to moles as:

1.00mg Aspartame(\frac{1g}{1000mg})(\frac{1mole}{294.31g})

= 3.40*10^-^6 moles of aspartame

So, there would be 3.40*10^-^6 moles of aspartame in 1.00 mg of it.

3 0
4 years ago
1.12g H2 is allowed to react with 9.60 g N2, producing 1.23 g NH3.
andriy [413]

Answer:

A. m_{NH_3}^{theo} =1.50gNH_3

B. Y=82.2\%

Explanation:

Hello!

In this case, since the undergoing chemical reaction between nitrogen and hydrogen is:

N_2+3H_2\rightarrow 2NH_3

Thus we proceed as follows:

A. Here, we first need to compute the moles of ammonia yielded by each reactant, in order to identify the limiting one:

n_{NH_3}^{by \ H_2}=1.12gH_2*\frac{1molH_2}{2.02gH_2}*\frac{2molNH_3}{3molH_2}=0.370molNH_3\\\\  n_{NH_3}^{by \ N_2}=1.23gN_2*\frac{1molN_2}{28.02gN_2}*\frac{2molNH_3}{1molN_2}=0.0878molNH_3

Thus, since nitrogen yields the fewest moles of ammonia, we realize it is the limiting reactant, so the theoretical yield, in grams, of ammonia is:

m_{NH_3}^{theo}=0.0878mol*\frac{17.04gNH_3}{1molNH_3} =1.50gNH_3

B. Finally, since the actual yield of ammonia is 1.23, the percent yield turns out:

Y=\frac{1.23gNH_3}{1.50gNH_3} *100\%\\\\Y=82.2\%

Best regards!

5 0
3 years ago
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