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Tju [1.3M]
3 years ago
5

Now take a fresh 60 mL of the 0.50 M pH 6.83 buffer and add 3.7 mL of 1.00 M NaOH. Using steps similar to those above, calculate

the new pH of the solution.
Chemistry
1 answer:
Digiron [165]3 years ago
4 0

Answer:

pH = pKa + log (33.7 mm / 26.3 mm)

Explanation:

Initial pH = 6.83

If we add a strong base, we expect that pH would be more alkaline

Initially we have: 60 mL . 0.50M =  30 mmoles of acid

Now, we add the base:

3.7 mL . 1.00 M = 3.7 mmmoles of base

Concentration of acid: 30 mmoles - 3.7 mmoles of base = 26.3 mm

Concentration of salt: 30 mmmoles + 3.7 mmoles of base = 33.7 mm

pH = pka + log (33.7 mm / 26.3 mm)

Information of pKa is missing

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Which of these processes removes carbon from
GenaCL600 [577]
Answer is c photosynthesis
3 0
3 years ago
SOMEONE PLEASE HELP
Zinaida [17]

Answer:

6. 7870 kg/m³ (3 s.f.)

7. 33.4 g (3 s.f.)

8. 12600 kg/m³ (3 s.f.)

Explanation:

6. The SI unit for density is kg/m³. Thus convert the mass to Kg and volume to m³ first.

1 kg= 1000g

1m³= 1 ×10⁶ cm³

Mass of iron bar

= 64.2g

= 64.2 ÷1000 kg

= 0.0642 kg

Volume of iron bar

= 8.16 cm³

= 8.16 ÷ 10⁶

= 8.16 \times 10^{ - 6} \:  kg

\boxed{density =  \frac{mass}{volume} }

Density of iron bar

=  \frac{0.0642}{8.16 \times 10^{ - 6} }

= 7870 kg/m³ (3 s.f.)

7.

\boxed{mass = density \:  \times volume}

Mass

= 1.16 ×28.8

= 33.408 g

= 33.4 g (3 s.f.)

8. Volume of brick

= 12 cm³

= 12  \times  10^{ - 6}  \: m^{3}  \\  = 1.2 \times 10^{ - 5}  \: m ^{3}

Mass of brick

= 151 g

= 151 ÷ 1000 kg

= 0.151 kg

Density of brick

= mass ÷ volume

=  \frac{0.151} {1.25 \times 10^{ - 5} }  \\  = 12600 \: kg/ {m}^{3}

(3 s.f.)

6 0
3 years ago
How many moles of potassium are contained in 449 g of<br> potassium?<br> pls help im so lost
e-lub [12.9K]

Answer:

449 (g K) / 39.1 (g/mol K) = 11.5 mol K

Explanation:

Potassium has atomic number 39.1

amount of K in 449g sample = 449/39.1 = 11.5 mol

4 0
3 years ago
Read 2 more answers
Balance the equation (NH4)3 PO4 +NA0H arrow Na3P04 +3NH3 +3H20.
mr_godi [17]

Answer:

Explanation:

(NH4)3 PO4 +NaOH arrow Na3PO4 +3NH3 +3H2O

Start by seeing what happens with the Na. You need 3 on the left, so put a 3 in front of NaOH

(NH4)3 PO4 +3NaOH arrow Na3PO4 +3NH3 +3H2O  Next work with the nitrogens. YOu have 3 on the left and 3 on the right, so they are OK. Next Go to the stray oxygens.

You have 3 on left in (NaOH) and three on the right in 3H2O so they are fine as well. The last thing you should look at are hydrogens.

There are 12 + 3 on the left which is 15. There are 9 (in 3NH3) and 6 more in the water. They seem fine.

Why didn't I do something with the PO4^(-3)? The reason is a deliberately stayed away from them and balanced everything else. Since they were untouched with 1 on the left and 1 on the right, they are balanced.

Species      Na        H        O         N       PO4

Left             3          15        3         3          1

Right           3         15         3         3          1

8 0
3 years ago
Read 2 more answers
At equilibrium, the concentrations of the products and reactants for the reaction, H2 (g) + I2 (g)  2 HI (g), are [H2] = 0.106
lana [24]

Answer:

The new equilibrium concentration of HI: <u>[HI] = 3.589 M</u>          

Explanation:

Given: Initial concentrations at original equilibrium- [H₂] = 0.106 M; [I₂] = 0.022 M; [HI] = 1.29 M        

Final concentrations at new equilibrium- [H₂] = 0.95 M; [I₂] = 0.019 M; [HI] = ? M

<em>Given chemical reaction:</em> H₂(g) + I₂(g) → 2 HI(g)

The equilibrium constant (K_{c}) for the given chemical reaction, is given by the equation:

K_{c} = \frac {[HI]^{2}}{[H_{2}]\: [I_{2}]}

<u><em>At the original equilibrium state:</em></u>

K_{c} = \frac {(1.29\: M)^{2}}{(0.106\: M) \times (0.022\: M)}

K_{c} = \frac {1.6641}{0.002332} = 713.59

<u><em>Therefore, at the new equilibrium state:</em></u>

K_{c} = \frac {[HI]^{2}}{(0.95\: M) \times (0.019\: M)}

\Rightarrow K_{c} = 713.59 = \frac {[HI]^{2}}{0.01805}

\Rightarrow [HI]^{2} = 713.59 \times 0.01805 = 12.88

\Rightarrow [HI] = \sqrt {12.88} = 3.589 M

<u>Therefore, the new equilibrium concentration of HI: [HI] = 3.589 M</u>

6 0
3 years ago
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