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3 years ago

Name the particle discovered by James Chadwick in 1932.

Chemistry

1 answer:

LenKa [72]3 years ago

4 0

Answer:

He discovered neutrons in 1932

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Which element is chemically similar to chlorine?

ad-work [718]

bromine

Explanation:

halogens are a group of elemnts simlar to eachother

flourine, chlorine, and bromine

6 0

3 years ago

Calculate the pKa of lactic acid (CH3CH(OH)COOH) given the following information. 3.005 grams of potassium lactate are added to

snow_lady [41]

Answer:

$\displaystyle \text{p} K_a \approx 3.856$

Explanation:

Because 3.005 grams of potassium lactate is added to 100. mL of solution, its concentration is:

$\displaystyle \begin{aligned} \left[ \text{KC$_3$H_$_5$O$_3$}\right] & = \frac{3.005\text{ g KC$_3$H_$_5$O$_3$}}{100.\text{ mL}} \cdot \frac{1\text{ mol KC$_3$H_$_5$O$_3$}}{128.17 \text{ g KC$_3$H_$_5$O$_3$}} \cdot \frac{1000\text{ mL}}{1\text{ L}} \\ \\ &= 0.234\text{ M}\end{aligned}$

By solubility rules, potassium is completely soluble, so the compound will dissociate completely into potassium and lactate ions. Therefore, [KC₃H₅O₃] = [C₃H₅O₃⁺]. Note that lactate is the conjugate base of lactic acid.

Recall the Henderson-Hasselbalch equation:

$\displaystyle \begin{aligned}\text{pH} = \text{p}K_a + \log \frac{\left[\text{Base}\right]}{\left[\text{Acid}\right]} \end{aligned}$

[Base] = 0.234 M and [Acid] = 0.500 M. We are given that the resulting pH is 3.526. Substitute and solve for p<em>Kₐ</em>:

$\displaystyle \begin{aligned} (3.526) & = \text{p}K_a + \log \frac{(0.234)}{(0.500)} \\ \\ 3.526 & = \text{p}K_a + (-0.330) \\ \\ \text{p}K_a & = 3.856\end{aligned}$

In conclusion, the p<em>Kₐ </em>value of lactic acid is about 3.856.

5 0

1 year ago

This is very difficult

Oliga [24]

Answer: C

Explanation:

6 0

3 years ago

What is the significance of the Roman Numeral when naming binary ionic compounds?

serious [3.7K]

Answer:

Roman numbers are oxidation state of the metals

Because some elements of metals show more than one oxidation state like iron Fe2+ in ferrous and Fe3+ in ferric.

Explanation:

8 0

2 years ago

How much heat is required to raise the temperature of 10.35g of CCl4 from 32.1°c to 56.4°c

Travka [436]

We are going to use this formula:

Q = M*C*ΔT

when Q is the heat required

M is the mass of CCl4 = 10.35 g

C is the specific heat capacity of CCl4 = 0.874J/g.c

and ΔT the change in temperature = 56.4 - 32.1 °C =24.3 °C

∴ Q = 10.35g * 0.874 * 24.3 °C

= 219.8 J

4 0

3 years ago