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3 years ago

Why might Marie Curie have been motivated to teach the children of beet workers?

Chemistry

1 answer:

pav-90 [236]3 years ago

6 0

Marie curie worked as a governess and she was awarded noble prize for her discoveries.

Explanation:

When Marie curie worked as a governess for three years her employer allowed her to teach reading to the children of peasant workers at his beet sugar factory. This was banned under Russian rule. During that time she studied chemistry lessons from her father by mail.

Marie curie was awarded the Noble prize in chemistry for her discoveries and studies of elements radium and polonium. She is the only women who is awarded the Noble prize twice.

She made her first research into the treatment of tumors with radiation and founded a Curie institutes, that are important medical research centers.

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calculate the number of coulombs of positive charge in 250cm 3 of (neutral) water. (hint: a hydrogen atom contains one proton; a

Sidana [21]

Number of coulombs of positive charge in 250cm^3 water is 1.3×10^7 C

The volume of 250 cm^3 corresponds to a mass of 250 g since the density of water is 1.0 g/cm^3

This mass corresponds to 250/18 = 14 moles since the molar mass of water is 18. There are ten proton (each with charge q = +e) in each molecule of $H_{2}O$ So,

Q = 14NA q =14(6.02×10^23)(10)(1.60×10^−19C) = 1.3×10^7 C.

Mass is the quantity of matter in a physical body. It is also a measure of the body's inertia, the resistance to acceleration when a net force is applied. An object's mass also determines the strength of its gravitational attraction to other bodies.

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1 year ago

Angiosperms produce brightly colored blooms and sweet-smelling flowers. Why have angiosperms developed these adaptations? A. to

blsea [12.9K]

Answer:

Explanation:

Angiosperms have developed these adaptations because it attracts pollinators which helps the ecosystem grow.

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2 years ago

What are land plants used for?

kotegsom [21]

Answer:

Land plants are also known as Embryophytes. Embryophytes are complex multicellular eukaryotes with specialized reproductive organs. The name derives from their innovative characteristic of nurturing the young embryo sporophyte during the early stages of its multicellular development within the tissues of the parent gametophyte.

Explanation:

4 0

3 years ago

Read 2 more answers

Calculate the ph of a dilute solution that contains a molar ratio of potassium acetate to acetic acid (pka ???? 4.76) of (a) 2:1

givi [52]

According to Hasselbach-Henderson equation:

$pH=pK_{a}+log\frac{[A^{-}]}{[HA]}$

Here, $[A^{-}]$ is concentration of conjugate base and [HA] is concentration of acid.

In the given problem, conjugate base is $CH_{3}COOK$ and acid is $CH_{3}COOH$ thus, Hasselbach-Henderson equation will be as follows:

$pH=pK_{a}+log\frac{[CH_{3}COOK]}{[CH_{3}COOH]}$ ...... (1)

(a) Ratio of concentration of potassium acetate and acetic acid is 2:1 thus,

$\frac{[CH_{3}COOK]}{[CH_{3}COOH]}=2$

Also, $pK_{a}=4.76$

Putting the values in equation (1),

$pH=4.76+log\frac{2}{1}=5.06$

Therefore, pH of solution is 5.06.

(b) Ratio of concentration of potassium acetate and acetic acid is 1:3 thus,

$\frac{[CH_{3}COOK]}{[CH_{3}COOH]}=\frac{1}{3}$

Also, $pK_{a}=4.76$

Putting the values in equation (1),

$pH=4.76+log\frac{1}{3}=4.28$

Therefore, pH of solution is 4.28.

(c)Ratio of concentration of potassium acetate and acetic acid is 5:1 thus,

$\frac{[CH_{3}COOK]}{[CH_{3}COOH]}=\frac{5}{1}$

Also, $pK_{a}=4.76$

Putting the values in equation (1),

$pH=4.76+log\frac{5}{1}=5.45$

Therefore, pH of solution is 5.45.

(d) Ratio of concentration of potassium acetate and acetic acid is 1:1 thus,

$\frac{[CH_{3}COOK]}{[CH_{3}COOH]}=1$

Also, $pK_{a}=4.76$

Putting the values in equation (1),

$pH=4.76+log\frac{1}{1}=4.76$

Therefore, pH of solution is 4.76.

(e) Ratio of concentration of potassium acetate and acetic acid is 1:10 thus,

$\frac{[CH_{3}COOK]}{[CH_{3}COOH]}=\frac{1}{10}$

Also, $pK_{a}=4.76$

Putting the values in equation (1),

$pH=4.76+log\frac{1}{10}=3.76$

Therefore, pH of solution is 3.76.

4 0

3 years ago

A radioactive element reduces to 5.00% of its initial mass in

Stolb23 [73]

The half-life in months of a radioactive element that reduce to 5.00% of its initial mass in 500.0 years is approximately 1389 months

To solve this question, we'll begin by calculating the number of half-lives that has elapsed. This can be obtained as follow:

Amount remaining (N) = 5%

Original amount (N₀) = 100%

<h3>Number of half-lives (n) =?</h3>

N₀ × 2ⁿ = N

5 × 2ⁿ = 100

2ⁿ = 100/5

2ⁿ = 20

Take the log of both side

Log 2ⁿ = log 20

nlog 2 = log 20

Divide both side by log 2

n = log 20 / log 2

Thus, 4.32 half-lives gas elapsed.

Finally, we shall determine the half-life of the element. This can be obtained as follow.

Number of half-lives (n) = 4.32

Time (t) = 500 years

t½ = t / n

t½ = 500 / 4.32

t½ = 115.74 years

Multiply by 12 to express in months

t½ = 115.74 × 12

<h3>t½ ≈ 1389 months </h3>

Therefore, the half-life of the radioactive element in months is approximately 1389 months

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8 0

2 years ago