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4 years ago

When it is spring in China, it is ______ in South America

Chemistry

1 answer:

Vanyuwa [196]4 years ago

4 0

I would say fall/ autumn in South America because they are in opposite hemispheres.

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Which experiment lead to the discovery that the tiny, dense nucleus contained most of

oksian1 [2.3K]

I think it was from los griegos but a long time a go because he was making and experiment with and apple and he notice that he couldn’t cut more so he name it atoms (sorry for my bad English I don’t speak English)

7 0

3 years ago

0.10 M potassium chromate is slowly added to a solution containing 0.20 M AgNO3 and 0.20 M Ba(NO3)2. What is the Ag+ concentrati

erastova [34]

Answer:

$[Ag^{+}]=4.2\times 10^{-2}M$

Explanation:

Given:

[AgNO3] = 0.20 M

Ba(NO3)2 = 0.20 M

[K2CrO4] = 0.10 M

Ksp of Ag2CrO4 = 1.1 x 10^-12

Ksp of BaCrO4 = 1.1 x 10^-10

$BaCrO_4 (s)\leftrightharpoons Ba^{2+}(aq)\;+\;CrO_{4}^{2-}(aq)$

$Ksp=[Ba^{2+}][CrO_{4}^{2-}]$

$1.2\times 10^{-10}=(0.20)[CrO_{4}^{2-}]$

$[CrO_{4}^{2-}]=\frac{1.2\times 10^{-10}}{(0.20)}= 6.0\times 10^{-10}$

Now,

$Ag_{2}CrO_4(s) \leftrightharpoons 2Ag^{+}(aq)\;+\;CrO_{4}^{2-}(aq)$

$Ksp=[Ag^{+}]^{2}[CrO_{4}^{2-}]$

$1.1\times 10^{-12}=[Ag^{+}]^{2}](6.0\times 10^{-10})$

$[Ag^{+}]^{2}]=\frac{1.1\times 10^{-12}}{(6.0\times 10^{-10})}= 1.8\times 10^{-3}$

$[Ag^{+}]=\sqrt{1.8\times 10^{-3}}=4.2\times 10^{-2}M$

So, BaCrO4 will start precipitating when [Ag+] is 4.2 x 1.2^-2 M

7 0

3 years ago

Which of the following is NOT the same?

Inessa [10]

All are the same. It equals to the same thing.

4 0

3 years ago

Read 2 more answers

When an atom undergoes beta decay, wich of the following is true?

tigry1 [53]

<span> The atomic number increases by one and the element becomes a different element. </span>

3 0

3 years ago

How many atoms are in 1.204x10^22 moles of arsenic?

ra1l [238]

Answer:

<h2>0.02 moles </h2>

Explanation:

To find the number of moles in a substance given it's number of entities we use the formula

$n = \frac{N}{L} \\$

where n is the number of moles

N is the number of entities

L is the Avogadro's constant which is

6.02 × 10²³ entities

We have

$n = \frac{1.204 \times {10}^{22} }{6.02 \times {10}^{23} } \\ = 0.02$

We have the final answer as

<h3>0.02 moles</h3>

Hope this helps you

7 0

3 years ago