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LuckyWell [14K]
3 years ago
6

I gave brainlest:: answer this .....​

Chemistry
1 answer:
andreev551 [17]3 years ago
3 0
There’s no word bank or anything to go off of
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What happens when an object has a density greater than that of the surrounding liquid?
vekshin1

Answer:

An object that has a higher density than the liquid it's in will sink

Explanation:

hopefully I helped

4 0
3 years ago
Read 2 more answers
Oxidation unit test
3241004551 [841]

In this case, according to the given information about the oxidation numbers anf the compounds given, it turns out possible to figure out the oxidation number of manganese in both MnI2, manganese (II) iodide and MnO2, manganese (IV) oxide, by using the concept of charge balance.

Thus, we can define the oxidation state of iodine and oxygen as -1 and -2, respectively, since the former needs one electron to complete the octet and the latter, two of them.

Next, we can write the following x, since manganese has five oxidation states, and it is necessary to calculate the appropriate ones:

Mn^xI_2^-\\\\Mn ^xO_2^{-2}

Next, we multiply each anion's oxidation number by the subscript, to obtain the following:

Mn^xI_2^-\rightarrow x-2=0;x=+2\\\\Mn ^xO_2^{-2}\rightarrow x-4=0;x=+4

Thus, the correct choice is Manganese has an oxidation number of +2 in Mnl2 and +4 in MnO2.

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  • brainly.com/question/15167411
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2 years ago
In sexual reproduction, an offspring is produced with genes from both parents. When the offspring has a new genetic variation th
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Answer:c a mutation

Explanation:

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3 years ago
Fine the mole. 2.41 x 10^24 molecules CO2
vlada-n [284]

Answer:

<h2>4 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

From the question we have

n =  \frac{2.41 \times  {10}^{24} }{6.02 \times  {10}^{23} }  \\  = 4.00322...

We have the final answer as

<h3>4 moles</h3>

Hope this helps you

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3 years ago
When 0.514 g of biphenyl (C12H10) undergoes combustion in a bomb calorimeter, the temperature rises from 25.8 °C to 29.4 °
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