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Jlenok [28]
3 years ago
7

How many protons does the neutral atom pictured have? A) 8 B) 18 C) 2 D) 20

Chemistry
2 answers:
elena-14-01-66 [18.8K]3 years ago
7 0
18 because well I took the the test
Shalnov [3]3 years ago
3 0

Answer:

b 18

Explanation:

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What is the balanced equation for Fe+O2—->Fe3O4
DaniilM [7]

Answer:

3Fe2O2---> Fe3O4

Explanation:

You need to add three to the left side to make it equal to the one on the right, and add a 2 since you can't change the subscripts

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3 years ago
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Whay does a solid change into liquid<br>on<br>heating? Expalin in brief​
Reptile [31]

Answer:

It changes into a liquid because when it gets heated up to a certain temperature, it reaches it's "boiling point", which is when it will start to transform into a liquid because the temperature is too hot for it to remain a solid.

Explanation:

Just think of when icicles melt because of the sun. The sun is heating them up and it gets too warm to the point that it liquidizes.

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3 years ago
How many oxygen atoms are found in one molecule of nitric acid, HNO3? (2 points) 1 2 3 4
Korolek [52]
There are 3 atoms of O2
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How many liters of fluorine gas, at standard temperature and pressure, will react with 23.5 grams of potassium metal? Show all o
natka813 [3]
The balanced chemical reaction is written as:

2K + F2 = 2KF

We are given the amount of potassium to be used in the reaction. THis will be the starting point. We do as follows:

23.5 g K ( 1 mol / 39.10 g) (1 mol F2 / 2 mol K ) ( 22.4 L / 1 mol ) = 6.73 L F2 gas needed
7 0
3 years ago
Suppose you heat a metal object with a mass of 31.7 g to 96.5 oC and transfer it to a calorimeter containing 100.0 g of water at
OverLord2011 [107]

Answer: The specific heat of the metal in 1.34J/g^0C

Explanation:

Q_{absorbed}=Q_{released}

As we know that,  

Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})

m_1\times c\times (T_{final}-T_1)=-[m_2\times c\times (T_{final}-T_2)]      

where,

m_1 = mass of metal = 31.7 g

m_2 = mass of water = 100.0 g

T_{final} = final temperature = 24.6^oC

T_1 = temperature of metal = 96.5^oC

T_2 = temperature of water = 17.3^oC

c_1 = specific heat of metal= ?

c_2 = specific heat of water =  4.184J/g^0C

Now put all the given values in equation (1), we get

m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]

-(31.7\times c_1\times (24.6-96.5)^0C)=(100.0\times 4.184\times (24.6-17.3)]

c_1=1.34J/g^0C

Therefore, the specific heat of the metal in 1.34J/g^0C

4 0
3 years ago
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