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KatRina [158]
2 years ago
11

An isotope of an element has the following symbol: 131 I How many electrons does it

Chemistry
1 answer:
WINSTONCH [101]2 years ago
5 0

Answer:

D

Explanation:

sorry bout ealier

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3 natural chemicals a part of our life
Mademuasel [1]
Sodium chloride, methane gas and water
4 0
3 years ago
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What is true when an element is oxidized? It bonds with the hydroxide ion. It loses electrons to another element. It reacts with
Mama L [17]

Answer: It loses electrons to another element.

Explanation:- Oxidation is the process in which an element loses electrons and there is an increase in the oxidation state. On losing electrons it combines with a electronegative element such as oxygen, sulphur or nitrogen etc.

Fe\rightarrow Fe^{2+}+2e^-

Reduction is the process in which an element gains electrons and there is a decrease in the oxidation state.

\frac{1}{2}O_2+2e^-\rightarrow O^{2-}

8 0
3 years ago
"how many grams of calcium are consumed when 156.8 ml of oxygen"
Akimi4 [234]
Your question is incomplete. However, I found a similar problem fromanother website as shown in the attached picture.

To solve this problem, you must know that at STP, the volume for any gas is 22.4 L/mol. So,

Moles O₂: 156.8 mL * 1 L/1000 mL* 1 mol/22.4 L = 0.007 moles
Mass calcium: 0.007 mol O₂ * 2 mol Ca/1 mol O₂ * 40 g/mol Ca =<em> 0.56 g Ca</em>

3 0
3 years ago
A 45-g aluminium spoon(specific heat 0.80 / J/gdegree Celsius) at 24 degree celsius is placed in 180 ml(180 grams) of coffee at
Vlad [161]

Explanation:

a) The amount of heat released by coffee will be absorbed by aluminium spoon.

Thus, heat_{absorbed}=heat_{released}

To calculate the amount of heat released or absorbed, we use the equation:  

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

Also,

m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]    ..........(1)

where,

q = heat absorbed or released

m_1 = mass of aluminium = 45 g

m_2 = mass of coffee = 180 g

T_{final} = final temperature = ?

T_1 = temperature of aluminium = 24^oC

T_2 = temperature of coffee = 85^oC

c_1 = specific heat of aluminium = 0.80J/g^oC

c_2 = specific heat of coffee= 4.186 J/g^oC

Putting all the values in equation 1, we get:

45 g\times 0.80J/g^oC\times (T_{final}-24^oC)=-[180 g\times 4.186J/g^oC\times (T_{final}-83^oC)]

T_{final}=80.30^oC

80.30 °C is the final temperature.

b) Energy flows from higher temperature to lower temperature.Whenever two bodies with different energies and temperature come in contact. And the resulting temperature of both bodies will less then the body with high temperature and will be more then the body with lower temperature.

So, is our final temperature of both aluminium and coffee that is 80°C less than initial temperature of coffee and more than the initial temperature of the aluminum.

8 0
3 years ago
Just need a letter so no that does not help
il63 [147K]
Like A B C D or like a mailed letter

3 0
3 years ago
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