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Vsevolod [243]
3 years ago
14

What is the name of sc(oh)3

Chemistry
2 answers:
ella [17]3 years ago
8 0

Answer:

Scandium(III) hydroxide

Explanation:

May I have brainliest please? :)

Mnenie [13.5K]3 years ago
7 0

Answer:

kidtavo13

Explanation:

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Question 2 of 50
wolverine [178]

The thermal decomposition of calcium carbonate will produce 14 g of calcium oxide. The stoichiometric ratio of calcium carbonate to calcium oxide is 1:1, therefore the number of moles of calcium carbonate decomposed is equal to the number of moles of calcium oxide formed.

Further Explanation:

To solve this problem, follow the steps below:

  1. Write the balanced chemical equation for the given reaction.
  2. Convert the mass of calcium carbonate into moles.
  3. Determine the number of moles of calcium oxide formed by using the stoichiometric ratio for calcium oxide and calcium carbonate based on the coefficient of the chemical equation.
  4. Convert the number of moles of calcium oxide into mass.

Solving the given problem using the steps above:

STEP 1: The balanced chemical equation for the given reaction is:

CaCO_{3} \rightarrow \ CaO \ + \ CO_{2}

STEP 2: Convert the mass of calcium carbonate into moles using the molar mass of calcium carbonate.

mol \ CaCO_{3} \ = 25 \ g \ CaCO_{3} \ (\frac{1 \ mol \ CaCO_{3}}{100.0869 \ g \ CaCO_{3}})\\ \\\boxed {mol \ CaCO_{3} \ = 0.2498 \ mol}

STEP 3: Use the stoichiometric ratio to determine the number of moles of CaO formed.

For every mole of calcium carbonate decomposed, one more of a calcium oxide is formed. Therefore,

mol \ CaO \ = 0.2498 \ mol

STEP 4: Convert the moles of CaO into mass of CaO using its molar mass.

mass \ CaO \ = 0.2498 \ mol \ CaO \ (\frac{56.0774 \ g \ CaO}{1 \ mol \ CaO})\\ \\mass \ CaO \ = 14.008 \ g

Since there are only 2 significant figures in the given, the final answer must have the same number of significant figures.

Therefore,

\boxed {mass \ CaO \ = 14 \ g}

Learn More

  1. Learn more about stoichiometry brainly.com/question/12979299
  2. Learn more about mole conversion brainly.com/question/12972204
  3. Learn more about limiting reactants brainly.com/question/12979491

Keywords: thermal decomposition, stoichiometry

5 0
3 years ago
Organic Chem Rxn Question
NemiM [27]

Answer:

a, g, c

Explanation:

The conversion of the stable cyclopentane into Trans-1, 2dibromocyclopentane will require three step reactions.

The first is to convert the compound into a cyclopentene, through the addition of Bromine water under heat and photons (light). So option A is the first in the order. This will generate 1 bromocyclopentane through halogenation of the alkane. Secondly, a hot and strong base should be added like the NaOEt, EtOH to remove the added bromine and one atom of hydrogen from the resulting 1 bromocyclopentane in the previous reaction. This will yield cyclopentene, thus making the compound more electrophilic. So option g is required. Thirdly, bromine molecules will be added (C) to take up their places at the two electrophilic regions of the compound to produce Trans-1, 2dibromocyclopentane.

8 0
3 years ago
There are 6 oxygen atoms in a half dozen water (h2o) molecules. how many hydrogen atoms are in the same half dozen molecules?
dlinn [17]
There are 12 hydrogen atoms in 6(H2O)
5 0
3 years ago
Which of these mixtures is a suspension?
natita [175]

Answer:

D. apple juice

Explanation:

8 0
2 years ago
In an electroplating process, copper (ionic charge +2e, atomic weight 63.6 g/mol) is deposited using a current of 10.0 A. What m
salantis [7]

Answer : The mass of copper deposit is, 1.98 grams

Explanation :

First we have to calculate the charge.

Formula used : Q=I\times t

where,

Q = charge = ?

I = current = 10 A

t = time = 10 min = 600 sec      (1 min = 60 sec)

Now put all the given values in this formula, we get

Q=10A\times 600s=6000C

Now we have to calculate the number of atoms deposited.

As, 1 atom require charge to deposited = 2\times (1.6\times 10^{-19})  

Number of atoms deposited = \frac{(6000)}{2\times(1.6\times 10^{-19})}=1.875\times 10^{22} atoms

Now we have to calculate the number of moles deposited.

Number of moles deposited = \frac{(1.875\times 10^{22})}{(6.022\times 10^{23})}=0.03113 moles

Now we have to calculate the mass of copper deposited.

1 mole of Copper has mass = 63.5 g  

Mass of Copper Deposited = 63.5\times 0.03113 =1.98g

Therefore, the mass of copper deposit is, 1.98 grams

5 0
3 years ago
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