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Gnoma [55]
3 years ago
15

9. Elements that are characterized by the filling of p orbitals are classified as

Chemistry
1 answer:
Bogdan [553]3 years ago
6 0

Answer:

p block elements

Explanation:

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You carefully weigh out 16.00 g of CaCO3 powder and add it to 64.80 g of HCl solution. You notice bubbles as a reaction takes pl
bulgar [2K]

Answer: Mass of CO_2  produced in this reaction was 6.56 grams

Explanation:

According to the law of conservation of mass, mass can neither be created nor be destroyed. Thus the mass of products has to be equal to the mass of reactants. The number of atoms of each element has to be same on reactant and product side. Thus chemical equations are balanced.

CaCO_3(s)+2HCl(aq)\rightarrow H_2O(l)+CO_2(g)+CaCl_2(aq)

Mass or reactants =  Mass of CaCO_3+ mass of HCl = 16.00 + 64.80 = 80.80 g

Mass of products  = mass of aqueous solution + mass of CO_2 + = 74.24 + x g

Mass or reactants = Mass of products

80.80 g = 74.24 + x g

x = 6.56 g

Thus mass of CO_2  produced in this reaction was 6.56 grams

7 0
3 years ago
7.<br> How many grams are contained in 3.9 x 1023 sulfur atoms?<br> atoms → moles<br> grams<br> I
Anni [7]

Answer: 20.775 g S

Explanation: 3.9x10^23 atoms = 0.648 mol

Atomic mass S = 32.08

S in grams = 20.775

4 0
3 years ago
Ayyyyyy<br> todays my birthday<br> hmu plzzz
SVEN [57.7K]

Answer:

happy bday

Explanation:

plz give this brainliest ;)

3 0
3 years ago
A compound is found to have 55.7% hafnium+and+44.3%+chlorine.+what+is+the+empirical+formula?
uranmaximum [27]

The empirical formula of a compound found to have 55.7% hafnium and 44.3% chlorine is HfCl4.

<h3>How to calculate empirical formula?</h3>

The empirical formula of a compound is a notation indicating the ratios of the various elements present in a compound, without regard to the actual numbers.

The empirical formula of the given compound can be calculated as follows:

  • Hafnium = 55.7% = 55.7g
  • Chlorine = 44.3% = 44.3g

First, we convert mass values to moles by dividing by the molar mass of each element

  • Hafnium = 55.7g ÷ 178.49g/mol = 0.312mol
  • Chlorine = 44.3g ÷ 35.5g/mol = 1.25mol

Next, we divide each mole value by the smallest

  • Hafnium = 0.312 ÷ 0.312 = 1
  • Chlorine = 1.25 ÷ 0.312 = 4

Therefore, the empirical formula of a compound found to have 55.7% hafnium and 44.3% chlorine is HfCl4.

Learn more about empirical formula at: brainly.com/question/14044066

#SPJ1

7 0
1 year ago
What is the total ionic equation for the following reaction?
schepotkina [342]

Answer: Option (d) is the correct answer.

Explanation:

An equation in which electrolytes are represented in the form of ions is known as an ionic equation.

Strong electrolytes easily dissociate into their corresponding ions. Hence, they form ionic equation.

H_{2}CrO_{4} is a strong acid and Ba(OH)_{2} is a strong bases, therefore, both of them will dissociate into ions.

Thus, total ionic equation will be as follows.

2H^{+} + CrO_{4}^{-} + Ba^{2+} + 2OH^{-} \rightarrow Ba^{2+} + CrO_{4}^{-} + 2H_{2}O

5 0
3 years ago
Read 2 more answers
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