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Luda [366]
3 years ago
11

What are the Group 1A and Group 7A elements examples of? my last wuestion

Chemistry
2 answers:
avanturin [10]3 years ago
8 0

Group 1A  and Group 7A  elements are examples  of  representative  elements

    <u><em>Explanation</em></u>

  • Representative element  - are element in the first two families that is group 1 and 11 element on the far left  of periodic table.
  • In addition  representative element include  the last 6 families  on  the right  side of periodic table.
  •   The representative element include s  and p  blocks element of periodic table.

alisha [4.7K]3 years ago
4 0
Representative elements
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A student placed 15.5 g of glucose (C6H12O6) in a volumetric flask, added enough water to dissolve the glucose by swirling, then
Ede4ka [16]

<u>Answer:</u> The mass of glucose in final solution is 1.085 grams

<u>Explanation:</u>

To calculate the molarity of solution, we use the equation:

\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}      ......(1)

Given mass of glucose = 15.5 g

Molar mass of glucose = 180.2 g/mol

Volume of solution = 100 mL

Putting values in equation 1, we get:

\text{Molarity of glucose solution}=\frac{15.5\times 1000}{180.2\times 100}\\\\\text{Molarity of glucose solution}=0.860M

To calculate the molarity of the diluted solution, we use the equation:

M_1V_1=M_2V_2

where,

M_1\text{ and }V_1 are the molarity and volume of the concentrated glucose solution

M_2\text{ and }V_2 are the molarity and volume of diluted glucose solution

We are given:

M_1=0.860M\\V_1=35.0mL\\M_2=?M\\V_2=0.500L=500mL

Putting values in above equation, we get:

0.860\times 35.0=M_2\times 500\\\\M_2=\frac{0.860\times 35.0}{500}=0.0602M

Now, calculating the mass of glucose by using equation 1, we get:

Molarity of glucose solution = 0.0602 M

Molar mass of glucose = 180.2 g/mol

Volume of solution = 100 mL

Putting values in equation 1, we get:

0.0602=\frac{\text{Mass of glucose solution}\times 1000}{180.2\times 100}\\\\\text{Mass of glucose solution}=\frac{0.0602\times 180.2\times 100}{1000}=1.085g

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4 0
3 years ago
Please help me with number 4
quester [9]

Answer:

Similar propoties is the answer

6 0
3 years ago
I need help please ASAP
Yuliya22 [10]

Answer:

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6 0
3 years ago
Suppose you are provided with a 30.86 g sample of potassium chlorate to perform this experiment. What is the mass of oxygen you
oee [108]

Answer:

The mass of oxygen is 12.10 g.

Explanation:

The decomposition reaction of potassium chlorate is the following:

2KClO₃(s) → 2KCl(s) + 3O₂(g)

We need to find the number of moles of KClO₃:

\eta_{KClO_{3}} = \frac{m}{M}

Where:

m: is the mass = 30.86 g

M: is the molar mass = 122.55 g/mol

\eta_{KClO_{3}} = \frac{30.86 g}{122.55 g/mol} = 0.252 moles                                      

Now, we can find the number of moles of O₂ knowing that the ratio between KClO₃ and O₂ is 2:3

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m = 0.378 moles*32 g/mol = 12.10 g

Therefore, the mass of oxygen is 12.10 g.

I hope it helps you!

6 0
3 years ago
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