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Akimi4 [234]
3 years ago
11

Which of the following classifications is NOT correct? Group 1 and Group 2 - Metals Group 7 - Halogens Group 8 - Noble Gases Gro

up 5 - Metalloids
Chemistry
1 answer:
serious [3.7K]3 years ago
6 0

Answer: -

Group 5 - Metalliods.

Group 7 - Halogens Group

Group 8 - Noble Gases

Explanation: -

Group 1 are the alkali metals.

Group 2 are the alkaline earth metals.

Group 5 are a part of transition metals which range from Group 3 to Group 12.

Group 17 are the halogens.

Group 18 are the noble gases.

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What is the molar concentration of Pb+ in a solution that contains 6.73 ppm of
Katarina [22]

Answer:

2.03 × 10⁻⁵ M

Explanation:

Step 1: Given data

Concentration of Pb(NO₃)₂: 6.73 ppm = 6.73 mg/L

Step 2: Convert 6.73 mg/L to mol/L

The molar mass of 331.2 g/mol.

6.73 × 10⁻³ g/L × 1 mol/331.2 g = 2.03 × 10⁻⁵ mol/L = 2.03 × 10⁻⁵ M

Step 3: Calculate the molar concentration of Pb²⁺

Let's consider the ionization of Pb(NO₃)₂.

Pb(NO₃)₂(aq) ⇒ Pb²⁺(aq) + 2 NO₃⁻(aq)

The molar ratio of Pb(NO₃)₂ to Pb²⁺ is 1:1. The molar concentration of Pb²⁺ is 1/1 × 2.03 × 10⁻⁵ M = 2.03 × 10⁻⁵ M.

8 0
3 years ago
Which quantity could best be measured with a meter stick?
Ilya [14]

Answer:

B

Explanation:

A meter stick measures distances ( like centimeters)   so....B  

  it does NOT measure mass , volume or VERY large distances (Like earth's diameter)

3 0
2 years ago
Read 2 more answers
The radiator in a car is filled with a solution of 60% antifreeze and 40% water. The manufacturer of the antifreeze suggests tha
lakkis [162]

Answer:

0.6 Liters of coolant should be drained and 0.6 Liter of water should be added.

Explanation:

Volume of the radiator = 3.6 L

Percentage of antifreeze = 60%

Total volume of anti freeze in radiator = 60% of 3.6 L :

\frac{60}{100}\times 3.6 = 2.16 L

Percentage of water= 40%

Given,optimal cooling of the engine is obtained with only 50% antifreeze.

So, now we want to reduce the percentage of antifreeze from 60% to 50 %

Volume of coolant removed = x

Volume of water added = x

Volume of anti freeze removed = 60% of(x) = 0.6x

Volume of antifreeze left in radiator  =50% of 3.6 L = \frac{50}{100}\times 3.6=1.8 L

1.8 Liter is the desired volume of the antifreeze.

Total volume - Removed volume = desired  volume

2.16 L - 60% of( x) = 1.8 L

2.16 L-0.6x=1.8 L

x=\frac{2.16 l- 1.8 L}{0.6}=0.6 L

0.6 Liters of coolant should be drained and 0.6 Liter of water should be added.

4 0
4 years ago
How do the circulatory, respiratory, or digestive systems work together?
Alexandra [31]

Answer:

by pumping blood throughout the body.

Explanation:

7 0
3 years ago
A 2.00 L sample of gas at 35C is to be heated at constant pressure until it reaches a volume of 5.25 L. To what Kelvin temperatu
Anastaziya [24]

Answer:

The sample will be heated to 808.5 Kelvin

Explanation:

Step 1: Data given

Volume before heating = 2.00L

Temperature before heating = 35.0°C = 308 K

Volume after heating = 5.25 L

Pressure is constant

Step 2: Calculate temperature

V1 / T1 = V2 /T2

⇒ V1 = the initial volume = 2.00 L

⇒ T1 = the initial temperature = 308 K

⇒ V2 = the final volume = 5.25 L

⇒ T2 = The final temperature = TO BE DETERMINED

2.00L / 308.0 = 5.25L / T2

T2 = 5.25/(2.00/308.0)

T2 = 808.5 K

The sample will be heated to 808.5 Kelvin

7 0
4 years ago
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