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vichka [17]
3 years ago
12

Which of the following is NOT true about ionic compounds?

Chemistry
1 answer:
Usimov [2.4K]3 years ago
7 0

Answer:

C.it has a lower melting temperature

Explanation:

A good way to solve this problem may be to think about a simple ionic compound, such as salt, NaCl:

  • NaCl easily disolves in water, so options A and B are true about ionic compounds.
  • Salt powder does not conduct electricity, however a concentrated solution of salt in water does. Meaning that option D is true as well.
  • NaCl has a melting point of 801 °C, not particularly a low value. Thus the correct option is C.
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How is the human heart adapted to its function?
Tresset [83]

It receives blood from the whole body. Blood from the lungs enters the heart through the pulmonary vein into the left atrium. ... - Has ventricles that has thicker walls thus more muscular than auricles generating higher pressure to pump blood over long distances.

8 0
3 years ago
To investigate the reactivity of metals a student uses four metals. Each time they added 1 g of the metal to 25 cm³ of sulfuric
Mazyrski [523]

Answer:

Volume of the sulfuric acid (25cm³), same mass of each metal (1g)

Explanation:

In an experiment, the CONTROL VARIABLE also known as constant is the variable that is kept unchanged for all groups in an experiment. This is done in order not to influence the outcome of the experiment.

In this case, students are trying to investigate the reactivity of four different metals. They added 1 g of each metal to 25cm³ of sulfuric acid and recorded the temperature change. Based on the explanation of control variable above, the VOLUME OF SULFURIC ACID (25cm³) and the MASS OF EACH METAL (1g) are the CONTROL VARIABLES because they are the same or unchanged in this experiment.

5 0
3 years ago
A 700 mL sample of gas at STP is compressed to 300 mL, and the
bogdanovich [222]

Answer:

P₂ =  3.19 atm

Explanation:

Given data:

Initial volume = 700 mL

Initial pressure = 1 atm

Initial temperature = 273K

Final temperature =100°C (100+273= 373 K)

Final volume = 300 mL

Final pressure = ?

Formula:  

P₁V₁/T₁ = P₂V₂/T₂  

P₁ = Initial pressure

V₁ = Initial volume

T₁ = Initial temperature

P₂ = Final pressure

V₂ = Final volume

T₂ = Final temperature

Solution:

P₂ = P₁V₁ T₂/ T₁ V₂  

P₂ = 1 atm × 700 mL × 373K / 273 K ×300 mL  

P₂ = 261100 atm .mL. K   /81900 K .mL

P₂ =  3.19 atm

5 0
3 years ago
What is the mass of six of these marbles? What is the volume? What is the<br> density?
chubhunter [2.5K]

Answer:

All right. So let's calculate the density of a glass marble. Remember that the formula for density is mass over volume. So if I know that the masses 18.5 g. And I know that the um volume is 6.45 cubic centimeters. I can go ahead and answer this to three significant figures. So it's going to be 2.87 grams per cubic centimeter. Okay, that's our density. Now, density is an intensive process. Okay. We're an intensive property. I really should say. It doesn't depend on how much you have. Mhm. If I have one marble, its density is going to be 2.87 g per cubic centimeter. If I have two marbles, the density will be the same because I'll double the mass and I'll also double the volume. So when I divide them I'll get the same number. Okay, that's what makes it an intensive property. No matter how many marbles I have, they'll have the same density. Mass though is not an intensive property. So if I have six marbles and I want to know what the massive six marbles is. Well, I know the mass of each marble is 18.5 g. So the mass of six marbles Is going to be 100 11 g. Because mass is an extensive property. It depends on how much you have. If I change the number of marbles, I'm going to change the mass. That's an extensive property. All right. So we've calculated the density. We've calculated the mass and then what happens to the density of one marble compared to six marbles as we mentioned before. Since densities and intensive property, the densities will be the same, no matter how may.

Explanation:

5 0
3 years ago
Calculate the solubility of ( = ) in moles per liter. Ignore any acid–base properties. s = mol/L Calculate the solubility of ( =
BaLLatris [955]

This is an incomplete question, here is a complete question.

Calculate the solubility of each of the following compounds in moles per liter. Ignore any acid-base properties.

CaCO₃, Ksp = 8.7 × 10⁻⁹

Answer : The solubility of CaCO₃ is, 9.33\times 10^{-5}mol/L

Explanation :

As we know that CaCO₃ dissociates to give Ca^{2+} ion and CO_3^{2-} ion.

The solubility equilibrium reaction will be:

CaCO_3\rightleftharpoons Ca^{2+}+CO_3^{2-}

The expression for solubility constant for this reaction will be,

K_{sp}=[Ca^{2+}][CO_3^{2-}]

Let solubility of CaCO₃ be, 's'

K_{sp}=(s)\times (s)

K_{sp}=s^2

8.7\times 10^{-9}=s^2

s=9.33\times 10^{-5}mol/L

Therefore, the solubility of CaCO₃ is, 9.33\times 10^{-5}mol/L

4 0
3 years ago
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