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salantis [7]
4 years ago
13

Salt acts as an antifreeze. a. True b. False

Chemistry
2 answers:
Tomtit [17]4 years ago
8 0

Answer:

It's true

Explanation:

An example can be observed when the people add salt on the road surface to avoid icing. The salt water mixture is frozen at approximately a temperature of -21°C and not at a temperature of 0°C as it would be in the case of water. Salt when dissolved in water releases heat, thereby contributing to defrost ice. This is the same principle that is used for car antifreeze, which is to lower the freezing point.

r-ruslan [8.4K]4 years ago
5 0
A. I think that is the a answer. Not to sure
You might be interested in
What is the mass (in grams) of 6.78 moles of dinitrogen pentoxide
WINSTONCH [101]
<h2>Hello!</h2>

The answer is:

There are 732.24 grams of Dinitrogen Pentoxide  N_{2}O_{5} in 6.78 moles of the same compound.

<h2>Why?</h2>

The dinitrogen pentoxide, or nitrogen pentoxide, is a rare salt that consists of anion and cations, and it's an important compound when preparing some kinds of explosives.

So, to answer the question, first,  need to look for the molecular formula of the dinitrogen pentoxide, then, use it to calculate the mass in gram of 6.78 moles of the same compound.

The dinitrogen pentoxide formula is:

N_{2}O_{5}

Where, the molar mass of each element are:

N=\frac{14g}{mol}

O=\frac{16g}{mol}

Calculating the molar mass  of the compound, we have:

N_{2}O_{5}=2*(\frac{14g}{mol})+5*(\frac{16g}{mol})

N_{2}O_{5}=(\frac{28g}{mol})+(\frac{80g}{mol})=(\frac{108g}{mol}

Now, calculating what is the mass of 6.78 of N_{2}O_{5}, we have:

mass=moles*MolarMass=6.78mol(N_{2}O_{5})*\frac{108g}{mol}=732.24g(N_{2}O_{5})

So, there are 732.24 grams of Dinitrogen Pentoxide  N_{2}O_{5} in 6.78 moles of the same compound.

Have a nice day!

3 0
4 years ago
What is the molarity of a solution that has 23.0 g of NaOH dissolved in 0.500 L of solution?
madreJ [45]

Answer: 1.15M

Explanation:

Given that,

Amount of moles of NaOH (n) = ?

Mass of NaOH in grams = 23.0g

For molar mass of NaOH, use the atomic masses: Na = 23g; O = 16g; H = 1g

NaOH = (23g + 16g + 1g)

= 40g/mol

Since, n = mass in grams / molar mass

n = 23.0g / 40.0g/mol

n = 0.575 mole

Volume of NaOH solution (v) = 0.5 L

Concentration of NaOH solution (c) = ?

Since concentration (c) is obtained by dividing the amount of solute dissolved by the volume of solvent, hence

c = n / v

c = 0.575 mole / 0.5L

c = 1.15 M (M stands for molarity and gives the concentration in moles per litres)

Thus, the molarity of the solution is 1.15M

5 0
3 years ago
How are cellular respiration and photosynthesis different? (4 points)
avanturin [10]
Answer b, cellular and photosynthesis are reversible
3 0
3 years ago
An element contains 25 electrons. How many empty orbitals are in its valence energy level?
frez [133]

Answer:

No empty orbitals but five unpaired electrons.

Explanation:

Hello there!

In this case, when going over the empty orbitals in valence energy levels, it is required to develop the electron configuration in order to appropriately describe the element. In such a way, for the element whose atomic number is 25 because it equals the number of electrons, we obtain:

1s^2,2s^2,2p^6,3s^2,3p^6,4s^2,3d^5

It means that the d orbital is half filled which means that it does not have any empty orbital wherein two electrons are contained per orbital. Moreover, we can find 5 unpaired electrons according to:

↑  ↑  ↑  ↑  ↑

Which is the orbital notation.

Best regards!

4 0
3 years ago
A compound is found to contain 9.224 % boron and 90.74 % chlorine by mass. To answer the question, enter the elements in the ord
motikmotik

Answer:

Empirical formula of a compound means that it provides simplest ratio of whole number.

Explanation:

QUESTION-1

Given,

Mass of boron and chlorine is 9.224% and 90.74%

Then, Boron- 9.224×\frac{1}{10.811}=0.85 mol   (Mass value of boron=10.811)

      Chlorine- 90.74×\frac{1}{35.453}=2.559 mol.(Mass value of chlorine=35.453)

By dividing the mole value with smallest number, then we will get-

Boron-\frac{0.85}{0.68}=1

Chlorine-\frac{2.559}{0.68}=3

The empirical formula of the compound is -BCl₃

QUESTION-2

BCl₃⇄B+ 3.Cl  (Putting the actual mass number of boron and chlorine we will get-)

           1×(10.811)+ 3×(35.453)=117.17g/mol

Molecular formula=\frac{molar mass}{empirical formula}

                               = \frac{117.17}{117}≈1

  The molecular formula is same with the empirical formula.

3 0
3 years ago
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