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2 years ago

4. An atom is neutral because it has the same number of protons as it has:

Chemistry

1 answer:

stealth61 [152]2 years ago

7 0

Neutrons the way this question is setup the atom would not be able to be neutral

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What happens to the pressure if the gas volume is cut in half while n and I are held constant?

N76 [4]

Answer: You can use Boyle's law, which states that pressure is inversely related to volume when other variables are held constant. If the final pressure of a gas is half of the initial, the volume must double if temperature is to remain the same.

Explanation:

5 0

3 years ago

Select all that apply. which of the following are density labels? kg/l g/m g/ml cm^3/g

Strike441 [17]

I would say that only the only density label is cm^3/g

4 0

2 years ago

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If the freezing point of an aqueous 0.10 m glucose solution is −x°c, what is the approximate freezing point of a 0.10 m nacl sol

maks197457 [2]

Answer is: the approximate freezing point of a 0.10 m NaCl solution is -2x°C.
Van't Hoff factor (i) for NaCl solution is approximately 2.
Van't Hoff factor (i) for glucose solution is 1.
Change in freezing point from pure solvent to solution: ΔT = i · Kf · m.
Kf - molal freezing-point depression constant for water is 1,86°C/m.
m - molality, moles of solute per kilogram of solvent.
Kf and molality for this two solutions are the same, but Van't Hoff factor for sodium chloride is twice bigger, so freezing point is twice bigger.

8 0

3 years ago

The air bags in cars are inflated when a collision triggers the explosive, highly exothermic decomposition of sodium azide (NaN3

Oksanka [162]

Answer : The mass of $NaN_3$ required is, 166.4 grams.

Explanation :

First we have to calculate the moles of nitrogen gas.

Using ideal gas equation:

$PV=nRT$

where,

P = Pressure of $N_2$ gas = 1.00 atm

V = Volume of $N_2$ gas = 113 L

n = number of moles $N_2$ = ?

R = Gas constant = $0.0821L.atm/mol.K$

T = Temperature of $N_2$ gas = $85^oC=273+85=358K$

Putting values in above equation, we get:

$1.00atm\times 113L=n\times (0.0821L.atm/mol.K)\times 358K$

$n=3.84mol$

Now we have to calculate the moles of sodium azide.

The balanced chemical reaction is,

$2NaN_3(s)\rightarrow 2Na(s)+3N_2(g)$

From the balanced reaction we conclude that

As, 3 mole of $N_2$ produced from 2 mole of $NaN_3$

So, 3.84 moles of $N_2$ produced from $\frac{2}{3}\times 3.84=2.56$ moles of $NaN_3$

Now we have to calculate the mass of $NaN_3$

$\text{ Mass of }NaN_3=\text{ Moles of }NaN_3\times \text{ Molar mass of }NaN_3$

Molar mass of $NaN_3$ = 65 g/mole

$\text{ Mass of }NaN_3=(2.56moles)\times (65g/mole)=166.4g$

Therefore, the mass of $NaN_3$ required is, 166.4 grams.

3 0

3 years ago

If x is a string, then x = new String("OH"); and x = "OH"; will accomplish the same thing. Group of answer choices True False

Vinvika [58]

Answer:

True is the correct answer to the above question.

Explanation:

If x is a string then it can be assigned by the help of two ways in java:
By the help of constructor:- When we write " x = new String("OH");", then it will create a pass a string "OH" into the constructor. It is because the String is a class in java and x is an object created by the constructor of the String class.
With the help of assigning: The "x= OH", which assigns the value of x which is an object of String class it can also use the constructor to initialize the "OH" string on the class.
The above question states the two scenarios which are defined above. Hence the question statement is true.

5 0

3 years ago

Read 2 more answers