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3 years ago

6

Which factor below does NOT affect how fast a solute dissolves in a solvent? A) Quantity of solute B) Particle size of solvent C

) Temperature Eliminate D) Stirring

2 answers:

tamaranim1 [39]3 years ago

5 0

Quantity of solute, I hope I am right

prohojiy [21]3 years ago

4 0

I think the particle size is irrelevant or maybe quantity of solute. either f the two might be the answer. I could also be wrong on both opinions.

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Can anyone help me solve this chemistry question?

ollegr [7]

139.850 there ya go hope u ace it

4 0

4 years ago

Automobile airbags contain solid sodium azide, NaN 3 , that reacts to produce nitrogen gas when heated, thus inflating the bag.

Yuri [45]

<u>Answer:</u> The value of work for the system is -935.23 J

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Given mass of sodium azide = 16.5 g

Molar mass of sodium azide = 65 g/mol

Putting values in above equation, we get:

$\text{Moles of sodium azide}=\frac{16.5g}{65g/mol}=0.254mol$

The given chemical equation follows:

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

By Stoichiometry of the reaction:

2 moles of sodium azide produces 3 moles of nitrogen gas

So, 0.254 moles of sodium azide will produce = $\frac{3}{2}\times 0.254=0.381mol$ of nitrogen gas

To calculate volume of the gas given, we use the equation given by ideal gas, which follows:

$PV=nRT$

where,

P = pressure of the gas = 1.00 atm

V = Volume of the gas = ?

T = Temperature of the gas = $22^oC=[22+273]K=295K$

R = Gas constant = $0.0821\text{ L. atm }mol^{-1}K^{-1}$

n = number of moles of nitrogen gas = 0.381 moles

Putting values in above equation, we get:

$1.00atm\times V=0.381mol\times 0.0821\text{ L atm }mol^{-1}K^{-1}\times 295K\\\\V=\frac{0.381\times 0.0821\times 295}{1.00}=9.23L$

To calculate the work done for expansion, we use the equation:

$W=-P\Delta V$

We are given:

P = pressure of the system = $1atm=1.01325\times 10^5Pa$ (Conversion factor: 1 atm = 101325 Pa)

$\Delta V$ = change in volume = $9.23L=9.23\times 10^{-3}m^3$ (Conversion factor: $1m^3=1000L$ )

Putting values in above equation, we get:

$W=-1.01325\times 10^5Pa\times 9.23\times 10^{-3}m^3\\\\W=-935.23J$

Hence, the value of work for the system is -935.23 J

7 0

4 years ago

How many grams of aluminum hydroxide are obtained from 13.7 g of aluminum sulfide??

Vladimir [108]

<span>For that we need to know something about the chemistry aluminum sulfide and aluminum hydroxide. Aluminum sulfide will react with water to make insoluble aluminum hydroxide and hydrogen sulfide gas. Therefore, we need a balanced chemical equation.

Al2S3(s) + 6HOH --> 2Al(OH)3(s) + 3H2S(g)
17.7g ........................ ???g

17.7g Al2S3 x (1 mol Al2S3 / 150.3 g Al2S3) x (2 mol Al(OH)3 / 1 mol Al2S3) x (78.0g Al(OH)3 / 1 mol Al(OH)3) = 18.4 g Al(OH)3 ...... to three significant digits.</span>

8 0

3 years ago

Read 2 more answers

How to balance the equation HNO3 + Ca(OH)2 --> Ca(NO3)2 + H20

Colt1911 [192]

2HNO3 + Ca(OH)2 —> Ca(NO3)2 + 2H2O

3 0

3 years ago

If 455 J of heat is transferred to 25.0g of water at 45.0 degrees Celsius, what is the final temperature of the water?

masya89 [10]

Answer:49.3

Explanation:4.1j/g c * 25g * (t2-45c)=455j

T2-45c = 455j/4.1j/g c * 25g

455/104.6

45+4.3= 49.3 celsius

3 0

3 years ago