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Afina-wow [57]
3 years ago
5

An object has a mass of 441 g and a volume of 10cm3

Chemistry
1 answer:
expeople1 [14]3 years ago
3 0

Answer:

44.1g/cm3

Explanation:

d=m/v

=441/10

=44.1 g/cm3

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NH3 is a weak alkali that does not dissociate fully into its solution. Which of the following is true about NH3?
ad-work [718]
<h2>NH3 is a weak alkali that does not dissociate fully into its solution. Which of the following is true about NH3? </h2><h2> </h2><h2>A. It has a very low pH. </h2><h2>B. It's dissociation is a reversible reaction. </h2><h2>C. It has a high H+ concentration. </h2><h2>D. It will release all of its OH- ions.</h2>

Explanation:

<h3>NH3 is a weak alkali that does not dissociate fully into its solution: It's dissociation is a reversible reaction. </h3><h3></h3>

Reactions are also :

  • Reversible
  • Irreversible

Reversible reaction

A reaction in which products can combine back to give reactants under same given condition .

Example : N₂+H₂-------NH₃

Irreversible reaction

A reaction in which the products cant combine back to give reactants under same set of conditions .

Example : Burning of paper

3 0
3 years ago
Read 2 more answers
These are 3 questions in one but i really need help i put the most points i could give and i’ll give Brainliest !!
kap26 [50]

Answer:

1) 950 mL

2) 625 mmHg

3) 426 mL

Explanation:

1) This is the relationship between pressure and volume. This relationship looks like this:

P1*V1 = P2*V2

This means the first pressure times the initial volume is equal to the second pressure times the second volume. We are solving for the second volume. First, convert the mmHg to atm and the mL to L.

1 L * 1 atm = 1.053 atm * X

X = 0.95 L or 950 mL

2) This is the same concept as the last one. :) We don't have to convert the mmHg to atm since the answer wants it in mmHg.

750 mmHg * 0.25 L = 0.3 L * X

X = 625 mmHg

3) The relationship between volume and temperature is similar to the one between pressure and temperature (like the problem in your last question). Remember to convert degrees C to Kelvin and mL to L.

V1 / T1 = V2 / T2

0.4 L / 303 K = X / 323 K

X = 0.426 L pr 426 mL

These problems become much easier once you learn the relationships between the different variables (temp, pressure, volume, etc.) When you have a problem like this, I like to first determine what relationship I am dealing with and then write out what I have and what I am solving for. This helps with organizing the problem. Then just solve it like a normal algebra problem. Always remember to convert temp to Kelvin, mL to L, and pressure to atm (unless it wants it in a different unit, then just make sure all the units match).

Good luck with you studies! :)

4 0
4 years ago
The reaction below shows how silver chloride can be synthesized.
Vesna [10]
AgNO_{3} _{(aq)}  +  NaCl_{(aq)}  ----\ \textgreater \   NaNO_{3} _{(aq)}   +  AgCl _{(aq)}
mole ration of AgCl _{(aq)}  :  AgNO_{3} _{(aq)} =  1 : 1

∴ if moles of  AgNO_{3} _{(aq)} =  15.0 mol
then   "      "     AgCl _{(aq)}  =  15.0 mol

∴  the Option  2 is the answer.

5 0
4 years ago
Read 2 more answers
A mixture of 15.0 g of the anesthetic halothane (C2HBrClF3 197.4 g/mol) and 22.6 g of oxygen gas has a total pressure of 862 tor
AlexFokin [52]

Answer : The partial pressure of C_2HBrClF_3 and O_2 are, 84 torr and 778 torr respectively.

Explanation : Given,

Mass of C_2HBrClF_3 = 15.0 g

Mass of O_2 = 22.6 g

Molar mass of C_2HBrClF_3 = 197.4 g/mole

Molar mass of O_2 = 32 g/mole

First we have to calculate the moles of C_2HBrClF_3 and O_2.

\text{Moles of }C_2HBrClF_3=\frac{\text{Mass of }C_2HBrClF_3}{\text{Molar mass of }C_2HBrClF_3}=\frac{15.0g}{197.4g/mole}=0.0759mole

and,

\text{Moles of }O_2=\frac{\text{Mass of }O_2}{\text{Molar mass of }O_2}=\frac{22.6g}{32g/mole}=0.706mole

Now we have to calculate the mole fraction of C_2HBrClF_3 and O_2.

\text{Mole fraction of }C_2HBrClF_3=\frac{\text{Moles of }C_2HBrClF_3}{\text{Moles of }C_2HBrClF_3+\text{Moles of }O_2}=\frac{0.0759}{0.0759+0.706}=0.0971

and,

\text{Mole fraction of }O_2=\frac{\text{Moles of }O_2}{\text{Moles of }C_2HBrClF_3+\text{Moles of }O_2}=\frac{0.706}{0.0759+0.706}=0.903

Now we have to partial pressure of C_2HBrClF_3 and O_2.

According to the Raoult's law,

p^o=X\times p_T

where,

p^o = partial pressure of gas

p_T = total pressure of gas

X = mole fraction of gas

p_{C_2HBrClF_3}=X_{C_2HBrClF_3}\times p_T

p_{C_2HBrClF_3}=0.0971\times 862torr=84torr

and,

p_{O_2}=X_{O_2}\times p_T

p_{O_2}=0.903\times 862torr=778torr

Therefore, the partial pressure of C_2HBrClF_3 and O_2 are, 84 torr and 778 torr respectively.

6 0
3 years ago
Mrs. Johnson's science class perform an experiment. They combine two substances that react with lots of fizzing and bubbling. Th
s2008m [1.1K]

Chris is correct because the reactants and products do not have to have the same mass, but they do have to weigh the same.  Is the correct answer:) Hannah is right because the mass of the reactants was different than the mass of the products. is incorrect

3 0
3 years ago
Read 2 more answers
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