Is H2SO4 + NaOH > Na2SO + H2O balanced

Examine the images.

gavmur [86]

Answer:

4 th image

Explanation:

water is being boiled n the 4th image indicating that the change between liquids and gasses is water vapor.

5 0

3 years ago

Read 2 more answers

You carefully weigh out 11.00 g of caco3 powder and add it to 44.55 g of hcl solution. you notice bubbles as a reaction takes pl

Zinaida [17]

The bubbles that were observed after the mixing of the two substances is one of the products of the reaction. It is the carbon dioxide that is produced. To determine the mass of this gas produced, we need to remember the Law of conservation of mass where mass cannot be created or destroyed. With this, we can say that the total mass that goes in a process should be equal to the mass that is goes out of the process no matter what the reaction is. We do as follows:

Mass of reactants = mass of products
11.00 + 44.55 = 51.04 + mass of carbon dioxide
mass of carbon dioxide = 4.51 g

8 0

3 years ago

Which has a larger molar mass, salt (NaCl) or sugar (C12H22011) – how can you tell?

uysha [10]

Answer:

Explanation:

Molar mass of NaCl = (23+35.5)

Molar mass of NaCl = 58.5g/mol

Molar mass of C12H22011

= 12(12) + 22(1) + 16(11)

= 144 +22 + 176

= 342g/mol

4 0

3 years ago

Which of the following solutions is a good buffer system?A solution that is 0.10 M HCN and 0.10 M NaClA solution that is 0.10 M

patriot [66]

Answer:

A solution that is 0.10 M HCN and 0.10 M LiCN

Explanation:

A good buffer system contains a weak acid and its salt or a weak base and its salt.
In this case; A solution that is 0.10 M HCN and 0.10 M LiCN, would make a good buffer system.
HCN is a weak acid, while LiCN is a salt of the weak acid, that is, CN- conjugate of the acid.

8 0

3 years ago

I need help with 1,2,3, and 4

Schach [20]

Answer:

Problem 1: 1.85atm

Problem 2: 110mL

Problem 3: 290 mL

Problem 4: 1.14 atm

Explanation:

Problem 1

1. Data

a) P₁ = 3.25atm

b) V₁ = 755mL

c) P₂ = ?

d) V₂ = 1325 mL

r) T = 65ºC

2. Formula

Since the temeperature is constant you can use Boyle's law for idial gases:

$PV=constant\\\\P_1V_1=P_2V_2$

3. Solution

Solve, substitute and compute:

$P_1V_1=P_2V_2\\\\P_2=P_1V_1/V_2$

$P_2=3.25atm\times755mL/1325mL=1.85atm$

Problem 2

1. Data

a) V₁ = 125 mL

b) P₁ = 548mmHg

c) P₁ = 625mmHg

d) V₂ = ?

2. Formula

You assume that the temperature does not change, and then can use Boyl'es law again.

$P_1V_1=P_2V_2$

3. Solution

This time, solve for V₂:

$P_1V_1=P_2V_2\\\\V_2=P_1V_1/P_2$

Substitute and compute:

$V_2=548mmHg\times 125mL/625mmHg=109.6mL$

You must round to 3 significant figures:

$V_2=110mL$

Problem 3

1. Data

a) V₁ = 285mL

b) T₁ = 25ºC

c) V₂ = ?

d) T₂ = 35ºC

2. Formula

At constant pressure, Charle's law states that volume and temperature are inversely related:

$V/T=constant\\\\\\\dfrac{V_1}{T_1}=\dfrac{V_2}{T_2}$

The temperatures must be in absolute scale.

3. Solution

a) Convert the temperatures to kelvins:

T₁ = 25 + 273.15K = 298.15K

T₂ = 35 + 273.15K = 308.15K

b) Substitute in the formula, solve for V₂, and compute:

$\dfrac{V_1}{T_1`}=\dfrac{V_2}{T_2}\\\\\\\\\dfrac{285mL}{298.15K}=\dfrac{V_2}{308.15K}\\\\\\V_2=308.15K\times285mL/298.15K=294.6ml$

You must round to two significant figures: 290 ml

Problem 4

1. Data

a) P = 865mmHg

b) Convert to atm

2. Formula

You must use a conversion factor.

1 atm = 760 mmHg

Divide both sides by 760 mmHg

$\dfrac{1atm}{760mmHg}=\dfrac{760mmHg}{760mmHg}\\\\\\1=\dfrac{1atm}{760mmHg}$

3. Solution

Multiply 865 mmHg by the conversion factor:

$865mmHg\times \dfrac{1atm}{760mmHg}=1.14atm\leftarrow answer$

3 0

3 years ago