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3 years ago

Is the law of conservative mass observed in this equation CaCO3 + 2HCI -->CaCI2 +H2O + CO2

Chemistry

1 answer:

pychu [463]3 years ago

8 0

Answer:

The law is observed in the given equation.

Explanation:

CaCO₃ + 2HCI → CaCI₂ +H₂O + CO₂

In order to find out if the law of conservative mass is followed, we need to count how many atoms of each element are there in both sides of the equation:

Ca ⇒ 1 on the left, 1 on the right.

C ⇒ 1 on the left, 1 on the right.

O ⇒ 3 on the left, 3 on the right.

H ⇒ 2 on the left, 2 on the right.

Cl ⇒ 2 on the left, 2 on the right.

As the numbers for all elements involved are the same, the law is observed in the given equation.

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3 years ago

Read 2 more answers

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:

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

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3 years ago

Is Lilium lancifolium Heterotroph, Autotroph, or Both

Sphinxa [80]

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3 years ago

The average rate of disappearance of ozone in the reaction 2o3(g) → 3o2(g) is found to be 7.25×10–3 atm over a certain interval

worty [1.4K]

<h3>Answer;</h3>

1.0875 x 10-2 atm

<h3>Explanation;</h3>

2O3(g) → 3O2(g)

rate = -(1/2)∆[O3]/∆t = +(1/3)∆[O2)/∆t

The average rate of disappearance of ozone ... is found to

be 7.25 × 10–3 atm over a certain interval of time.

This means (ignoring time)

∆[O3]/∆t = -7.25 × 10^–3 atm

(it is disappearing, thus the negative sign)

rate = -(1/2)∆[O3]/∆t

rate = -(1/2)*(-7.25 × 10^–3 atm)

= 3.625 × 10^–3 atm

Now use the other part of the expression:

rate = +(1/3)∆[O2)∆t

3.625 × 10–3 atm = +(1/3)∆[O2)/t

∆[O2)/∆t = (3)*(3.625× 10^–3 atm)

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2 years ago