Which integration phase moves the recovered person to the continental united states

A fine of 50.0 mL of 0.0900 M CaCl2 reacts with excess sodium carbonate to give 0.366 g of calcium carbonate precipitate. What i

In-s [12.5K]

Answer:

81.26% is the percent yield

Explanation:

Based on the reaction:

CaCl₂ + Na₂CO₃ → 2NaCl + CaCO₃

Where 1 mole of CaCl₂ in excess of sodium carbonate produces 1 mole of calcium carbonate.

To solve this question we must find the moles of CaCl2 added = Moles CaCO₃ produced (Theoretical yield). The percent yield is:

Actual yield (0.366g) / Theoretical yield * 100

Moles CaCl₂ = Moles CaCO₃:

0.0500L * (0.0900moles / L) = 0.00450 moles of CaCO₃

Theoretical mass -Molar mass CaCO₃ = 100.09g/mol-:

0.00450 moles of CaCO₃ * (100.09g / mol) = 0.450g of CaCO₃

Percent yield = 0.366g / 0.450g * 100

81.26% is the percent yield

3 0

2 years ago

The equilibrium 2NO(g)+Cl2(g)⇌2NOCl(g) is established at 500 K. An equilibrium mixture of the three gases has partial pressures

QveST [7]

Answer:

For A: The $K_p$ for the given reaction is $4.0\times 10^1$

For B: The $K_c$ for the given reaction is 1642.

Explanation:

The given chemical reaction follows:

$2NO(g)+Cl_2(g)\rightleftharpoons 2NOCl(g)$

For A:

The expression of $K_p$ for the above reaction follows:

$K_p=\frac{(p_{NOCl})^2}{(p_{NO})^2\times p_{Cl_2}}$

We are given:

$p_{NOCl}=0.24 atm\\p_{NO}=9.10\times 10^{-2}atm=0.0910atm\\p_{Cl_2}=0.174atm$

Putting values in above equation, we get:

$K_p=\frac{(0.24)^2}{(0.0910)^2\times 0.174}\\\\K_p=4.0\times 10^1$

Hence, the $K_p$ for the given reaction is $4.0\times 10^1$

For B:

Relation of $K_p$ with $K_c$ is given by the formula:

$K_p=K_c(RT)^{\Delta ng}$

where,

$K_p$ = equilibrium constant in terms of partial pressure = $4.0\times 10^1$

$K_c$ = equilibrium constant in terms of concentration = ?

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

T = temperature = 500 K

$\Delta ng$ = change in number of moles of gas particles = $n_{products}-n_{reactants}=2-3=-1$

Putting values in above equation, we get:

$4.0\times 10^1=K_c\times (0.0821\times 500)^{-1}\\\\K_c=\frac{4.0\times 10^1}{(0.0821\times 500)^{-1})}=1642$

Hence, the $K_c$ for the given reaction is 1642.

7 0

2 years ago

How much precipitation is shown in the rain gauge below?

VARVARA [1.3K]

Answer: b

Explanation:

7 0

2 years ago

Products: 1-methylcyclohexene, 3-methylcyclohexene, methylenecyclohexane

Lostsunrise [7]

Answer:

See explanation

Explanation:

The reaction that we are considering here is quite a knotty reaction. It is difficult to decide if the mechanism is actually E1 or E2 since both are equally probable based on the mass of scientific evidence regarding this reaction. However, we can easily assume that the methylenecyclohexane was formed by an E1 mechanism.

Looking at the products, one could convincingly assert that the reaction leading to the formation of the two main products proceeds via an E1 mechanism with the formation of a carbocation intermediate as has been shown in mechanism attached to this answer. Possible rearrangement of the carbocation yields the 3-methylcyclohexene product.

8 0

2 years ago

3. Indicate the type of reaction represented by each equation:

BartSMP [9]

3. Is double replacement

6 0

3 years ago