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4 years ago

Aqueous solutions of copper (II) bromide and silver (1) acetate react to form solid

Chemistry

1 answer:

Sladkaya [172]4 years ago

4 0

Answer:

5.63 g

Explanation:

Step 1: Write the balanced equation

CuBr₂(aq) + 2 AgCH₃CO₂(aq) ⇒ 2 AgBr(s) + Cu(CH₃CO₂)₂(aq)

Step 2: Calculate the reacting moles of copper (II) bromide

30.0 mL of 0.499 M CuBr₂ react. The reacting moles of CuBr₂ are:

$0.0300L \times \frac{0.499mol}{L} = 0.0150mol$

Step 3: Calculate the moles formed of silver (I) bromide

The molar ratio of CuBr₂ to AgBr is 1:2. The moles formed of AgBr are 2/1 × 0.0150 mol = 0.0300 mol.

Step 4: Calculate the mass corresponding to 0.0300 mol of AgBr

The molar mass of AgBr is 187.77 g/mol.

$0.0300 mol \times \frac{187.77g}{mol} = 5.63 g$

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How many moles of CO2 are produced

Sergio039 [100]

Answer:

30.0 mol CO₂

Explanation:

C₃H₈ + 5O₂ → 3CO₂ + 4H₂O

To answer this problem we need to convert moles of C₃H₈ into moles of CO₂: We'll do that by using the <u>stoichiometric coefficients</u>, using a conversion factor that has C₃H₈ moles in the denominator and CO₂ moles in the numerator:

10.0 mol C₃H₈ * $\frac{3molCO_2}{1molC_3H_8}$ = 30.0 mol CO₂

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3 years ago

How does the body react when the outside temperature gets too hot

mr_godi [17]

When temperatures rise, the body reacts by increasing blood flow to the skin's surface, taking the heat from within the body to the surface. This means sweat. ... If, as in the UK, our skin temperature is warmer than the external temperature we are also able to lose heat to the environment, termed 'dry heat loss'.

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3 years ago

Calculate the density of O2(g) at 415 K and 310 bar using the ideal gas and the van der Waals equations of state. Use a numerica

Lera25 [3.4K]

Answer:

Explanation:

From the given information:

The density of O₂ gas = $d_{ideal} = \dfrac{P\times M}{RT}$

here:

P = pressure of the O₂ gas = 310 bar

= $310 \ bar \times \dfrac{0.987 \ atm}{1 \ bar}$

= 305.97 atm

The temperature T = 415 K

The rate R = 0.0821 L.atm/mol.K

molar mass of O₂ gas = 32 g/mol

∴

$d_{ideal} = \dfrac{305.97 \ \times 32}{0.0821 \times 415}$

$d_{ideal}$ = 287.37 g/L

To find the density using the Van der Waal equation

Recall that:

the Van der Waal constant for O₂ is:

a = 1.382 bar. L²/mol² &

b = 0.0319 L/mol

The initial step is to determine the volume = Vm

The Van der Waal equation can be represented as:

$P =\dfrac{RT}{V-b}-\dfrac{a}{V^2}$

where;

R = gas constant (in bar) = 8.314 × 10⁻² L.bar/ K.mol

Replacing our values into the above equation, we have:

$310 =\dfrac{0.08314\times 415}{V-0.0319}-\dfrac{1.382}{V^2}$

$310 =\dfrac{34.5031}{V-0.0319}-\dfrac{1.382}{V^2}$

$310V^3 -44.389V^2+1.382V-0.044=0$

After solving;

V = 0.1152 L

∴

$d_{Van \ der \ Waal} = \dfrac{32}{0.1152}$

$d_{Van \ der \ Waal}$ = 277.77 g/L

We say that the repulsive part of the interaction potential dominates because the results showcase that the density of the Van der Waals is lesser than the density of ideal gas.

5 0

3 years ago

A piece of metal with a mass of 142 g is placed in a 50 mL graduated cylinder. The water level rises from 20 mL to 40 mL. What i

zheka24 [161]

Explanation:

The end goal of this question is to find density. Density can be found using the following formula:

D=m/v

D=Density

m=mass

v=volume

Therefore we need mass and volume to find density. Mass is already given in the question as 142g. All we have left is volume. When a solid is placed in water in a graduated cylinder the volume goes up. This means if you subtract the volume after the metal was placed from before it was placed, you can find the volume of the metal. Therefore the volume of the metal is 40mL-20mL=20mL. However typically, volume is found in liters so the volume should be .04L-.02L=0.02L. Now that we have volume and mass we can find density using the formula d=m/v. So 142/.02=7,100 grams per liget or 7.1 kilograms per liter. This answer seems unrealistic so I would double check with your teacher about the question.

Hope I helped, 2Trash4U

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3 years ago

4. When is distillation used?

Arlecino [84]

Answer:

Uses of Distillation

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3 years ago