All categories

3 years ago

List three possible results of a double replacement reaction

Chemistry

1 answer:

Naddik [55]3 years ago

8 0

1. two compounds form
2. precipitate forms
3. ions exchanged between the two reactants

You might be interested in

What is an example of a soultion

Marrrta [24]

Answer:

Did you mean solution??

In chemistry, a solution is considered as a special type of homogeneous mixture composed of two or more substances. In such a mixture, a solute is a substance dissolved in another substance, known as a solvent.

Eg: sweet tea or coffee (sugar added to solution)

bleach (sodium hypochlorite dissolved in water)

HOPE IT HELPS :)

3 0

3 years ago

How many moles of HCl are present in 1.30 L of a 0.50 M solution

Shtirlitz [24]

Answer:

0.65moles of HCl was produced

4 0

3 years ago

What is the coefficient of the H2O in the balanced chemical

Ivahew [28]

Answer:

Al2O3 + H2SO4 = Al2(SO4)3 + H2O

Explanation:

4 0

3 years ago

An ideal gas (C}R), flowing at 4 kmol/h, expands isothermally at 475 Kfrom 100 to 50 kPa through a rigid device. If the power pr

Zina [86]

<u>Answer:</u> The rate of heat flow is 3.038 kW and the rate of lost work is 1.038 kW.

<u>Explanation:</u>

We are given:

$C_p=\frac{7}{2}R\\\\T=475K\\P_1=100kPa\\P_2=50kPa$

Rate of flow of ideal gas , n = 4 kmol/hr = $\frac{4\times 1000mol}{3600s}=1.11mol/s$ (Conversion factors used: 1 kmol = 1000 mol; 1 hr = 3600 s)

Power produced = 2000 W = 2 kW (Conversion factor: 1 kW = 1000 W)

We know that:

$\Delta U=0$ (For isothermal process)

So, by applying first law of thermodynamics:

$\Delta U=\Delta q-\Delta W$

$\Delta q=\Delta W$ .......(1)

Now, calculating the work done for isothermal process, we use the equation:

$\Delta W=nRT\ln (\frac{P_1}{P_2})$

where,

$\Delta W$ = change in work done

n = number of moles = 1.11 mol/s

R = Gas constant = 8.314 J/mol.K

T = temperature = 475 K

$P_1$ = initial pressure = 100 kPa

$P_2$ = final pressure = 50 kPa

Putting values in above equation, we get:

$\Delta W=1.11mol/s\times 8.314J\times 475K\times \ln (\frac{100}{50})\\\\\Delta W=3038.45J/s=3.038kJ/s=3.038kW$

Calculating the heat flow, we use equation 1, we get:

[ex]\Delta q=3.038kW[/tex]

Now, calculating the rate of lost work, we use the equation:

$\text{Rate of lost work}=\Delta W-\text{Power produced}\\\\\text{Rate of lost work}=(3.038-2)kW\\\text{Rate of lost work}=1.038kW$

Hence, the rate of heat flow is 3.038 kW and the rate of lost work is 1.038 kW.

4 0

3 years ago

How does this look? (Its for a science project about animal life

konstantin123 [22]

Answer:

why is there a cherry on its head?

Explanation:

You may want to add details/description to the photo as it is a little confusing

3 0

3 years ago

Read 2 more answers