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3 years ago

Complete the following single replacement reaction. If they don’t react, just write “NR”

Chemistry

1 answer:

stiv31 [10]3 years ago

4 0

We have to complete all the given reactions.

1. Fe(s) + CuCl₂ → Cu + FeCl₂

2. Cu(s) + FeCl₂(aq) → NR (no reaction takes place)

3. K(s) + NiBr2(aq) → NR (no reaction takes place)

4. Ni(s) + KBr(aq) → K + NiBr₂

5. Zn(s) + Ca(NO₃)₂(aq) → NR (no reaction)

6. Ca(s) + Zn(NO₃)₂(aq) → Zn(s) + Ca(NO₃)₂(aq)

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How many phosphorus atoms are contained in 158 kg of phosphorus? how many phosphorus atoms are contained in 158 kg of phosphorus

zheka24 [161]

To determine the number of phosphorus atoms from a given mass, we need to determine the number of moles of the substance by dividing the molar mass which for in this case is equal to 123.88 g/mol for P4. Then, we multiply Avogadro's number. It <span>represents the number of units in one mole of any substance. This has the value of 6.022 x 10^23 units / mole.

mole P4 = 158 kg P4 ( 1000 g / 1 kg ) ( 1 mol / 123.88 g ) = 1275.43 mol P4
# of P4 atoms = 1275.43 mol P4 ( 6.022 x 10^23 atoms P4 / 1 mol P4 ) = 7.68x10^26 atoms P4</span>

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3 years ago

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Your friend had an upset stomach caused by indigestion. His mother explained to him

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2 years ago

You have a 5-liter container with 1.30×10^24 molecules of ammonia gas (NH3) at STP.

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3 years ago

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A 99.8 mL sample of a solution that is 12.0% KI by mass (d: 1.093 g/mL) is added to 96.7 mL of another solution that is 14.0% Pb

andre [41]

Answer:

$m_{PbI_2}=18.2gPbI_2$

Explanation:

Hello,

In this case, we write the reaction again:

$Pb(NO_3)_2(aq) + 2 KI(aq)\rightarrow PbI_2(s) + 2 KNO_3(aq)$

In such a way, the first thing we do is to compute the reacting moles of lead (II) nitrate and potassium iodide, by using the concentration, volumes, densities and molar masses, 331.2 g/mol and 166.0 g/mol respectively:

$n_{Pb(NO_3)_2}=\frac{0.14gPb(NO_3)_2}{1g\ sln}*\frac{1molPb(NO_3)_2}{331.2gPb(NO_3)_2} *\frac{1.134g\ sln}{1mL\ sln} *96.7mL\ sln\\\\n_{Pb(NO_3)_2}=0.04635molPb(NO_3)_2\\\\n_{KI}=\frac{0.12gKI}{1g\ sln}*\frac{1molKI}{166.0gKI} *\frac{1.093g\ sln}{1mL\ sln} *99.8mL\ sln\\\\n_{KI}=0.07885molKI$

Next, as lead (II) nitrate and potassium iodide are in a 1:2 molar ratio, 0.04635 mol of lead (II) nitrate will completely react with the following moles of potassium nitrate:

$0.04635molPb(NO_3)_2*\frac{2molKI}{1molPb(NO_3)_2} =0.0927molKI$

But we only have 0.07885 moles, for that reason KI is the limiting reactant, so we compute the yielded grams of lead (II) iodide, whose molar mass is 461.01 g/mol, by using their 2:1 molar ratio:

$m_{PbI_2}=0.07885molKI*\frac{1molPbI_2}{2molKI} *\frac{461.01gPbI_2}{1molPbI_2} \\\\m_{PbI_2}=18.2gPbI_2$

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