IMPORTANT IMPORTANT NEED TO GET OFF SOON BC ITS MY BROTHERS BIRTHDAY PLEASEEEE NEED THIS EXTREMELY BAD ❤️ FOR A BIG GRADE TOO

Chemistry

2 answers:

frosja888 [35]3 years ago

8 0

its d because it has to be diagnoly to the right

love history [14]3 years ago

8 0

Answer:

I think C

happy birthday to your brother though

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What is the mass of 6 atom(s) of copper in grams?

USPshnik [31]

Answer:

6.33×10¯²² g

Explanation:

From the question given above, the following data were obtained:

Number of atoms = 6 atoms

Mass of copper (Cu) =?

From Avogadro's hypothesis, we understood that:

6.02×10²³ atoms = 1 mole of Cu

But 1 mole of Cu = 63.5 g

Thus,

6.02×10²³ atoms = 63.5 g of Cu

Finally, we shall determine the mass of 6 atoms of copper. This can be obtained as illustrated below:

6.02×10²³ atoms = 63.5 g of Cu

Therefore,

6 atoms = (6 × 63.5) / 6.02×10²³

6 atoms = 6.33×10¯²² g of Cu

Therefore, the mass of 6 atoms of copper is 6.33×10¯²² g.

4 0

3 years ago

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:

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