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3 years ago

A tablespoon of very hot water is added to a pint of very cold water. What will happen to the temperature of the water?

1 answer:

6 0

Well although the amount of hot water is slight, and there's much more cold water, I would assume that the water would more than likely increase in temperature.... even though it may be slight

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Which statement holds true when the molar mass of hydrogen is 2.0 grams and that of helium is 4.0 grams?

ehidna [41]

I think you forgot to give the choices along with your question. I am answering the question based on my research and knowledge. "Helium effuses 0.71 times faster than hydrogen" is the statement that holds true when the molar mass of hydrogen is 2.0 grams and that of helium is 4.0 grams. I hope the answer has helped you.

7 0

3 years ago

ASAP will give brainliest!!!

Ahat [919]

Answer:

For me

Explanation:

It's Cu because other compound contains negative radicals

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2 years ago

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Which statement about the orinoco river is true.

WITCHER [35]

Hello,

I think the answer is B) It is second longest river in south america

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3 years ago

Which type of reaction produces the most dangerous radioactive waste?

k0ka [10]

The correct answer to the question above is fussion of uranium. The fussion of uranium is the only type of reaction in which produces the most dangerous radioactive waste. The fusion of uranium means the binding of their atoms and produce a radioactive waste.

7 0

3 years ago

Write the balanced equation for the reaction of aqueous Pb ( ClO 3 ) 2 Pb(ClO3)2 with aqueous NaI . NaI. Include phases. chemica

FinnZ [79.3K]

Answer: The mass of precipitate (lead (II) iodide) that will form is 119.89 grams

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Molarity of NaI solution = 0.130 M

Volume of solution = 0.400 L

Putting values in above equation, we get:

$0.130M=\frac{\text{Moles of NaI}}{0.400L}\\\\\text{Moles of NaI}=(0.130mol/L\times 0.400L)=0.52mol$

The balanced chemical equation for the reaction of lead chlorate and sodium iodide follows:

$Pb(ClO_3)_2(aq.)+2NaI(aq.)\rightarrow PbI_2(s)+2NaClO_3(aq.)$

The precipitate (insoluble salt) formed is lead (II) iodide

By Stoichiometry of the reaction:

2 moles of NaI produces 1 mole of lead (II) iodide

So, 0.52 moles of NaI will produce = $\frac{1}{2}\times 0.52=0.26mol$ of lead (II) iodide

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Moles of lead (II) iodide = 0.26 moles

Molar mass of lead (II) iodide = 461.1 g/mol

Putting values in above equation, we get:

$0.26mol=\frac{\text{Mass of lead (II) iodide}}{461.1g/mol}\\\\\text{Mass of lead (II) iodide}=(0.26mol\times 461.1g/mol)=119.89g$

Hence, the mass of precipitate (lead (II) iodide) that will form is 119.89 grams

4 0

3 years ago