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3 years ago

How many atoms of hydrogen-1 are in a 1.007-g sample?

Chemistry

1 answer:

Greeley [361]3 years ago

8 0

I think it's 2 I tried looking it up because I was not sure.

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riadik2000 [5.3K]

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3 years ago

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What is the specific latent heat of fusion of ice if it takes 863 kJ to convert 4.6 kg of ice into water at 0 C?

allochka39001 [22]

Answer:

$1^{f} =187.7 \frac{J}{kg}$

Explanation:

SO in order to calculate the specific latent heat of fusion, you need to remember the formula:

$1^{f} =\frac{Q}{m}$

Where $1^{f}$ representes the specific latent heart of fusion.

$Q$ represents the heat energy added, usually represented in kJ

$m$ represents the mass of the object, in kg.

Now that we have our formula we just have to put our values into the formula:

$1^{f} =\frac{Q}{m}$

$1^{f} =\frac{863kJ}{4.6kg}$

$1^{f} =187.7 \frac{J}{kg}$

SO our answer would be $1^{f} =187.7 \frac{J}{kg}$

7 0

3 years ago

I need help ASAP

vova2212 [387]

Answer:

1.4 g/cm3

Explanation:

Density = Mass/Volume

Mass = 21g

Volume = 15cm3

Density = 21/15 = 1.4

8 0

2 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]

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

<u>Explanation:</u>

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

Which sub-atomic particle is NOT found at the center of the atom?

Ostrovityanka [42]

All the positive charge of an atom is contained in the nucleus, and originates from the protons. Neutrons are neutrally-charged. Electrons, which are negatively-charged, are located outside of the nucleus.

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3 years ago

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