5. How many miles are there in 3.44 x 10^8 inches?

How much heat does your body lose when 6.35 g of sweat evaporates from your skin at 25 ∘C ? (Assume that the sweat is only water

ziro4ka [17]

Answer:

15.52 kj

Explanation:

H2 O mole wt ~ 18 gm

6.35 gm is 6.35 gm /18 gm/mole = .353 moles of H2O

.353 moles * 44 kj/mole =

5 0

1 year ago

A student mixes 33.0 mL of 2.70 M Pb ( NO 3 ) 2 ( aq ) with 20.0 mL of 0.00157 M NaI ( aq ) . How many moles of PbI 2 ( s ) prec

GalinKa [24]

Answer: The moles of precipitate (lead (II) iodide) produced is $1.57\times 10^{-5}$ moles

Explanation:

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

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$ .....(1)

For lead (II) nitrate:

Molarity of lead (II) nitrate solution = 2.70 M

Volume of solution = 33.0 mL = 0.033 L (Conversion factor: 1 L = 1000 mL)

Putting values in equation 1, we get:

$2.70M=\frac{\text{Moles of lead (II) nitrate}}{0.033L}\\\\\text{Moles of lead (II) nitrate}=(2.70mol/L\times 0.0330L)=0.0891mol$

For NaI:

Molarity of NaI solution = 0.00157 M

Volume of solution = 20.0 mL = 0.020 L

Putting values in equation 1, we get:

$0.00157M=\frac{\text{Moles of NaI}}{0.020L}\\\\\text{Moles of NaI}=(0.00157mol/L\times 0.0200L)=3.14\times 10^{-5}mol$

For the given chemical reaction:

$Pb(NO_3)_2(aq.)+2NaI(aq.)\rightarrow PbI_2(s)+2NaNO_3(aq.)$

By Stoichiometry of the reaction:

2 moles of NaI reacts with 1 mole of lead (II) nitrate

So, $3.14\times 10^{-5}$ moles of NaI will react with = $\frac{1}{2}\times 3.14\times 10^{-5}=1.57\times 10^{-5}mol$ of lead (II) nitrate

As, given amount of lead (II) nitrate is more than the required amount. So, it is considered as an excess reagent.

Thus, NaI is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

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

So, $3.14\times 10^{-5}$ moles of NaI will produce = $\frac{1}{2}\times 3.14\times 10^{-5}=1.57\times 10^{-5}moles$ of lead (II) iodide

Hence, the moles of precipitate (lead (II) iodide) produced is $1.57\times 10^{-5}$ moles

4 0

2 years ago

Calculate the number of moles in 8 g of water.

polet [3.4K]

Answer:

refer the above attachment

4 0

1 year ago

A. Describe the molecule chlorine dioxide, CIO in terms of three possible resonance structures.

cupoosta [38]

Answer:

See explanation

Explanation:

The compound ClO2 has 19 valence electrons. ClO2 is a bent molecule with tetrahedral electron pair geometry but has two lone pairs of electrons. This is indicated by the presence of four electron pairs on the outermost shell of the central atom.

The molecule has an odd number of valence electrons, hence, it is generally regarded as a paramagnetic radical. None of the proposed Lewis structures for the molecule is satisfactory because none of them obeys the octet rule.

From the images attached, one can easily see that the electron dots around the oxygen and chlorine atoms does not satisfy the octet rule in all the resonance structures shown.

5 0

2 years ago

Help me plz, I need help on this.

kirza4 [7]

Omg i lost everything ugh
To do it again

1. 12g+2(16g)= 44g/mol
25.01/ 44g/mol= .... mol

2. 14g+3(1g)= 17g/mol
34.05g/ 17g/mol=.... mol

3. 23g+1g+ 12g+ 3(16g)= 84g/mol
17.31g/ 84g/mol=.... mol

4. 6(12g)+12(1g)+6(16g)= 180g/mol
123.44g/ 180g/mol=.... mol

5. 23g+16g+1g= 40g/mol
2.2mol x 40g/mol= .... g

6. 2(35g)= 71g/mol
4.5mol x 71g/mol= .... g

7. 137g+ 2(14g)+ 6(16g)= 261g/mol
0.002mol x 261g/mol= ....g

8. 2(56g)+ 3(32g)+ 12(16g)= 400g/mol
5.4mol x 400g/mol=.... g

I cant believe i had to do this all over

4 0

2 years ago