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2 years ago

10

Merchants usually sell small nuts, washers, and bolts by weight (like jelly beans!) rather than by individually counting the ite

ms. Suppose a particular type of washer weighs 0.110g on the average. What would 100 such washers weigh? How many washers would there be in 100. g of washers?

1 answer:

PolarNik [594]2 years ago

6 0

Answer:

1. 11g

2. 909

Explanation:

To get the Weight of a 109 washers who weigh 0.110g each we have to find the ratio of Weight of one washer to 1 washer and multiply by 100

= 100 washers x 0.110g/1 washer

= 11g

Therefore the mass of 100 washers is 11 grams

2.

Then the number of washers =

100g x 1washer/0.110g

= 909

Therefore number of washers = 909

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Calculate the ratio of the velocity of helium atoms to the velocity of neon atoms at the same temperature.

o-na [289]

Answer:

vHe / vNe = 2.24

Explanation:

To obtain the velocity of an ideal gas you must use the formula:

v = √3RT / √M

Where R is gas constant (8.314 kgm²/s²molK); T is temperature and M is molar mass of the gas (4x10⁻³kg/mol for helium and 20,18x10⁻³ kg/mol for neon). Thus:

vHe = √3×8.314 kgm²/s²molK×T / √4x10⁻³kg/mol

vNe = √3×8.314 kgm²/s²molK×T / √20.18x10⁻³kg/mol

The ratio is:

vHe / vNe = √3×8.314 kgm²/s²molK×T / √4x10⁻³kg/mol / √3×8.314 kgm²/s²molK×T / √20.18x10⁻³kg/mol

vHe / vNe = √20.18x10⁻³kg/mol / √4x10⁻³kg/mol

<em>vHe / vNe = 2.24</em>

<em />

I hope it helps!

8 0

3 years ago

Boron sulfide, B2S3(s), reacts violently with water to form dissolved boric acid (H3BO3) and hydrogen sulfide gas.Express your a

ioda

Answer:

Explanation:

From the statement of the problem,

B₂S₃ $_{s}$ + H₂O $_{l}$ → H₃BO₃ $_{aq}$ + H₂S $_{g}$

B₂S₃ + H₂O → H₃BO₃ + H₂S

We that the above expression does not conform with the law of conservation of mass:

To obey the law, we need to derive a balanced reaction equation:

Let us use the mathematical method to obtain a balanced equation.

let the balanced equation be:

aB₂S₃ + bH₂O → cH₃BO₃ + dH₂S

where a, b, c and d will make the equation balanced.

Conservating B: 2a = c

S: 3a = d

H: 2b = 3c + 2d

O: b = 3c

if a = 1,

c = 2,

b = 6,

2d = 2(6) - 3(2) = 6, d = 3

Now we can input this into our equation:

B₂S₃ + 6H₂O → 2H₃BO₃ + 3H₂S

B₂S₃ $_{s}$ + 6H₂O $_{l}$ → 2H₃BO₃ $_{aq}$ + 3H₂S $_{g}$

4 0

2 years ago

What is the mass, in grams, of 28.58 mL of acetone?

Vedmedyk [2.9K]

<span>If you look up the density of Acetone (Propanone in IUPAC names) you will find it is 0.7925g/cm3. This is the same as 0.7925g/ml.

You can calculate mass using the equation:- mass = density x volume
In your example mass = 0.7925 x 28.40 = 22.51g</span><span>
I think That's right. Hope this helps!!! Good luck!</span>

7 0

2 years ago

Read 2 more answers

Complete and balance the chemical equations for the precipitation reactions, if any, between the following pairs of reactants, a

Crank

Explanation:

a. $Pb(NO_3)_2(aq) + Na_2SO_4(aq)$ → ?

$Pb(NO_3)_2(aq) + Na_2SO_4(aq)\rightarrow PbSO_4(s)+2NaNO_3(aq)$

$Pb(NO_3)_2(aq)\rightarrow Pb^{2+}(aq)+2NO_3^{-}(aq)$

$Na_2SO_4(aq)\rightarrow 2Na^++SO_4^{2-}(aq)$

$Pb^{2+}(aq)+2NO_3^{-}(aq)+2Na^++SO_4^{2-}(aq)\rightarrow PbSO_4(s)+2Na^++2NO_3^{-}(aq)$

Removing common ions from both sides, we get the net ionic equation:

$Pb^{2+}(aq)+SO_4^{2-}(aq)\rightarrow PbSO_4(s)$

b. $NiCl_2(aq) + NH_4NO_3(aq)$ →

$NiCl_2(aq) + NH4NO_3(aq) \rightarrow Ni(NO_3)_2+NH_4Cl(aq)$

No precipitation is occuring.

c. $Fe_Cl2(aq) + Na_2S(aq)$ →

$FeCl_2(aq) + Na_2S(aq)\rightarrow FeS(s)+2NaCl(aq)$

$FeCl_2(aq)\rightarrow Fe^{2+}(aq)+2Cl^{-}(aq)$

$Na_2S(aq)\rightarrow 2Na^++S{2-}(aq)$

$Fe^{2+}(aq)+2Cl^{-}(aq)+2Na^++S^{2-}(aq)\rightarrow FeS(s)+2Na^++2Cl^{-}(aq)$

Removing common ions from both sides, we get the net ionic equation:

$Fe^{2+}(aq)+S^{2-}(aq)\rightarrow FeS(s)$

d. $MgSO_4(aq) + BaCl_2(aq)$ →

$MgSO4(aq) + BaCl2(aq)\rightarrow BaSO_4(s)+MgCl_2$

$MgSO_4(aq)\rightarrow Mg^{2+}(aq)+SO_4^{2-}(aq)$

$BaCl_2(aq)\rightarrow Ba^{2+}+2Cl^{-}(aq)$

$Mg^{2+}(aq)+SO_4^{2-}(aq)+Ba^{2+}+2Cl^{-}(aq)(aq)\rightarrow BaSO_4(s)+Mg^{2+}(aq)+2Cl^{-}(aq)$

Removing common ions from both sides, we get the net ionic equation:

$Ba^{2+}(aq)+SO_4^{2-}(aq)\rightarrow BaSO_4(s)$

5 0

2 years ago

A 72.0 mL aliquot of a 1.40 M solution is diluted to a total volume of 248 mL. A 124 mL portion of that solution is diluted by a

Aliun [14]

Answer: 0.20 M

Explanation:

According to the dilution law,

$M_1V_1=M_2V_2$

where,

$M_1$ = molarity of stock solution = 1.40 M

$V_1$ = volume of stock solution = 72.0 ml

$M_2$ = molarity of diluted solution = m

$V_2$ = volume of diluted solution = 248 ml

$1.40\times 72.0=m\times 248$

$m=0.41M$

Now 124 mL portion of this prepared solution is diluted by adding 133 mL of water.

According to the dilution law,

$M_1V_1=M_2V_2$

where,

$M_1$ = molarity of stock solution = 0.41 M

$V_1$ = volume of stock solution = 124 ml

$M_2$ = molarity of diluted solution = m

$V_2$ = volume of diluted solution = (124 +133) ml = 257 ml

$0.41\times 124=m\times 257$

$m=0.20M$

Thus the final concentration of the solution is 0.20 M.

8 0

3 years ago