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olganol [36]
3 years ago
7

The density of gold is 19.3 g/cm3. What is the mass of 15cm3 of gold?

Chemistry
1 answer:
Ulleksa [173]3 years ago
7 0
Just multiple 19.3 by 15 because cm^3 will cancel outand you are left with g with represents mass
You might be interested in
Using the van der Waals equation, the pressure in a 22.4 L vessel containing 1.00 mol of neon gas at 100 °C is ________ atm. (a
svetoff [14.1K]

Answer:

The answer to your question is        P = 1.357 atm

Explanation:

Data

Volume = 22.4 L

1 mol

temperature = 100°C

a = 0.211 L² atm

b = 0.0171 L/mol

R = 0.082 atmL/mol°K

Convert temperature to °K

Temperature = 100 + 273

                      = 373°K

Formula

               (P + \frac{a}{v^{2}} )(v - b) = RT

Substitution

               (P + \frac{0.211}{22.4})(22.4 - 0.0171) = (0.082)(373)

Simplify

               (P + 0.0094)(22.3829) = 30.586

Solve for P

                           P + 0.0094 = \frac{30.586}{22.3829}

                           P + 0.0094 = 1.366

                                 P = 1.336 - 0.0094

                                P = 1.357 atm

7 0
3 years ago
31) give an example of one renewable and one non renewable resource. be sure to state which is renewable and which is nonrenewab
Yakvenalex [24]
31) A renewable resource is qualified as things like, Solar, Wind, Hydro-electric, and thermal.
A non-renewable source can be like coal, trees, Gasoline (oil), and petroleum.

32) I don't know what you mean, sorry.

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5 0
3 years ago
How many grams of CO are needed to react with an excess of Fe 2 O 3 to produce 209.7 g Fe?
gavmur [86]

Answer:

Mass = 157.5 g

Explanation:

Given data:

Mass of CO needed  = ?

Mass of Fe formed = 209.7 g

Solution:

Chemical equation:

3CO + F₂O₃   →       2Fe + 3CO₂

Number of moles of Fe:

Number of moles = mass/ molar mass

Number of moles = 209.7 g/ 55.85 g/mol

Number of moles = 3.75 mol

Now we will compare the moles of iron and carbon monoxide.

                               Fe            :              CO

                                 2            :              3

                                3.75         ;             3/2×3.75 = 5.625 mol

Mass of CO:

Mass = number of moles × molar mass

Mass = 5.625 mol × 28 g/mol

Mass = 157.5 g

4 0
2 years ago
As shown in table 15.2, kp for the equilibrium n21g2 + 3 h21g2 δ 2 nh31g2 is 4.51 * 10-5 at 450 °c. for each of the mixtures lis
Setler79 [48]
Our reaction balanced equation at equilibrium N2(g) + 3 H2(g) ↔ 2 NH3(g)
and we have the Kp value at equilibrium = 4.51 X 10^-5
A) 98 atm NH3, 45 atm N2, 55 atm H2

when Kp = [P(NH3)]^2 / [P(N2)] * [P(H2)]^3
                = 98^2 / (45 * 55^3) = 1.28 x 10^-3
by comparing the Kp by the Kp at equilibrium(the given value) So,
Kp > Kp equ So the mixture is not equilibrium,
 it will shift leftward (to decrease its value) towards the reactants to achieve equilibrium.
B) 57 atm NH3, 143 atm N2, no H2
∴ Kp = [P(NH3)]^2 / [P(N2)]
         = 57^2 / 143 = 22.7
∴Kp> Kp equ (the given value) 
∴it will shift leftward (to decrease its value) towards reactants to achieve equilibrium.

c) 13 atm NH3, 27 atm N2, 82 atm H2
∴Kp = [P(NH3)]^2 / [P(N2)] * [P(H2)]^3
       =  13^2 / (27* 82^3) = 1.14 X 10^-5
∴ Kp< Kp equ (the given value)
∴it will shift rightward (to increase its value) towards porducts to achieve equilibrium.


3 0
3 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
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