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kari74 [83]
3 years ago
13

The density of gold is 19.3 g/cm³. Which of the following shows the mass of a gold bar that is 4.50 cm × 8.00 cm × 20.00 cm?

Chemistry
1 answer:
MakcuM [25]3 years ago
4 0

Answer:

13896g

Explanation:

volume = 4.50×8.00×20.00 = 720 cm³

mass = density × volume

mass= 19.3 × 720 = 13896g

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Question 43 please. i’m not sure if i got the right answer
hram777 [196]

a) The reaction is exothermic since the overall enthalpy change is negative. this means that the system has lost energy to the environment, namely, the apparatus and due to drought.

b) We first calculate the number of moles in 3.55 grams of magnesium.

number of moles= mass/ atomic mass

                            =3.55/24

                            =0.1479 moles(to 4sf)

now, if 2 moles of magnesium give -1204kJ

        How much energy is given by 0.1479 moles

       = (0.1479×-1204kJ)

        =-89.0358kJ (don't forget the negative sign)

c) two molesof MgO produces -1204kJ of energy

then -234kJ will be produced by

=(-234kJ×2moles)/1204kJ

=0.3887moles

one mole of MgO weighs 24+16=40

therefore the mass produced is 0.3887moles×40=15.548grams

(d) we first find the number of moles of MgO in 40.3 grams

number of moles=mass/RFM

=40.3g/40= 1.0075moles

if 2 moles of MgO give 1204 kJ then decomposing 1.0075 moles requires

(1.0075 moles×1204kJ)/2=606.515kJ

4 0
3 years ago
!!!!!!!!!!!!!!!????<br> hElP!!!!!!!!!!!!
kakasveta [241]
I think the answer is transparent and translucent
5 0
3 years ago
Read 2 more answers
Can matter every be destroyed? Why or why not?
REY [17]

Hey

Matter can neither be created nor destroyed.

Explanation:

This is the law of conservation of matter (mass). One can prove this by performing a simple experiment at home.

Have a great day

4 0
3 years ago
The electron configuration of a neutral atom is 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁵. An electron in which of the following orbitals
mezya [45]

Answer:

Provide an image

Explanation:

5 0
2 years ago
Liquid octane CH3CH26CH3 will react with gaseous oxygen O2 to produce gaseous carbon dioxide CO2 and gaseous water H2O. Suppose
Sholpan [36]

Answer:

10 g of CO2

Explanation:

Equation of the reaction:

CH3(CH2)6CH3 + 17O2 ----> 18H2O + 8CO2

Fom the above balanced equation,

1 mole of Octane gas reacts with 17 moles of oxygen gas to produce 8 moles of CO2

Molar mass of Octane = 114 g/mol

Molar mass of oxygen gas = 32 g/mol

Molar mass of CO2 = 44 g/mol

Therefore, 114 g of Octane reacts completely with 17 * 32g (= 544 g) of oxygen to produce 8 * 44 g(=352g) of CO2.

From the given mass of reactants;

3.4 g of Octane will react with (544 * 3.4)/114 g of oxygen = 16.22g of oxygen.

Therefore oxygen is the limiting reactant.

15.6 g of oxygen will react with (114 * 15.6)/544 g of CO2 = 3.27 g of octane.

Mass of CO2 produced will be

(352 * 15.6)/544 = 10 g of CO2

4 0
3 years ago
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