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3 years ago

Please help me with these I'll give 15 points

Chemistry

2 answers:

VLD [36.1K]3 years ago

5 0

1.d
2.g
3.j
4.i
5.c
6.h
7.e
8.b
9.a
10.f

sveticcg [70]3 years ago

4 0

1) D
2) G
3) J
4) I
5) C
6) H
7) E
9) A
10) F

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If the pressure changes from 1 atm to 3 atm, what does the volume of 30 L change to

DiKsa [7]

Answer: 10L

Explanation:

Given that:

Initial pressure P1 = 1 atm

New pressure P2 = 3 atm

Initial volume V2 = 30 L

New volume V2 = ?

Since pressure and volume are involved, apply the formula for Boyle's law

P1V1 = P2V2

1 atm x 30L = 3 atm x V2

30 atm L = 3 atm x V2

V2 = (30 atm L / 3 atm)

V2 = 10L

Thus, volume changed to 10 liters

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3 years ago

Predicting the products CuCl2 (aq) + Al (s)

Flauer [41]

The reaction will produce solid copper and aluminium chloride salt.

Explanation:

Copper chloride (CuCl₂) in solution will react with aluminium to form solid cooper and aluminium chloride (AlCl₃).

3 CuCl₂ (aq) + 2 Al (s) → 3 Cu (s) + 2 AlCl₃ (aq)

Learn more about:

numerical problems with copper chloride and aluminium

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3 years ago

Iron has a density of 7.86 g/cm3. Calculate the volume (in dL) of a piece of iron having a mass of 4.07 kg . Note that the densi

elena-s [515]

Answer:

The volume of the piece of iron is 5.18dL.

Explanation:

The density (ρ) is equal to the mass (m) divided the volume (V).

$\rho = \frac{m}{V}$

If we rearrange it, we have:

$V=\frac{m}{\rho }$

To express the volume in dL we will need the following relations:

1 dL = 0.1 L
1 kg = 10³ g
1 cm³ = 1 mL
1mL = 10⁻³L

Then,

$\rho = \frac{7.86g}{cm^{3} } .\frac{1cm^{3} }{1mL} .\frac{1mL}{10^{-3}L } .\frac{1kg}{10^{3}g } =7.86kg/L$

Finally,

$V=\frac{m}{\rho }=\frac{4.07kg}{7.86 kg/L} =0.518L.\frac{1dL}{0.1L} =5.18dL$

5 0

3 years ago

Read 2 more answers

When 15.3 g of sodium nitrate, NaNO3,was dissolved in water in a calorimeter, the temperature fell from 25.00oC to 21.56oC. If t

satela [25.4K]

Answer:

20.468 kilo Joules is the enthalpy change when one mole of sodium nitrate dissolves.

Explanation:

Heat lost by solution ad calorimeter = Q

Heat capacity of solution ad calorimeter = C = 1071 J/°C

Change in temperature = ΔT = 21.56°C - 25.00°C = -3.44°C

$Q=C\times Delta T$

$Q=1071 J/^oC\times (-3.44^oC)=-3,684.24 J$

Heat gained by sodium nitrate = -Q = -(-3,684.24 J)=3,684.24 J

Moles of sodium nitrate = $\frac{15.3 g}{85 g/mol}=0.18 mol$

When 0.18 mole of sodium nitrate was dissolved in water 3,684.24 joulesof heat was absorbed by it.

Then heat absorbed by 1 mole of sodium nitrate :

$\frac{3,684.24 J}{0.18}=20,468 J=20.468 kJ$

1 J = 0.001 kJ

20.468 kilo Joules is the enthalpy change when one mole of sodium nitrate dissolves.

8 0

3 years ago

Stronger acids are those that —

Harlamova29_29 [7]

F. hold on to their protons more strongly

7 0

3 years ago