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suter [353]
2 years ago
7

A solid is 5 cm tall 3 cm wide and 2 cm thick it has a mass of 129 g what is the density

Chemistry
2 answers:
belka [17]2 years ago
7 0
First you would multiply 5,3, and 2 to get a volume of 30. Density is mass over volume so you would then divide 129 by 30. You would get 4.3. The answer would be 4.3 g/cm^3
dlinn [17]2 years ago
4 0

<u>Answer:</u> The density of solid is 4.3g/cm^3

<u>Explanation:</u>

To calculate the volume of cuboid, we use the equation:

V=lbh

where,

V = volume of cuboid

l = length of cuboid = 5 cm

b = breadth of cuboid = 3 cm

h = height of cuboid = 2 cm

Putting values in above equation, we get:

V=5\times 3\times 2=30cm^3

To calculate density of a substance, we use the equation:

Density=\frac{Mass}{Volume}

Putting values in above equation, we get:

\text{Density of solid}=\frac{129g}{30cm^3}\\\\\text{Density of solid}=4.3g/cm^3

Hence, the density of solid is 4.3g/cm^3

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Lina20 [59]

Answer:

true

Explanation:

because the atoms make up the elements which forms the compound

8 0
2 years ago
Silver chloride is formed by mixing silver nitrate and barium chloride solutions. What volume of 1.50 M barium chloride solution
konstantin123 [22]

Answer:

1.22 mL

Explanation:

Let's consider the following balanced reaction.

2 AgNO₃ + BaCl₂ ⇄ Ba(NO₃)₂ + 2 AgCl

The molar mass of silver chloride is 143.32 g/mol. The moles corresponding to 0.525 g are:

0.525 g × (1 mol/143.32 g) = 3.66 × 10⁻³ mol

The molar ratio of AgCl to BaCl₂ is 2:1. The moles  of BaCl₂ are 1/2 × 3.66 × 10⁻³ mol = 1.83 × 10⁻³ mol.

The volume of 1.50 M barium chloride containing 1.83 × 10⁻³ moles is:

1.83 × 10⁻³ mol × (1 L/1.50 mol) = 1.22 × 10⁻³ L = 1.22 mL

8 0
2 years ago
Five darts strike near the center of the target. Who ever threw the darks is?
Burka [1]

Better than i am and very precice


7 0
2 years ago
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Question 17 In the Haber reaction, patented by German chemist Fritz Haber in 1908, dinitrogen gas combines with dihydrogen gas t
mars1129 [50]

Answer:

Explanation:

N₂       + 3H₂     =     2 NH₃

1 vol                         2 vol

786 liters               1572 liters

786 liters of dinitrogen will result in the production of 1572 liters of ammonia

volume of ammonia V₁ = 1572 liters

temperature T₁ = 222 + 273 = 495 K

pressure = .35 atm

We shall find this volume at NTP

volume V₂ = ?

pressure = 1 atm

temperature T₂ = 273

\frac{P_1V_1}{T_1} =\frac{P_2V_2}{T_2}

\frac{.35\times 1572}{495} =\frac{1\times V_2}{ 273 }

V_2 =303.44 liter .

mol weight of ammonia = 17

At NTP mass of 22.4 liter of ammonia will have mass of 17 gm

mass of 303.44 liter of ammonia will be equal to (303.44 x 17) / 22.4 gm

= 230.28 gm

=.23 kg / sec .

Rate of production of ammonia = .23 kg /s .

5 0
2 years ago
. A saturated solution of Ag2SO4 at 25°C contains 0.032 M Ag+ ions. From this information, estimate the ΔG°rxn for the dissoluti
Gnoma [55]

Answer:

20 kJ/mol

Explanation:

From ∆G°= -RTlnK

But

Ag2SO4(s)<----------->2Ag+(aq) + SO4^2-(aq)

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But Ag+ = 0.032M

Ksp= (2×0.032)^2 (0.032)

Ksp= 1.31072×10^-4

∆G°= -RTlnK

∆G°= -(8.314× 298×(-8.93976))= 20KJmol-1( to the nearest KJ)

4 0
2 years ago
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