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Blizzard [7]
3 years ago
15

which occupies a larger volume, 600 g of water (with a density of 0.995 g/cm3 ) or 600 g of lead (with a density of 11.35 g/cm3

)? energy units
Chemistry
1 answer:
Montano1993 [528]3 years ago
5 0
Water will occupy larger volume
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murzikaleks [220]

Answer:

7.  b

8. d

9. c

10. c

11. d

Explanation:

4 0
3 years ago
How many Joules of heat would be required
mote1985 [20]

Heat would be required  : 1,670 J

<h3>Further explanation</h3>

Given

mass of H₂O=5 g

Required

Heat to melt

Solution

The heat to change the phase can be formulated :

Q = m.Lf (melting/freezing)

Lf=latent heat of fusion

The heat of fusion for water at 0 °C : 334 J/g

Input given values in formula :

\tt Q=5\times 334=\boxed{\bold{1,670~J}}

4 0
3 years ago
Help on this question please
oee [108]

Answer:

First choice: 2

Explanation:

There are 2 phosphorous (P) in the substance.

Ignore the strontium (Sr3) part because you are looking to isolate the P from (PO4)2.

Break the chemical equation apart to get 1 Phosphorous atom, and 4 Oxygen atoms.

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8 0
3 years ago
if 3.26 g of FeNO33 is dissolved in enough water to make exactly what is the molar concentration of nitrate ion g
Tanzania [10]

Answer:

0.404M

Explanation:

...<em>To make exactly 100.0mL of solution...</em>

Molar concentration is defined as the amount of moles of a solute (In this case, nitrate ion, NO₃⁻) in 1 L of solution.

To solve this question we need to convert the mass of Fe(NO₃)₃ to moles. As 1 mole of Fe(NO₃)₃ contains 3 moles of nitrate ion we can find moles of nitrate ion in 100.0mL of solution, and we can solve the amount of moles per liter:

<em>Moles Fe(NO₃)₃ -Molar mass: 241.86g/mol-:</em>

3.26g * (1mol / 241.86g) =

0.01348 moles Fe(NO₃)₃ * (3 moles of NO₃⁻ / 1mole Fe(NO₃)₃) =

<em>0.0404 moles of NO₃⁻</em>

In 100mL = 0.1L, the molar concentration is:

0.0404 moles of NO₃⁻ / 0.100L =

<h3>0.404M</h3>
5 0
3 years ago
What volume will 145 g of butane occupy at 745 torr and 35 ∘c?
Margaret [11]

First let us compute for the number of moles of butane (molar mass = 58.12 g/mol)

number of moles = 145 g / (58.12 g/mol) = 2.49 mol

<span>We  use the ideal gas equation to calculate the volume:</span>

<span> V = n R T / P</span>

V = 2.49 mol * 62.36367 L torr / mol K * 308.15 K / 745 torr

<span>V = 64.35 L</span>

8 0
3 years ago
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