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Readme [11.4K]
3 years ago
8

Refer to the following table for information about gold and mercury. Then choose which statements are true about equal sample si

zes of the metals. Element Specific Heat (cal/g°C) Density (g/cm3) gold 0.031 19.3 mercury 0.033 13.5
Chemistry
1 answer:
babymother [125]3 years ago
7 0
Below are the choices:

A The mercury will change temperature at a much faster rate under the same heating conditions. 
<span>B The two metal samples will change temperature at about the same rate. </span>
<span>C The gold would float if placed in the mercury. </span>
<span>D The gold would sink to the bottom if placed in the mercury.
</span>
  <span>a = false, it will take 0.031 cal to raise 1g Au 1degree while it will take 0.033 cal to raise 1g Hg 1 degree so, although Au will heat up faster, it will not be discernably faster so... 
b = true 
c = false, Au density > Hg 
d = true</span>
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Hard water often contains dissolved Ca2+ and Mg2+ ions. One way to soften water is to add phosphates. The phosphate ion forms in
katrin2010 [14]

Answer:

22.4269 grams of sodium phosphate must be added to 1.4 L of this solution to completely eliminate the hard water ions

Explanation:

We will first write the balanced equation for this scenario

3 CaCl2 + 2 Na3PO4 ----> 6 NaCl + Ca3 (PO4)2

3 Mg(NO3)2 + 2 Na3PO4 -----> 6 NaNO3 + Mg3 (PO4)2

The ratio here for both calcium chloride and magnesium nitrate is 3:2

The number of moles of each compound is equal to

0.054 * 1.4 = 0.0756\\0.093* 1.4 = 0.1302

Using the mole ratio of 3:2, convert each to moles of sodium phosphate.

0.0756 mole of CaCl2 is equal to 0.05\\ Na3PO4

0.1302 mole of CaCl2 is equal to 0.0868 Na3PO4

Converting moles of sodium phosphate to grams of sodium phosphate we get

(0.05 +0.0868) * 163.94 g/mol

22.4269 grams of sodium phosphate must be added to 1.4 L of this solution to completely eliminate the hard water ions

8 0
3 years ago
n unknown metal is either aluminum, iron or lead. If 150. g of this metal at 150.0 °C was placed in a calorimeter that contains
Nitella [24]

Answer : The metal used was iron (the specific heat capacity is 0.449J/g^oC).

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

q_1=-q_2

m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)

where,

c_1 = specific heat of unknown metal = ?

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m_1 = mass of unknown metal = 150 g

m_2 = mass of water = 200 g

T_f = final temperature of water = 34.3^oC

T_1 = initial temperature of unknown metal = 150.0^oC

T_2 = initial temperature of water = 25.0^oC

Now put all the given values in the above formula, we get

150g\times c_1\times (34.3-150.0)^oC=-200g\times 4.184J/g^oC\times (34.3-25.0)^oC

c_1=0.449J/g^oC

Form the value of specific heat of unknown metal, we conclude that the metal used in this was iron (Fe).

Therefore, the metal used was iron (the specific heat capacity is 0.449J/g^oC).

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Hey there!

<span>What activates a convection current, starting the flow of a fluid ?

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