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2 years ago

A Metal mass of 71.68g occupies a volume of 8cm^3. Calculate the density

Chemistry

1 answer:

Damm [24]2 years ago

5 0

Density = mass
--------------
volume

= 71.68
-------------
8

= 8.96 g/cm3.

Hope this helps!

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A solution may contain Ag+, Pb2+, and/or Hg22+. A white precipitate forms when 6 M HCl is added. The precipitate is partially so

omeli [17]

Answer:

All three are present

Explanation:

Addition of 6 M HCl would form precipitates of all the three cations, since the chlorides of these cations are insoluble: $AgCl (s), PbCl_2 (s), Hg_2Cl_2 (s)$ .

Firstly, the solid produced is partially soluble in hot water. Remember that out of all the three solids, lead(II) choride is the most soluble. It would easily completely dissolve in hot water. This is how we separate it from the remaining precipitate. Therefore, we know that we have lead(II) cations present, as the two remaining chlorides are insoluble even at high temperatures.
Secondly, addition of liquid ammonia would form a precipitate with silver: $AgCl (s) + 2 NH_3 (aq) + H_2O (l)\rightarrow [Ag(NH_3)_2]OH (s) + HCl (aq)$ ; Silver hydroxide at higher temperatures decomposes into black silver oxide: $2 [Ag(NH_3)_2]OH (s)\rightarrow Ag_2O (s) + H_2O (l) + 4 NH_3 (g)$ .
Thirdly, we also know we have $Hg_2^{2+}$ in the mixture, since addition of potassium chromate produces a yellow precipitate: $Hg_2^{2+} (aq) + CrO_4^{2-} (aq)\rightarrow Hg_2CrO_4 (s)$ . The latter precipitate is yellow.

3 0

3 years ago

Dissolve 30 g of sodium sulphate into 300 mL of water

Aneli [31]

Answer:

number of moles = 0.21120811

Explanation:

To find the number of moles, given the mass of the solute, we use the formula:

$\mathrm{n = \dfrac{ m }{ M } }$

$\mathrm{n = number\:of\:moles\:(mol)}$

$\mathrm{m = mass\:of\:solute\:(g)}$

$\mathrm{M = molar\:mass\:of\:solute\:( \dfrac{ g }{ mol } )}$

Label the variables with the numbers in the problem:

$\mathrm{n =\:?}$

$\mathrm{m =30\:g }$

$\mathrm{M =\:?\:Calculate\:the\:molar\:mass }$

The first thing we have to do is find the molar mass of sodium sulfate, in order for us to use the formula for finding the number of moles:

Formula for finding the molar mass of sodium sulfate:

$M({ \left Na \right }_{ 2 } { \left So \right }_{ 4 }) = m \left( Na \right) +m \left( S \right) +m \left( O \right)$

For the variables and what they mean are below for finding the molar mass of sodium sulfate:

$\mathrm{M =molar\:mass }$

$\mathrm{m =moles=2\:moles\:for\:Na\:,1\:mole\:for\:S,\:and\:4\:moles\:for\:O}$

$\mathrm{Na =sodium=22.99\:g }$

$\mathrm{S =sulfur=32.06\:g }$

$\mathrm{O =oxygen=16.00\:g }$

Plug the numbers into the formula, to find the molar mass of sodium sulfate:

$M({ \left Na \right }_{ 2 } { \left So \right }_{ 4 }) = m \left( Na \right) +m \left( S \right) +m \left( O \right)$

$\mathrm{Substitute\:the\:values\:into\:the\:formula}$

$M = 2 \left( 22.99 \right) +1 \left( 32.06 \right) +4 \left( 16.00 \right)$

$\mathrm{Multiply\:2\:by\:22.99\:to\:get\:45.98\:and\:1\:by\:32.06\:to\:get\:32.06}$

$\mathrm{M = 45.98+32.06+4\:(16)}$

$\mathrm{Multiply\:4\:by\:16\:to\:get\:64}$

$\mathrm{M = 45.98+32.06+64}$

$\mathrm{Add\:45.98\:and\:32.06\:to\:get\:78.04}$

$\mathrm{M = 78.04+64}$

$\mathrm{Add\:78.04\:and\:64\:to\:get\:142.04}$

$\mathrm{M = 142.04}$

Now that we have found the molar mass, we can calculate the number of moles in the solution of sodium sulfate with the formula:

$\mathrm{n = \dfrac{ m }{ M } }$

$\mathrm{n =\:?}$

$\mathrm{m =30\:g }$

$\mathrm{M = 142.04\:g/mol}$

$\mathrm{Substitute\:the\:values\:into\:the\:formula}$

$\mathrm{n = \dfrac{ 30 }{ 142.04 }}$

$\mathrm{Divide\:142.04\:by\:30\:to\:get\:0.21120811}$

$\mathrm{n = 0.21120811}$

0.21120811 rounded gives you 0.2112

or if you did the problem without decimals

30 grams of sodium sulfate divided by its molecular weight – which we found to be 142 – gives us a value of 0.2113 moles.

3 0

1 year ago

What would you need to do to calculate the molality of 10 g of NaCl in 2 kg of

jasenka [17]

Answer:

O B. Convert the 10 g of NaCl to moles of NaCl.

Explanation:

The formula for finding the molality is m=moles of solute/kg of solvent. The solute for this question is NaCl and the solvent is water.

(10g NaCl)(1 mol NaCl/58.44g NaCl)=0.1711 mol NaCl

58.44 is the molar mass of NaCl

m=0.1711 mol NaCl/2 kg H2O

m=0.085557837

7 0

2 years ago

How much heat is required to raise the temperature of 225 grams of ice from -26.8 °C to steam at 133 °C ?

lara [203]

<h3>Answer:</h3>

150000 J

<h3>General Formulas and Concepts:</h3>

Chemistry

Thermodynamics

Specific Heat Formula: q = mcΔT

q is heat (in J)
m is mass (in g)
c is specific heat (in J/g °C)
ΔT is change in temperature (in °C or K)

Math

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

<h3>Explanation:</h3>

Step 1: Define

Identify variables

[Given] m = 225 g

[Given] c = 4.184 J/g °C

[Given] ΔT = 133 °C - -26.8 °C = 159.8 °C

[Solve] q

Step 2: Solve for q

Substitute in variables [Specific Heat Formula]: q = (225 g)(4.184 J/g °C)(159.8 °C)
Multiply: q = (941.4 J/°C)(159.8 °C)
Multiply: q = 150436 J

Step 3: Check

Follow sig fig rules and round. We are given 3 sig figs.

150436 J ≈ 150000 J

Topic: AP Chemistry

Unit: Thermodynamics

Book: Pearson AP Chemistry

5 0

2 years ago

Ten pints of 15 % salt solution are mixed with 15 pints of 10 % salt solution. What is the

sergij07 [2.7K]

$Concentration\ rate=\frac{mass\ of\ product}{mass\ of\ substance}*100\%\\\\
15\%=\frac{m_1}{ms_1}*100\%\\ms_1=10pints\\
15\%=\frac{m_1}{10}*100\%\\
\frac{15}{100}=\frac{m_1}{10}\ \ |*10\\
m_1=\frac{150}{100}=1,5pints\\\\
10\%=\frac{m_2}{15}*100\%\\
0,1=\frac{m_2}{15}\ \ |*15\\
1,5=m_2\\
Resulting\ solution:\\
ms=ms_1+ms_2=15+10=25pints\\m=m_1+m_2=1,5+1,5=3pints
\\\frac{m}{ms}*100\%=\frac{3}{25}*100\%=12\%$

7 0

2 years ago