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fredd [130]
3 years ago
9

I need to check some chemistry questions. Help with any question is appreciated! :) Please include an explanation with your conc

lusion.

Chemistry
1 answer:
finlep [7]3 years ago
4 0

1. 4.67 kg; 2. 4.8 ×10^5 kg; 3. 0.106 cm^3; 4. 1.7 g/cm^3

<em>Q1. Mass of Hg </em>

Mass = 345 mL × (13.53 g/1 mL) = 4670 g = 4.67 kg

<em>Q2. Mass of Pb </em>

<em>Step 1</em>. Calculate the <em>volume of the Pb</em>.

<em>V = lwh</em> = 6.0 m × 3.5 m × 2.0 m = 42.0 m^3

<em>Step 2</em>. Calculate the <em>mass of the Pb</em>.

Mass = 42.0 m^3 × (11 340 kg/1 m^3) = 4.8 × 10^5 kg

<em>Q3. Volume of displaced water </em>

Volume of Ag = 0.987 g × (1 cm^3/9.320 g) = 0.106 cm^3

<em>Archimedes</em>: volume of displaced water = volume of Ag = <em>0.106 cm^3</em>

<em>4. Density of metal </em>

<em>Step 1</em>. Convert <em>ounces to grams </em>

Mass = 3.35 oz × (28.35 g/1 oz) = 94.97 g

<em>Step 2</em>. Calculate the <em>volume in cubic inches </em>

<em>V = lwh</em> = 3.0 in × 2.5 in × 0.45 in = 3.38 in^3

<em>Step 3</em>. Convert <em>cubic inches to cubic centimetres</em><em> </em>

<em>V</em> = 3.38 in^3 × (2.54 cm/1 in)^3 = 55.3 cm^3

<em>Step 4</em>. Calculate the <em>density</em>

ρ = <em>m</em>/<em>V</em> = (94.97 g/55.3 cm^3) = 1.7 g/cm^3 (magnesium?)

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Cars run on gasoline, where octane (C8H18) is the principle component. This combustion reaction is responsible for generating en
Bezzdna [24]

Answer:

  • 10.19 g CO₂
  • 4.69 g H₂O

Explanation:

The combustion reaction of Octane is:

  • C₈H₁₈ → 8CO₂ + 9H₂O

To calculate the mass of CO₂ and H₂O produced, we need to know the mass of octane combusted.

We calculate the mass of Octane from the given volume and density, using the following <em>conversion factors</em>:

  • 1 gallon = 3.785 L
  • 1 L = 1000 mL

Now we<u> convert 1.24 gallons to mL</u>:

  • 1.24 gallon * \frac{3.785L}{1gallon} *\frac{1000mL}{1L} = 4693.4 mL

We <u>calculate the mass of Octane</u>:

  • 4693.4 mL * 0.703 g/mL = 3.30 g Octane

Now we use the <em>stoichiometric ratios</em> and <em>molecular weights</em> to <u>calculate the mass of CO₂ and H₂O</u>:

  • CO₂ ⇒ 3.30 g Octane ÷ 114g/mol * \frac{8molCO_{2}}{1molOctane} * 44 g/mol =  10.19 g CO₂
  • H₂O ⇒ 3.30 g Octane ÷ 114g/mol * \frac{9molH_{2}O}{1molOctane} * 18 g/mol = 4.69 g H₂O

7 0
3 years ago
A scientist wants to make a solution of tribasic sodium phosphate, Na3PO4, for a laboratory experiment. How many grams of Na3PO4
balu736 [363]

Answer:

51.53 grams .

Explanation:

Na₃PO₄    ⇄  3Na⁺¹   +    PO₄⁻³ .

1 mole           3 mole

725 mL of 1.3 M Na⁺  ions

= .725 x 1.3 moles of Na⁺ ions

= .9425 moles

3 mole of Na⁺  is formed by 1 mole of Na₃PO₄

.9425 mole of Na⁺  is formed by .9425/3  mole of Na₃PO₄

Na₃PO₄ needed =  .9425/3  moles = .3142 moles

Molecular weight of Na₃PO₄ = 164

grams of Na₃PO₄ needed = .3142 x 164 = 51.53 grams .

8 0
2 years ago
Whoever can answer this I will give all of my points to and brainliest award.
ZanzabumX [31]

i dont see the picture.

7 0
3 years ago
Read 2 more answers
Calculate the change in pH when 71.0 mL of a 0.760 M solution of NaOH is added to 1.00 L of a solution that is 1.0O M in sodium
Eddi Din [679]

Explanation:

It is known that pK_{a} value of acetic acid is 4.74. And, relation between pH and pK_{a} is as follows.

                    pH = pK_{a} + log \frac{[CH_{3}COOH]}{[CH_{3}COONa]}

                          = 4.74 + log \frac{1.00}{1.00}

So, number of moles of NaOH = Volume × Molarity

                                                   = 71.0 ml × 0.760 M

                                                    = 0.05396 mol

Also, moles of  CH_{3}COOH = moles of CH_{3}COONa

                                          = Molarity × Volume

                                          = 1.00 M × 1.00 L

                                          = 1.00 mol

Hence, addition of sodium acetate in NaOH will lead to the formation of acetic acid as follows.

            CH_{3}COONa + NaOH \rightarrow CH_{3}COOH

Initial :    1.00 mol                                  1.00 mol

NaoH addition:               0.05396 mol

Equilibrium : (1 - 0.05396 mol)    0           (1.00 + 0.05396 mol)

                    = 0.94604 mol                       = 1.05396 mol

As, pH = pK_{a} + log \frac{[CH_{3}COONa]}{[CH_{3}COOH]}

               = 4.74 +  log \frac{0.94604}{1.05396}

               = 4.69

Therefore, change in pH will be calculated as follows.

                         pH = 4.74 - 4.69

                               = 0.05

Thus, we can conclude that change in pH of the given solution is 0.05.

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3 years ago
Compared to orange light, violet light
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Answer:

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