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Leya [2.2K]
2 years ago
6

Find the density of an 8 gram rock if the water in a graduated cylinder rises from 25.0 mL to

Chemistry
1 answer:
Olin [163]2 years ago
6 0

To measure the density of the stone placed in a graduated cylinder let us follow these steps bellow

  1. Measure the volume of water poured into a graduated cylinder
  2. Place the object in the water and remeasure the volume.
  3. The difference between the two volume measurements is the volume of the object.
  4. Divide the mass by the volume to calculate the density of the object.

<em>We know that the formula for density is given as </em>

Given data

Mass = 8gram

Initial Volume of water in cylinder = 25mL

Final Volume of water in cylinder  = 29mL

Hence the volume of the rock  = 29-25 = 4mL

Therefore the density of the rock = 8/4 = 2 g/mL

Learn more:

brainly.com/question/17336041

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⚠️HELP⚠️
morpeh [17]

Answer:

The balanced equation is :

Zn(s) + H2SO4(aq) → ZnSO4(aq) + H2(g)

7 0
3 years ago
A student mixes four reagents together, thinking that the solutions will neutralize each other. The solutions mixed together are
vaieri [72.5K]

Answer: Resulting solution will not be neutral because the moles of OH^-ions is greater. The remaining concentration of [OH^-]ions =0.0058 M.

Explanation:

Given,

[HCl]=0.100 M

[HNO_3] = 0.200 M

[Ca(OH)_2] =0.0100 M

[RbOH] =0.100 M

Few steps are involved:

Step 1: Calculating the total moles of H^+ ion from both the acids

moles of H^+ in HCl

HCl\rightarrow {H^+}+Cl^-

if 1 L of HClsolution =0.100 moles of HCl

then 0.05L of HCl solution= 0.05 \times0.1 moles= 0.005 moles    (1L=1000mL)

moles of H^+ in HCl = 0.005 moles

Similarliy

moles of H^+ in HNO_3

HNO_3\rightarrow H^++NO_3^-}

If 1L of HNO_3 solution= 0.200 moles

Then 0.1L of HNO_3 solution= 0.1 \times 0.200 moles= 0.02 moles

moles of H^+ in HNO_3 =0.02 moles

so, Total moles of H^+ ions  = 0.005+0.02= 0.025 moles     .....(1)

Step 2: Calculating the total moles of [OH^-] ion from both the bases

Moles of OH^-\text{ in }Ca(OH)_2

Ca(OH)_2\rightarrow Ca^2{+}+2OH^-

1 L of Ca(OH)_2= 0.0100 moles

Then in 0.5 L Ca(OH)_2 solution = 0.5 \times0.0100 moles = 0.005 moles

Ca(OH)_2 produces two moles of OH^- ions

moles of OH^- = 0.005 \times 2= 0.01 moles

Moles of OH^- in RbOH

RbOH\rightarrow Rb^++OH^-

1 L of RbOH= 0.100 moles

then 0.2 [RbOH] solution= 0.2 \times 0.100 moles = 0.02 moles

Moles of OH^- = 0.02 moles

so,Total moles of OH^- ions = 0.01 + 0.02=0.030 moles      ....(2)

Step 3: Comparing the moles of both H^+\text{ and }OH^- ions

One mole of H^+ ions will combine with one mole of OH^- ions, so

Total moles of H^+ ions  = 0.005+0.02= 0.025 moles....(1)

Total moles of OH^- ions = 0.01 + 0.02=0.030 moles.....(2)

For a solution to be neutral, we have

Total moles of H^+ ions = total moles of OH^- ions

0.025 moles H^+ will neutralize the 0.025 moles of OH^-

Moles of OH^- ions is in excess        (from 1 and 2)

The remaining moles of OH^- will be = 0.030 - 0.025 = 0.005 moles

So,The resulting solution will not be neutral.

Remaining Concentration of OH^- ions = \frac{\text{Moles remaining}}{\text{Total volume}}

[OH^-]=\frac{0.005}{0.85}=0.0058M

6 0
4 years ago
Which of these characteristics best describes a basic solution?
Liula [17]
Short Answer:  <span>Solutions of dilute, weak acids - the only kind that you might taste - are sour. Weak, dilute bases are bitter. Solutions of bases are slippery.

Hope I get brainliest!</span>
6 0
3 years ago
Read 2 more answers
50 POINTS BRAINLIEST - How many L of gas would 200.0 grams of CCl^4 occupy at STP? pls show work
Paladinen [302]

Answer:Oxygen has a molar mass of 16.00 g/mol. One mole of CO2 contains 32.00 grams of oxygen. One mole equals 22.4L at standard temperature and pressure. 0.60L CO2 equals 0.0268 mol at STP.

Explanation:

8 0
3 years ago
A helium gas balloon is expanded to 78.0 L, while the pressure is held constant at 0.37 atm. If the work done on the gas mixture
Elenna [48]

Answer:

77.248 L

Explanation:

From the question,

Work done on the gas mixture is given as,

W = PΔV.................. Equation 1

Where W = work done, P = pressure of the the gas, ΔV = Change in volume of the gas.

make ΔV the subject of the equation

ΔV = W/P..................... Equation 2

Given: W = 28.2 J, P = 0.37 atm = (0.37×101325) N/m² = 37490.25 N/m²

Substitute into equation 2

ΔV = 28.2/37490.25

ΔV = 0.000752 m³

ΔV = 0.752 L

But,

ΔV = V₂-V₁................. Equation 3

Where V₂ = Final volume of the helium gas, V₁ = Initial volume of the helium gas

make V₁ the subject of the equation

V₁ = V₂-ΔV................ Equation 4

Given: V₂ = 78 L.

Substitute into equation 4

V₁ = 78-0.752

V₁ = 77.248 L

8 0
3 years ago
Read 2 more answers
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