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djyliett [7]
3 years ago
10

The density of copper is 8.92 g/cm3. What is the volume in L of a 230.14 mg sample of copper?

Chemistry
1 answer:
aleksandr82 [10.1K]3 years ago
8 0
D=M/V so you plug in 8.92 for the density and 230.14 for the mass. then solve and get 25.8. you may need to do dimensional analysis with the different units of measures
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Why is the atom neutrally charged?
V125BC [204]
Atoms are electrically neutral because they have equal numbers of protons (positively charged) and electrons (negatively charged).
3 0
3 years ago
if 26.8g of copper (ll) chloride are dissolved in sufficient water to make 4.00 L of solution, what is the molarity of the solut
artcher [175]

Answer: Molarity of the solution is 4.97 \times 10^{-2} M and water is the solvent.

Explanation:

Given: Mass of solute = 26.8 g

Volume = 4.00 L

Now, moles of copper (II) chloride (molar mass = 134.45 g/mol) are calculated as follows.

No. of moles = \frac{mass}{molar mass}\\= \frac{26.8 g}{134.45 g/mol}\\= 0.199 mol

Molarity is the number of moles of a substance divided by volume of solution in liter.

Therefore, molarity of given solution is calculated as follows.

Molarity = \frac{no. of moles}{Volume (in L)}\\= \frac{0.199 mol}{4.00 L}\\= 0.04975 M\\= 4.97 \times 10^{-2} M

Solvent is defined as a component which is present in higher amount in a solution. Generally, a solvent is present in liquid state but it can also be a solid or gas.

In the given solution, copper (II) chloride is dissolved in water so copper (II) chloride is the solute and water is the solvent.

Thus, we can conclude that molarity of the solution is 4.97 \times 10^{-2} M and water is the solvent.

4 0
3 years ago
A sample of sand, 5cm in diameter and 10 cm long, was prepared at a porosity of 50% in a constant head apparatus. The total head
GarryVolchara [31]

Answer:

K = 2.037*10^{-3} m/s

V_s = 0.0122 \ m/s

Explanation:

Given that;

diameter (d) = 10cm/2 = 0.1m/2 = 0.05 m

length (l) = 10 cm = 0.1 m

porosity = 50%

height (h) = 30 cm = 0.3 m

time (t) = 5 s

volume (v) = 60 cm³ = 60 × 10⁻⁶ m³

Q (flow rate) = \frac{v}{t}

Q = \frac{60*10^{-6} m^3}{5}

Q = 12*10^{-6} m^3 /sec

From constant head method, we use the relation K = (\frac{Q*L}{A*h}) to determine the hydraulic conductivity ; we have:

K = \frac{12*10^{-6}*0.1}{\frac{\pi}{4}0.05^2*0.3}

K = 0.002037\\\\K = 2.037*10^{-3} m/s

Seepage velocity V_s = \frac{velocity }{porosity}

where; velocity = K*i

=(2.037*10^{-3}*)(\frac{0.3}{0.1})

= 6.111*10^{-3} m/s

V_s = \frac{6.111*10^{-3}}{0.5}

V_s = 0.0122 \ m/s

8 0
3 years ago
Find the density of Liquid A: Liquid A: Mass-138g Volume – 100 mL.
nlexa [21]

Answer:

The answer is

<h2>1.38 g/mL</h2>

Explanation:

The density of a substance can be found by using the formula

density =  \frac{mass}{volume}  \\

From the question

mass of liquid = 138 g

volume = 100 mL

The density of the liquid is

density =  \frac{138}{100}  = \frac{69}{5 0}

We have the final answer as

<h3>1.38 g/mL</h3>

Hope this helps you

7 0
3 years ago
Substance X is a compound containing 632mg of manganese and 368mg of oxygen. Substance X is shown
defon

The empirical formula : MnO₂.

<h3>Further explanation</h3>

Given

632mg of manganese(Mn) = 0.632 g

368mg of oxygen(O) = 0.368 g

M Mn = 55

M O = 16

Required

The empirical formula

Solution

You didn't include the pictures, but the steps for finding the empirical formula are generally the same

  • Find mol(mass : atomic mass)

Mn : 0.632 : 55 = 0.0115

O : 0.368 : 16 =0.023

  • Divide by the smallest mol(Mn=0.0115)

Mn : O =

\tt \dfrac{0.0115}{0.0115}\div \dfrac{0.023}{0.0115}=1\div 2

The empirical formula : MnO₂

8 0
3 years ago
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