Which word means the remains or imprints of once-living organisms found in layers of rock

10. Copper(i) bromide reacts with magnesium metal: 2 CuBr + Mg → 2 Cu + MgBrz

velikii [3]

Answer:

72.6 grams

Explanation:

I got this answer through stoichiometry. For every 1 mole of Mg, 2 moles of CuBr are consumed. Because of this, multiply the moles of Mg by 2. Then, convert moles to grams.

5 0

2 years ago

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What is the freezing point of a solution of 498mL of water (solute) dissolved in 2.50 L of ethanol (solvent), C2H5OH? The densit

jok3333 [9.3K]

Answer:

Freezing T° of solution is -142.4°C

Explanation:

This excersise is about colligative properties, in this case freezing point depression,

ΔT = Kf . m . i

Where ΔT = Freezing T° of solvent - Freezing T° of solution

Kf = Cryoscopic constant

m = mol/kg (molality)

i = Number of ions dissolved.

Water is not ionic, so i = 1

Let's find out m.

We determine mass of water, by density

498ml . 1 g/mL = 498 g

We convert the mass of water to moles → 498 g . 1mol/18g = 27.6 moles

We determine mass of solvent by density

2500 mL . 0.789 g/mL = 1972.5 g

Notice, we had to convert L to mL to cancel units.

1 cm³ = 1 mL

We convert the mass from g to kg → 1972.5 g . 1kg /1000g = 1.97kg

We determine m = mol/kg → 27.6mol / 1.97kg = 13.9 m

Kf for ethanol is: 1.99 °C/m

Freezing T° for ethanol is: -114.6°C

We replace at formula: - 114.6°C - Freezing T° solution = 1.99 °C/m . 13.9 m . 1

- 114.6°C - Freezing T° solution = 27.8 °C

- Freezing T° solution = 27.8°C + 114.6°C

Freezing T° Solution = - 142.4 °C

7 0

2 years ago

Determine how many millilitres of a 4.25 M HCl solution are needed to react completely with 8.75 g CaCO3?

Helen [10]

Answer:

41 mL

Explanation:

Given data:

Milliliter of HCl required = ?

Molarity of HCl solution = 4.25 M

Mass of CaCO₃ = 8.75 g

Solution:

Chemical equation:

2HCl + CaCO₃ → CaCl₂ + CO₂ + H₂O

Number of moles of CaCO₃:

Number of moles = mass/molar mass

Number of moles = 8.75 g / 100.1 g/mol

Number of moles = 0.087 g /mol

Now we will compare the moles of CaCO₃ with HCl.

CaCO₃ : HCl

1 : 2

0.087 : 2/1×0.087 = 0.174 mol

Volume of HCl:

Molarity = number of moles / volume in L

4.25 M = 0.174 mol / volume in L

Volume in L = 0.174 mol /4.25 M

Volume in L = 0.041 L

Volume in mL:

0.041 L×1000 mL/ 1L

41 mL

8 0

2 years ago

Consider two concentric spheres whose radii differ by t = 1 nm, the "thickness" of the interface. At what radius (in nm) of sphe

Gala2k [10]

Answer:

Radius of the interior sphere = 3.847 nm

Explanation:

The volume of the shell (Vs) is equal to the difference of the volume of the outer sphere (Vo) and the volume of the inner sphere (Vi). Then:

$V_s=V_o-V_i=V_i\\V_o=2*V_i\\$

If we express the radius of the outer sphere (ro) in function of the radius of the inner sphere (ri), we have (e being the shell thickness):

$r_o=r_i+e$

The first equation becomes

$\frac{4 \pi}{3}*r_o^{3} = 2 * \frac{4 \pi}{3}*r_i^{3} \\\\\frac{4 \pi}{3}*(r_i+e)^{3} = 2 * \frac{4 \pi}{3}*r_i^{3} \\\\(r_i+e)^{3} = 2 *r_i^{3} \\\\(r_i^{3}+3r_i^{2}e+3r_ie^{2}+e^{3})=2r_i^{3}\\\\-r_i^{3}+3r_i^{2}e+3r_ie^{2}+e^{3}=0\\\\-r_i^{3}+3r_i^{2}+3r_i+1=0$

To find ri that satisfies this equation we have to find the roots of the polynomial.

Numerically, it could be calculated that ri=3.847 nm satisfies the equation.

So if the radius of the interior sphere is 3.847 nm, the volume of the interior sphere is equal to the volume of the shell of 1nm.

8 0

2 years ago

Consider the condensed structure shown.

neonofarm [45]

Answer:

$C_{5}H_{10}\:C_{2}H_{6}$

Explanation:

Expanded Structure shows all the atoms and the bonds of the molecule. It is important to remember, each Carbon molecule have four bonds, while the Hydrogen just one. So:

1) $C_{5}H_{10}$

Pentene. 5 Carbon molecules, 10 Hydrogen molecules. It's an isomer, so this organic compound has multiple arrangements.

2) $C_{2}H_{6}$

Ethane. 2 Carbon and 6 Hydrogen Molecules. Check it below.

7 0

3 years ago