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3 years ago

I can’t find the answer, can you help me please?

Chemistry

1 answer:

Tatiana [17]3 years ago

7 0

Answer: Warm ocean air then rises into the storm, then forming an area of low pressure underneath. :)

Explanation:

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How many molecules are in 1.28 mols of silicon dioxide

coldgirl [10]

Answer:

7.71x10^23 molecules

Explanation:

Avogadro's # = 6.022x10^23

1.28 mol SiO2 x 6.022x10^23/ 1 mol SiO2 = 7.71x10^23 molecules

8 0

3 years ago

Ag2O(s) → 2Ag(s) + ½ O2(g) ΔH° = 31.05 kJ Which statements concerning the reaction above are true? (1) heat is released (2) heat

Sidana [21]

Answer:

D) 2, 4, and 5

Explanation:

In order to fully comprehend the answer choices we must take a close look at the value of ΔH° = 31.05. The enthalpy change of the reaction is positive. A positive value of enthalpy of reaction implies that heat was absorbed in the course of the reaction.

If heat is absorbed in a reaction, that reaction is endothermic.

Since ∆Hreaction= ∆H products -∆H reactants, a positive value of ∆Hreaction implies that ∆Hproducts >∆Hreactants, hence the answer choice above.

5 0

4 years ago

Calculate the fraction of lattice sites that are schottky defects for cesium chloride at its melting temperature (645oc). assume

baherus [9]

Answer : 7.87 X $10^{-6}$ .

Explanation : To find the fraction of schottky defects in the given lattice of CsCl,

we use the formula, $\frac{N_{s} }{N}} = exp ( \frac{-Q_{s}}{2KT} )$

on solving with the given values , $Q_{s}= 1.86 eV$ and T as 645 + 273 K and rest are the constants.

$\frac{N_{s} }{N}} = exp ( \frac{-1.86 eV}{2 X (8.62 X 10^{-5}) X (645 +273)} )$

we get the answer as 7.87 X $10^{-6}$ .

6 0

3 years ago

Read 2 more answers

How many cubic centimeters of an ore containing only 0.22% gold (by mass) must be processed to obtain $100.00 worth of gold? The

bezimeni [28]

Answer:

The cubic centimetres of the ore containing 0.22% gold (by mass) that must be processed to obtain the $100.00 worth of gold is approximately 216 cm³

Explanation:

The percentage by mass of gold in the ore = 0.22%

The density of the ore = 8.0 g/cm³

The price of the gold = $818 per troy ounce

14.6 troy oz = 1.0 pound

1 lb = 454 g

Given that one troy ounce = $818

$100 worth of gold = 1/818 ×100 troy ounce = 100/818 troy ounce

1 troy oz = 1.0/14.6 lb

100/818 troy oz = 100/818 × 1.0/14.6 lb = 250/29857 lb ≈ 0.0084 lb

1 lb = 454 g

250/29857 lb = 454 × 250/29857 g ≈ 3.8015 g

$100 = 3.8015 g worth of gold

The mass, M, of the ore containing 3.8015 g of gold is given as follows;

0.22% of M = 3.8015 g

0.22/100 × M = 3.8015 g

M = 3.8015 g × 100/0.22 = 1727.933 g

The volume, V, of the ore containing 3.8015 g of gold is given as follows;

Density of ore = Mass of ore/(Volume of ore)

Volume of ore = Mass of ore /(Density of ore)

The density of the ore = 8.0 g/cm³

Volume of ore = 1727.933 g /(8.0 g/cm³) = 215.99 cm³ ≈ 216 cm³

Therefore, the cubic centimetres of the ore containing 0.22% gold (by mass) that must be processed to obtain the $100.00 worth of gold ≈ 216 cm³.

5 0

4 years ago

How many liters are in 4.5 moles of CO2 gas at STP? Pls help :)

slavikrds [6]

Answer:

Explanation:

NCO 2= 4,5

VCO2= 4,5* 22,4=100,8

5 0

3 years ago