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4 years ago

Would the hardness of a mineral affect its performance in the streak test why or why not

Chemistry

1 answer:

insens350 [35]4 years ago

4 0

It would because diamonds are the hardest minerals and they do not streak unless it is on another diamond.

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To prevent tank rupture during deep-space travel, an engineering team is studying the effect of temperature on gases confined to

uranmaximum [27]

Answer: 63.88 atm

Explanation:

To answer this, we use the formula PV = nRT since the asumption is that the gas has an ideal behavior

where number of mole = 2.60 mol, R(gas constant) = 0.08205746 L atm/K mol,

T = 251 ∘C = (251 + 273) K = 524 K, Volume = 1.75 L

Making Pressure the subject of the formula, we have

P = nRT/V = 2.6 * 0.08205746 * 524/2.75 = 63.88 atm

6 0

3 years ago

Aluminum Oxide is formed when aluminum combines with oxygen in the air. How many

aleksandr82 [10.1K]

I think it’s something like 29.95g, sorry if it’s wrong long i haven’t done this in so long

7 0

3 years ago

A 112 gram sample of an unknown metal was heated from 0.0C to 20.0C. The sample absorbed 1004 J of energy. What was the specific

worty [1.4K]

Answer:

The specific heat of the sample unknown metal is approximately 0.45 J/g °C.

General Formulas and Concepts:
Thermodynamics

Specific Heat Formula: $\displaystyle q = mc \triangle T$

m is mass (g)
c is specific heat capacity (J/g °C)
ΔT is the change in temperature

Explanation:

Step 1: Define

Identify variables.

m = 112 g

ΔT = 20.0 °C

q = 1004 J

Step 2: Solve for c

Substitute in variables [Specific Heat Formula]: $\displaystyle 1004 \ \text{J} = (112 \ \text{g})(c)(20.0 \ ^{\circ} \text{C})$
Simplify: $\displaystyle 1004 \ \text{J} = (2240 \ \text{g} \ ^\circ \text{C})c$
Isolate c: $\displaystyle c = 0.448214 \ \text{J} / \text{g} \ ^\circ \text{C}$
Round [Sig Figs]: $\displaystyle c \approx 0.45 \ \text{J} / \text{g} \ ^\circ \text{C}$

∴ specific heat capacity c is equal to around 0.45 J/g °C.

---

Topic: AP Chemistry

Unit: Thermodynamics

5 0

3 years ago

A double replacement reaction occurs when aqueous potassium iodide and aqueous mercury (I) nitrate are mixed. What is the correc

pashok25 [27]

The answer should be 2KI(aq)+Hg(NO3)2(aq) = 2KNO3(aq) + HgI2(s). This is a double replacement reaction, the valence of elements will not change. The HgI2 is not soluble in water.

4 0

3 years ago

p32p32 is a radioactive isotope with a half-life of 14.3 days. if you currently have 63.163.1 g of p32p32 , how much p32p32 was

AlekseyPX

Answer:

92.93 g

Explanation:

Number of half lives that have elapsed in eight days =8/14.3 = 0.559

Fraction of the radioactive nuclide that remains after 0.559 half lives is given by

N/No=(1/2)^0.559

Where N= mass of radioactive nuclides remaining after a time t

No= mass of radioactive nuclides originally present

N/No=(1/2)^0.559= 0.679

Mass of nuclides present eight days before= 63.1g/0.679

Mass of nuclides present eight days before=92.93 g

6 0

4 years ago