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3 years ago

Particles are arranged in a crystal structure in a sample of:

1 answer:

6 0

The answer is number 4. Ag(s) or Solid Silver. Silver in a solid phase has a face centered cubic crystal structure. Among your choices, silver is the only example of particles that are arranged in their crystal structure while the other choices are gases and liquid. Liquid and gases have loose atoms that do not form crystal structures.

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A reaction in which a substance reacts with oxygen, emitting heat and forming oxygen-containing compounds is an example of a(n):

yaroslaw [1]

Answer: B) combustion reaction.

Explanation:

A) acid-base reaction: When an acid reacts with a base, to form metal salt and water, this type of reaction is Acid Base reaction.

Example: $HCl+NaOH\rightarrow NaCl+H_2O$

B) combustion reaction: When a hydrocarbon reacts with oxygen to produce carbon dioxide and water, this type of reaction is combustion reaction.

$CH_4+2O_2\rightarrow CO_2+2H_2O$

C) precipitation reaction: a reaction in which aqueous solution of two compounds on mixing react to form an insoluble compound which separate out as a solid are called precipitation reactions.

$Na_2SO_4(aq)+BaCl_2(aq)\rightarrow BaSO_4(s)+NaCl(aq)$

D) gas evolution reaction: a reaction in which one of the product is formed as a gas.

$CaCO_3(s)\rightarrow CaO(s)+CO_2(g)$

8 0

3 years ago

The stability of an isotope is based on its

sattari [20]

Option 4. ratio of electrons to protons

Isotope that has atomic number of 82 is stable. An element that has an atomic number ∠82 more stable except Tc and Pm. Also there is the concept that isotopes consisting a combination of even-even,even-odd,odd-even, and odd-odd are all stable. Many isotopes with no magic numbers of nucleons are stable

7 0

2 years ago

A rigid tank contains 0.66 mol of oxygen (O2). Find the mass of oxygen that must be withdrawn from the tank to lower the pressur

dsp73

Answer:

12.8 g of $O_{2}$ must be withdrawn from tank

Explanation:

Let's assume $O_{2}$ gas inside tank behaves ideally.

According to ideal gas equation- $PV=nRT$

where P is pressure of $O_{2}$ , V is volume of $O_{2}$ , n is number of moles of $O_{2}$ , R is gas constant and T is temperature in kelvin scale.

We can also write, $\frac{V}{RT}=\frac{n}{P}$

Here V, T and R are constants.

So, $\frac{n}{P}$ ratio will also be constant before and after removal of $O_{2}$ from tank

Hence, $\frac{n_{before}}{P_{before}}=\frac{n_{after}}{P_{after}}$

Here, $\frac{n_{before}}{P_{before}}=\frac{0.66mol}{43atm}$ and $P_{after}=17atm$

So, $n_{after}=\frac{n_{before}}{P_{before}}\times P_{after}=\frac{0.66mol}{43atm}\times 17atm=0.26mol$

So, moles of $O_{2}$ must be withdrawn = (0.66 - 0.26) mol = 0.40 mol

Molar mass of $O_{2}$ = 32 g/mol

So, mass of $O_{2}$ must be withdrawn = $(32\times 0.40)g=12.8g$

7 0

2 years ago

Can someone answer this?

yulyashka [42]

No se si aun necesitas ayuda o no

3 0

2 years ago

Help please I need it bad

natta225 [31]

= 9.1 × 10^6
(scientific notation)

= 9.1e6
(scientific e notation)

= 9.1 × 10^6
(engineering notation)
(million; prefix mega- (M))

= 9100000
<span>(real number)</span>

5 0

2 years ago