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4 years ago

1. Which one is the odd one out? 0 CaO o Cu o CO o CH

Chemistry

1 answer:

Sauron [17]4 years ago

4 0

Answer:

Yttrium barium copper oxide (YBCO) is a family of crystalline chemical compounds, famous for displaying high-temperature superconductivity.It includes the first material ever discovered to become superconducting above the boiling point of liquid nitrogen (77 K) at about 92 K.Many YBCO compounds have the general formula Y Ba 2 Cu 3 O 7−x (also known as Y123),

Explanation:

i hope this makes sense and helps man

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Identify the right

Basile [38]

.answer:

so the first one is analogous

the second is homologous

and the third is analogous

explanation:

analogous. this means they share a similar function (flight) but do not have the same embryonic origin.

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3 years ago

Three forms of radiation are also called ionizing radiation

Trava [24]

Answer:

Alpha Particle, Beta Particles and Gamma Rays

Explanation:

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3 years ago

"To determine the amount of heroin in the mixture, you dissolve 1.00 g of the white powdery mixture in water in a 100.0-mL volum

UkoKoshka [18]

Explanation:

Formula to calculate osmotic pressure is as follows.

Osmotic pressure = concentration × gas constant × temperature( in K)

Temperature = $25^{o} C$

= (25 + 273) K

= 298.15 K

Osmotic pressure = 531 mm Hg or 0.698 atm (as 1 mm Hg = 0.00131)

Putting the given values into the above formula as follows.

0.698 = $C \times 0.082 \times 298.15 K
$

C = 0.0285

This also means that,

$\frac{\text{moles}}{\text{volume (in L)}}$ = 0.0285

So, moles = 0.0285 × volume (in L)

= 0.0285 × 0.100

= $2.85 \times 10^{-3
}$

Now, let us assume that mass of $C_{12}H_{23}O_{5}N$ = x grams

And, mass of $C_{12}H{22}O_{11}$ = (1.00 - x)

So, moles of $C_{12}H_{23}O_{5}N = \frac{mass}{\text{molar mass}}$

= $\frac{x}{369}$

Now, moles of $C_{12}H_{22}O_{11} = \frac{(1.00 - x)}{342}$

= $\frac{x}{369} + \frac{(1.00 - x)}{342}$

= $2.85 \times 10^{-3}$

= x = 0.346

Therefore, we can conclude that amount of $C_{12}H_{23}O_{5}N$ present is 0.346 g and amount of $C_{12}H_{22}O_{11}$ present is (1 - 0.346) g = 0.654 g.