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3 years ago

NEED ANSWER ASAP

Chemistry

1 answer:

Solnce55 [7]3 years ago

8 0

The phase change is called deposition, when a gas transforms into a solid without passing through the liquid phase.

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A fossil would most likely be found in a(n): igneous rock sedimentary rock metamorphic rock

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I think it is Sedimentary rock, igneous is from fire/lava metamorphic is magma, so Sedimentary

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The relative equilibrium concentrations of two chemical species (a and

Vilka [71]

ΔG° = 14.1 kJ/mol

For the reaction A → B, K = [B]/[A].

If [A] = 240 and [B] = 1, then

K = 1/240 = 4.167 x 10^(-3)

The relationship between ΔG° and K is:

ΔG° = -RTlnK

where

R = the gas constant = 8.314 J·K^(-1)mol^(-1)

T = the Kelvin temperature

In this problem, T = (37 + 273.15) K = 310.15 K

∴ #ΔG° = -8.314 J·K^(-1)mol^(-1) × 310.15 K × ln(4.167× 10^(-3)

= -2579 × [-5.481 J·mol^(-1)] = 14 100 J·mol^(-1) = 14.1 kJ/mol

Note: We should expect ΔG° to be positive because K < 1.

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3 years ago

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What is the formula for Tetrachlorine pentaiodide

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Answer:

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Explanation:

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3 years ago

A 1.10 g sample contains only glucose and sucrose. When the sample is dissolved in water to a total solution volume of 25.0L, th

olga_2 [115]

Answer:

$\large \boxed{79 \, \%}$

Explanation:

I assume the volume is 2.50 L. A volume of 25.0 L gives an impossible answer.

We have two conditions:

(1) Mass of glucose + mass of sucrose = 1.10 g

(2) Osmotic pressure of glucose + osmotic pressure of sucrose = 3.78 atm

Let g = mass of glucose

and s = mass of sucrose. Then

g/180.16 = moles of glucose, and

s/342.30 = moles of sucrose. Also,

g/(180.16×2.50) = g/450.4 = molar concentration of glucose. and

s/(342.30×2.50) = s/855.8 = molar concentration of sucrose.

1. Set up the osmotic pressure condition

Π = cRT, so

$\begin{array}{rcl}\Pi_{\text{g}} +\Pi_{\text{s}}&=&\Pi_{\text{tot}}\\\dfrac{g}{450.4}\times8.314\times298 + \dfrac{s}{855.8}\times8.314\times298 & = & 3.78\\\\5.501g + 2.895s & = & 3.78\\\end{array}$

Now we can write the two simultaneous equations and solve for the masses.

2. Calculate the masses

$\begin{array}{lrcl}(1)& g + m & = & 1.10\\(2) &5.501g +2.895s & = & 3.78\\(3) & m & = &1.10 - g\\&5.501g + 2.895(1.10 - g) & = & 3.78\\&2.606g + 3.185 & = & 3.78\\ &2.606g & = & 0.595\\(4) & g & = & \mathbf{0.229}\\&0.229 + s & = & 1.10\\& s & = & \mathbf{0.871}\\\end{array}$

We have 0.229 g of glucose and 0.871 g of sucrose.

3. Calculate the mass percent of sucrose

$\text{Mass percent} = \dfrac{\text{Mass of component}}{\text{Total mass}} \times \, 100\%\\\\\text{Percent sucrose} = \dfrac{\text{0.871 g}}{\text{1.10 g}} \times \, 100\% = 79 \, \%\\\\\text{The mixture is $\large \boxed{\mathbf{79 \, \%}}$ sucrose}$

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3 years ago

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Assoli18 [71]

The answer would be A) Physical Change. When the silver tarnishes it is still remnants of silver. If it were B, the silver would change into a different substance.

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