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3 years ago

15 POINTS ASAP PLEASE!!

1 answer:

german3 years ago

8 0

Lymph<span> contains a variety of substances, including proteins, salts, glucose, fats, water, and white blood cells.</span>

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Based on the size and shape of the clasts in the sedimentary rock shown in the image, which classification would geologists mos

Veronika [31]

Answer:

D. Brecci

The pattern of the rock below resembles the distinct pattern of brecci.

The pattern of conglomerate usually appears to be more round peices in the rock. The pattern of brecci has a more shard like appearance.

Hope this helps,

One of the virtuosos on Brainly

3 0

3 years ago

Read 2 more answers

How many atoms are in 8.66 moles of barium? Round your answer to 2 decimal places.

yarga [219]

Answer:

52.15 × 10²³ atoms

Explanation:

Given data:

Number of moles = 8.66 mol

Number of atoms = ?

Solution:

Avogadro number:

The given problem will solve by using Avogadro number.

It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.

The number 6.022 × 10²³ is called Avogadro number.

For example,

1 mole = 6.022 × 10²³ atoms

8.66 mol × 6.022 × 10²³ atoms / 1mol

52.15 × 10²³ atoms

3 0

2 years ago

What is the volume of an object with a density of 2.6 g/cm³ and a mass of 30.5g?

inessss [21]

Answer:

The answer is

Explanation:

The volume of a substance when given the density and mass can be found by using the formula

$volume = \frac{mass}{density}$

From the question

mass of object = 30.5 g

Density = 2.6 g/cm³

The volume is

$volume = \frac{30.5}{2.6} \\ = 11.7307692...$

We have the final answer as

Hope this helps you

5 0

3 years ago

¿Qué hecho indujo a Heisenberg a establecer el principio de incertidumbre?

yaroslaw [1]

Answer:

Here's what I find

Explanation:

Heisenberg observed that if we want to locate a moving electron, we must bounce photons off it.

However, this makes it recoil. By the time the photon returns to our eye, the electron will no longer be in the same place.

He concluded that there is a limit to the precision with which we can simultaneously measure the position and speed (momentum) of a particle.

The more precisely we know the electron's speed, the less precisely we know its position and vice versa.

The uncertainty in the product of the two values cannot be less than a fixed small number.

7 0

3 years ago

CO2(g)+CCl4(g)⇌2COCl2(g) Calculate ΔG for this reaction at 25 ∘C under these conditions: PCO2PCCl4PCOCl2===0.140 atm0.185 atm0.7

padilas [110]

<u>Answer:</u> The $\Delta G$ for the reaction is 54.425 kJ/mol

<u>Explanation:</u>

For the given balanced chemical equation:

$CO_2(g)+CCl_4(g)\rightleftharpoons 2COCl_2(g)$

We are given:

$\Delta G^o_f_{CO_2}=-394.4kJ/mol\\\Delta G^o_f_{CCl_4}=-62.3kJ/mol\\\Delta G^o_f_{COCl_2}=-204.9kJ/mol$

To calculate $\Delta G^o_{rxn}$ for the reaction, we use the equation:

$\Delta G^o_{rxn}=\sum [n\times \Delta G_f(product)]-\sum [n\times \Delta G_f(reactant)]$

For the given equation:

$\Delta G^o_{rxn}=[(2\times \Delta G^o_f_{(COCl_2)})]-[(1\times \Delta G^o_f_{(CO_2)})+(1\times \Delta G^o_f_{(CCl_4)})]$

Putting values in above equation, we get:

$\Delta G^o_{rxn}=[(2\times (-204.9))-((1\times (-394.4))+(1\times (-62.3)))]\\\Delta G^o_{rxn}=46.9kJ=46900J$

Conversion factor used = 1 kJ = 1000 J

The expression of $K_p$ for the given reaction:

$K_p=\frac{(p_{COCl_2})^2}{p_{CO_2}\times p_{CCl_4}}$

We are given:

$p_{COCl_2}=0.735atm\\p_{CO_2}=0.140atm\\p_{CCl_4}=0.185atm$

Putting values in above equation, we get:

$K_p=\frac{(0.735)^2}{0.410\times 0.185}\\\\K_p=20.85$

To calculate the gibbs free energy of the reaction, we use the equation:

$\Delta G=\Delta G^o+RT\ln K_p$

where,

$\Delta G$ = Gibbs' free energy of the reaction = ?

$\Delta G^o$ = Standard gibbs' free energy change of the reaction = 46900 J

R = Gas constant = $8.314J/K mol$

T = Temperature = $25^oC=[25+273]K=298K$

$K_p$ = equilibrium constant in terms of partial pressure = 20.85

Putting values in above equation, we get:

$\Delta G=46900J+(8.314J/K.mol\times 298K\times \ln(20.85))\\\\\Delta G=54425.26J/mol=54.425kJ/mol$

Hence, the $\Delta G$ for the reaction is 54.425 kJ/mol

7 0

3 years ago