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4 years ago

Are metal ions larger or smaller than the neutral atoms they came from

Chemistry

1 answer:

NeX [460]4 years ago

7 0

Answer: the metal ions or smaller then the neutral

Explanation:

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Porcentaje en masa formula y unidad<br><br>

elena-14-01-66 [18.8K]

Explanation:

Expresa los gramos de soluto por cada 100 gramos de disolución. Porcentaje masa = masa de soluto___ x 100 masa de la disolución Cuando trabajamos con la masa, podemos sumar el soluto y el disolvente para obtener la disolución.

6 0

3 years ago

Arrange steps in order to describe what happens to a gas when it is cooled

AnnZ [28]

A) Particles of gas move slower.

B) Gas changes to liquid.

C) The gas loses thermal energy.

D) Gas particles decrease.

3 0

3 years ago

Octane has a density of 0.692 g/ml at 20∘c. how many grams of o2 are required to burn 17.0 gal of c8h18

german

156251.099 grams of O₂ are required to burn 17.0 gal of C₈H₁₈

<h3>Further explanation</h3>

Density is a quantity derived from the mass and volume

Density is the ratio of mass per unit volume

With the same mass, the volume of objects that have a high density will be smaller than objects with a smaller type of mass

The unit of density can be expressed in g / cm³ or kg / m³

Density formula:

$\large{\boxed{\bold{\rho~=~\frac{m}{V} }}}$

ρ = density

m = mass

v = volume

1 gal equal to = 3785.41 ml

then 17.0 gal = 17 x 3785.41 = 64351.97 ml Octane

grams Octane = ρ x ml

grams Octane = 0.692 g.ml x 64351.97

grams Octane = 44531.563

molar mass Octane (C₈H₁₈) = 114

mole Octane = grams : molar mass

mole Octane = 44531.563 : 114

mole Octane = 390.627

From the reaction

C₈H₁₈ + 25/2 O₂ ⇒ 8 CO₂ + 9H₂O

mole C₈H₁₈ : mole O₂ = 1 : 25/2

$mole\:O_2\:=\:\frac{25}{2} \times\:390.627$

mole O₂ = 4882.846

grams O₂ = mole x molar mass

grams O₂ = 4882.846 x 32

grams O₂ = 156251.099

<h3>Learn more </h3>

moles of water

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grams of oxygen required

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mass of copper required

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Keywords: Octane, mole, mass, gal, density

5 0

3 years ago

HELP ME PLSSSSS THIS TEST IS TIMED!!!! Write a conclusion statement that addresses the following questions: • How did your exper

SVETLANKA909090 [29]

Although the data for the experiment was not provided, we can offer a generalized answer in that when performing an experiment to achieve absolute zero temperatures, the value will never match the exact value.

<h3 /><h3>What is absolute zero?</h3>

Absolute zero is the lower limit of temperature. It is considered the coldest possible temperature that can exist. However, any attempt to reach this temperature in a controlled environment has failed, <u>scientists do not think it is possible to recreate this </u><u>temperature</u><u>. </u>

Therefore, we can confirm that the value of the absolute zero experiments did not match the accepted value. If the hypothesis was that it would be difficult or impossible to achieve, then the data would support the hypothesis, otherwise, it would fail to do so.

In summary, absolute zero is a temperature that cannot be recreated in a lab, so the value in this experiment does not match the accepted value and there is <u>no further exploration </u>to be done on this matter.

To learn more about absolute zero visit:

brainly.com/question/79835?referrer=searchResults

4 0

2 years ago

An excess of sodium carbonate, Na, CO3, in solution is added to a solution containing 17.87 g CaCl2. After performing the

Brrunno [24]

Answer:

Approximately $81.84\%$ .

Explanation:

Balanced equation for this reaction:

${\rm Na_{2}CO_{3}}\, (aq) + {\rm CaCl_{2}} \, (aq) \to 2\; {\rm NaCl}\, (aq) + {\rm CaCO_{3}}\, (s)$ .

Look up the relative atomic mass of elements in the limiting reactant, $\rm CaCl_{2}$ , as well as those in the product of interest, $\rm CaCO_{3}$ :

$\rm Ca$ : $40.078$ .
$\rm Cl$ : $35.45$ .
$\rm C$ : $12.011$ .
$\rm O$ : $15.999$ .

Calculate the formula mass for both the limiting reactant and the product of interest:

$\begin{aligned}& M({\rm CaCl_{2}}) \\ &= (40.078 + 2 \times 35.45)\; {\rm g \cdot mol^{-1}} \\ &= 110.978\; \rm g \cdot mol^{-1}\end{aligned}$ .

$\begin{aligned}& M({\rm CaCO_{3}}) \\ &= (40.078 + 12.011 + 3 \times 15.999)\; {\rm g \cdot mol^{-1}} \\ &= 100.086\; \rm g \cdot mol^{-1}\end{aligned}$ .

Calculate the quantity of the limiting reactant ( $\rm CaCl_{2}$ ) available to this reaction:

$\begin{aligned}n({\rm CaCl_{2}) &= \frac{m({\rm {CaCl_{2}})}}{M({\rm CaCl_{2}})} \\ &= \frac{17.87\; \rm g}{110.978\; \rm g \cdot mol^{-1}} \\ &\approx 0.161023\; \rm mol \end{aligned}$ .

Refer to the balanced equation for this reaction. The coefficients of the limiting reactant ( $\rm CaCl_{2}$ ) and the product ( ${\rm CaCO_{3}}$ ) are both $1$ . Thus:

$\displaystyle \frac{n({\rm CaCO_{3}})}{n({\rm CaCl_{2}})} = 1$ .

In other words, for every $1\; \rm mol$ of $\rm CaCl_{2}$ formula units that are consumed, $1\; \rm mol\!$ of $\rm CaCO_{3}$ formula units would (in theory) be produced. Thus, calculate the theoretical yield of $\rm CaCO_{3}\!$ in this experiment:

$\begin{aligned} & n(\text{${\rm CaCO_{3}}$, theoretical}) \\ =\; & n({\rm CaCl_{2}}) \cdot \frac{n({\rm CaCO_{3}})}{n({\rm CaCl_{2}})} \\ \approx \; & 0.161023\; {\rm mol} \times 1 \\ =\; & 0.161023\; \rm mol\end{aligned}$ .

Calculate the theoretical yield of this experiment in terms of the mass of $\rm CaCO_{3}$ expected to be produced:

$\begin{aligned} & m(\text{${\rm CaCO_{3}}$, theoretical}) \\ = \; & n(\text{${\rm CaCO_{3}}$, theoretical}) \cdot M(({\rm CaCO_{3}}) \\ \approx \; & 0.161023\; {\rm mol} \times 100.086\; {\rm g \cdot mol^{-1}} \\ \approx \; & 16.1161\; \rm g \end{aligned}$ .

Given that the actual yield in this question (in terms of the mass of $\rm CaCO_{3}$ ) is $13.19\; \rm g$ , calculate the percentage yield of this experiment:

$\begin{aligned} & \text{percentage yield} \\ =\; & \frac{\text{actual yield}}{\text{theoretical yield}} \times 100\% \\ \approx \; & \frac{13.19\; {\rm g}}{16.1161\; {\rm g}} \times 100\% \\ \approx \; & 81.84\%\end{aligned}$ .

6 0

3 years ago