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3 years ago

18. An example of an atom that has no

Chemistry

2 answers:

fgiga [73]3 years ago

7 0

Answer:

C. 2 protons, 2 electrons, and 1 neutron.

Explanation:

An atom with no charge or neutral charge is an atom with a charge of 0.

Neutrons have no charge so they do not affect it.

Electrons have a negative charge and protons have a positive charge. An equal number of protons and electrons will make the charge neutral.

allochka39001 [22]3 years ago

6 0

Answer:

2 protons, 2 electrons, and 1 neutron.

is correct answer

Explanation:

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0.65 moles of O2 originally at 85°C is cooled

soldier1979 [14.2K]

Answer:

2.09 atm

Explanation:

We can solve this problem by using the equation of state for an ideal gas, which relates the pressure, the volume and the temperature of an ideal gas:

$pV=nRT$

where

p is the pressure of the gas

V is its volume

n is the number of moles

R is the gas constant

T is the absolute temperature

In this problem we have:

n = 0.65 mol is the number of moles of the gas

V = 8.0 L is the final volume of the gas

$T=40C+273=313 K$ is the temperature of the gas

$R=0.082 atm L mol^{-1} K^{-1}$ is the gas constant

Solving for p, we find the final pressure of the gas:

$p=\frac{nRT}{V}=\frac{(0.65)(0.082)(313)}{8.0}=2.09 atm$

8 0

4 years ago

The heat of solution is found by adding a salt to water in a calorimeter and measuring the temperature change. The specific heat

melomori [17]

Answer:

The heat of the solution of salt is 1.66.11 J/g.

Explanation:

Mass of the water = m = 46.52 g

Initial temperature of the water = $T_1=22.83 ^oC$

Final temperature of the water = $T_2=18.98^oC$

The specific heat of water, c = 4.180 J/gºC

Heat associated with water on dissolving salt: q

$Q=mc(T_2-T_1)$

$Q=46.52 g\times 4.180 J/g&oC\times (18.98^oC-22.83 ^oC)$

$Q=-748.65 J$

Negative sign means that heat was lost by water on an addition of a salt.

Heat released on dissolving of salt = -Q = 748.65 J

Mass of salt added = 4.5069 g

Heat of the solution of salt :

= $\frac{-Q}{ 4.5069 g}=\frac{748.65 J}{ 4.5069 g}=1.66.11 J/g$

The heat of the solution of salt is 1.66.11 J/g.

5 0

4 years ago

Is cooking food a chemical change or physical change

Monica [59]

Explanation:

A chemical change results from a chemical reaction, while a physical change is when matter changes forms but not chemical identity. Examples of chemical changes are burning, cooking, rusting, and rotting. Examples of physical changes are boiling, melting, freezing, and shredding.

To understand why cooking is a chemical change, you should first understand what is a chemical change. Basically, all changes in this world can be classified as either physical changes or chemical changes. The difference is that chemical changes bring about new substances while physical changes don’t. Take the example of baking: when you bake a cake, the most immediately observable change is that it expands. This is because the baking soda in it has undergone a chemical change under heat to release carbon dioxide. Notice there is no carbon dioxide in the cake before we bake it. That is what I mean by bringing about new substances.

So why is cooking a chemical change? Because almost all cooking methods involving the rise of temperature (which is basically to say, all cooking methods) involve chemical changes. Once under heat, the antioxidants omnipresent in vegetables will get oxidized and the proteins in meats will get denatured. Among other things, the former process will mostly result in the change of color of the vegetables, and the latter the stiffening of the meats

4 0

4 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

PLEASE HELP this is chemistry 12 points

Inessa05 [86]

Answer:

6.791

Explanation:

For proper significant figures with addition, you would use the significant figures of the number with lowest decimal place. 6.298 goes to the 10⁻³ place. 0.492712 goes to the 10⁻⁶ place. You will go out to the 10⁻³ place.

6.298 + 0.492712 = 6.790712 ≈ 6.791

8 0

3 years ago