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3 years ago

The beta rays emitted from atomic nuclei are _____. an electron a neutron a proton

Chemistry

2 answers:

meriva3 years ago

7 0

Answer: Electrons I hope this helped! Your fellow Brainly user, GalaxyGamingKitty.

Sliva [168]3 years ago

6 0

The answer I believe is An Electron

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Why is important to only test one variable at a time?

Natalija [7]

Answer:

Explanation:

It is important to only test one variable at a time because you need to be able to disprove or prove a problem with just one independent variable. When you have several variables in the experiment, it would be impossible to know which variable honestly caused the end result.

4 0

2 years ago

The useful metal manganese can be extracted from the mineral rhodochrosite by a two-step process. In the first step, manganese(I

ale4655 [162]

Answer:

The answer is "6.52 kg and 13.1 kg"

Explanation:

For point a:

$Actual\ yield = 6.52 \ kg\\\\Percent \ yield= 66\%\\\\Percent \ yield = \frac{Actual \ yield}{Theoretical\ yield} \times 100 \%\\\\Theoretical\ yield \ of \ MnO_2 = \frac{Actual \ yield}{Percent \ yield} \times 100\%\\\\=\frac{6.52 \ kg}{66 \%} \times 100\% =9.88 \ kg\\\\$

Equation:

$3MnCO_3 +O_2 \longrightarrow 2MnO_2 + 2CO_2\\\\$

Calculating the amount of $MnCO_3$

$= 9.88 \ kg \times \frac{1000 \ g}{1 \ kg} \times \frac{1 \ mol \ MnO_2}{86.94 \ g} \times \frac{2 \ Mol \ MnCO_3}{2 \ mol \ MnO_2} \times \frac{114.95\ g}{1 \ mol \ MnCO_3 }\times \frac{1\ kg}{1000\ g}\\\\= 13.1 \ kg$

For point b:

$Actual\ yield = 4.0 \ kg\\\\Percent\ yield=97.0\%\\\\Percent \ yield = \frac{Actual \ yield}{Theoretical \ yield} \times 100 \% \\\\Theoretical \ yield\ of\ Mn = \frac{Actual \ yield}{Percent \ yield} \times 100\%\\\\$

$=\frac{4.0 \ kg}{97.0\%} \times 100\% =4.12 \ kg$

Equation:

$3MnO_2 +4AL \longrightarrow 3Mn + 2AL_2O_3\\\\$

Calculating the amount of $MnO_2:$

$= 4.12 \ kg \times \frac{1000 \ g}{1 \ kg} \times \frac{1 \ mol \ Mn}{54.94 \ g} \times \frac{3 \ Mol \ MnO_2}{3 \ mol \ Mn} \times \frac{86.94 \ g}{1 \ mol \ MnO_2 }\\\\= 6516 \ g \\\\=6.52 \ kg\\\\$

7 0

3 years ago

Determine the mass of the H2CO3 produced.

FrozenT [24]

Answer:

62.03 g/mol

Explanation:

7 0

3 years ago

What is the maximum number of moles of Al2O3 that can be produced by the reaction of .4 mol of Al with .4 mol of O2

Darya [45]

Given:

Moles of Al = 0.4

Moles of O2 = 0.4

To determine:

Moles of Al2O3 produced

Explanation:

4Al + 3O2 → 2Al2O3

Based on the reaction stoichiometry:

4 moles of Al produces 2 moles of Al2O3

Therefore, 0.4 moles of Al will produce:

0.4 moles Al * 2 moles Al2O3/4 moles Al = 0.2 moles Al2O3

Similarly;

3 moles O2 produces 2 moles Al2O3

0.4 moles of O2 will yield: 0.4 *2/3 = 0.267 moles

Thus Al will be the limiting reactant.

Ans: Maximum moles of Al2O3 = 0.2 moles

4 0

3 years ago

Read 2 more answers

Although both N2 and 02 are naturally present in the air we breathe, high levels of NO and NO2 in the atmosphere occur mainly in

joja [24]

Answer:

(a) Increasing the temperature adds heat which is a reactant shifting the equilibrium rightwards.

(b) Pressure has no effect since the change in the number of moles is zero.

Explanation:

Hello,

In this case, one could represent the given reaction as:

$N_2(g)+O_2(g)+ \Delta _rH \leftrightarrow 2NO$

Since it is endothermic. Thus, solving the (a) statement, one identifies the heat as a reagent, that is why the reaction cools down as it progress, therefore, by increasing the temperature, heat is added, that is, a reagent is added, which shifts the equilibrium rightwards, in other words, more NO is produced so its concentration increases.

Furthermore, for the (b) statement, since the change in the number of moles is zero, based on the stoichiometric coefficients as shown below:

$\Delta \nu =2-1-1=0$

Such value implies that the pressure has no effect on the concentration, taking into account the following form of the law of mass action:

$Kp=Kc(RT)^{\Delta \nu }$

Thus, since $\Delta \nu =0$ , $Kp=Kc$ , so no effect in concentration is due to the pressure.

Best regards.

6 0

3 years ago