How many significant figure are in the number 0.000850320

Silicon is prepared by the reduction of K₂SiF6 with Al. Write the equation for this reaction. (Hint: Can F⁻ be oxidized in this

BlackZzzverrR [31]

4Al + 3K2SiF6 = 6KF + 3Si + 4AIF3 is the reaction for preparation of silicon by the reduction of K₂SiF6 with Al.

AlF3xH2O-based inorganic compounds are referred to as aluminium fluoride. They are all solids without colour. Aluminium fluoride is a crystalline (sand-like), odourless, white, or colourless powder. In addition to being used to make aluminium, it also functions as a flux in welding processes and in ceramic glazes and enamels.

Silicon (Si) is created by reducing potassium silicofluoride with aluminium as the reducing agent (K2SIF6). While K2SiF6 is reduced to Si in this equation, aluminium is oxidised to aluminium fluoride. As a result, the balanced equation describing aluminum's reduction of K2SiF6 to silicon non-metal is as follows: 4Al + 3K2SiF6 = 6KF + 3Si + 4AIF3

Learn more about aluminium fluoride here:

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3 0

11 months ago

If 100.0 grams of na3n decompose to form sodium and nitrogen, how many moles of sodium are formed? write and balance equation be

Yakvenalex [24]

The balanced equation is:
$2Na_3N-\ \textgreater \ 6Na + N_2$

Then proceed with the following equations.

$100g Na_3N*(\frac{1molNa_3N}{82.98gNa_3N})*(\frac {6mol Na}{2molNa_3N})*(\frac {22.99gNa}{1molNa})=83.12gNa$

The answer is $83.12gNa$ .

7 0

3 years ago

Read 2 more answers

A sample of 250 g of water are heated from 40°C to 121°C, calculate the amount of heat energy absorbed.

Marysya12 [62]

Answer:

the anwser isn't in the choices

Explanation:

H=MC(change of temp.)

M=mass of water=250g

C=specific heat of water = 4.186 j/g

change in temperature is 121-40= 81

H= 250x4.186x81=84766.5J

7 0

3 years ago

A person with a body temperature of 37°C holds an ice cube with a temperature of 0°C in a room where the air temperature is 20.°

Arlecino [84]

The answer is (2). Heat always flows down the temperature gradient, from high temperature to low temperature. Therefore, since the person is the warmest, heat flows from the person to both the ice and the air. Additionally, since the air is warmer than the ice, heat flows from the air to the ice.

8 0

3 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