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2 years ago

How is bakelite made and what is it used for?

Chemistry

1 answer:

Ivan2 years ago

8 0

Answer:

Bakelite is made from the reaction between formaldehyde and phenolic materials at high temperatures. Bakelite is used for wire insulation, break pads and other automotive parts, and other electronics components.

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The standard enthapy of combustion of solid urea co(nh2)2 is -632kjmol-1 at 298k

Setler79 [48]

Answer:

The standard enthalpy of combustion of solid urea ((CO(NH2)2) is -632 kJ mol-1 at 298 K and its standard molar entropy is 104.60 J K-1 mol-1

Explanation:

8 0

3 years ago

What type of reaction is most likely to occur when barium reacts with fluorine? combustion synthesis decomposition single replac

Lana71 [14]

Answer is: synthesis.
Chemical reaction: Ba + F₂ → BaF₂.
Synthesis is type of reaction where two or more compounds (in this reaction barium and fluorine) react to form one product (in this reaction BaF₂).
BaF₂ - barium fluoride, salt, <span>white cubic crystals, soluble in methanol and ethanol.</span>

8 0

2 years ago

Read 2 more answers

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

2 years ago

Read 2 more answers

Nicotine, a component of tobacco, is composed of C, H, and N. A 7.875-mg sample of nicotine was combusted, producing 21.363 mg o

Gnom [1K]

Answer: The empirical formula for the given compound is $C_5H_7N$

Explanation:

The chemical equation for the combustion of compound having carbon, hydrogen, and nitrogen follows:

$C_xH_yN_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of carbon, hydrogen and nitrogen respectively.

We are given:

Mass of $CO_2=21.363mg=21.363\times 10^3g=21363g$

Mass of $H_2O=6.125g=6.125\times 10^3g=6125g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 21363 g of carbon dioxide, $\frac{12}{44}\times 21363=5826.27g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18 g of water, 2 g of hydrogen is contained.

So, in 6125 g of water, $\frac{2}{18}\times 6125=680.55$ of hydrogen will be contained.

Now we have to calculate the mass of nitrogen.

Mass of nitrogen in the compound = (7875) - (5826.27 + 680.55) = 1368.18 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{5826.27g}{12g/mole}=485.52moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{680.55g}{1g/mole}=680.55moles$

Moles of Nitrogen = $\frac{\text{Given mass of nitrogen}}{\text{Molar mass of nitrogen}}=\frac{1368.18g}{14g/mole}=97.73moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.0154 moles.

For Carbon = $\frac{485.52}{97.73}=4.96\approx 5$

For Hydrogen = $\frac{680.55}{97.73}=6.96\approx 7$

For Nitrogen = $\frac{97.73}{97.73}=1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : N = 5 : 7 : 1

Hence, the empirical formula for the given compound nicotine is $C_5H_7N_1=C_5H_7N$

7 0

2 years ago

James has three cubes with identical dimensions. He places the blocks in water, as shown below.

Fantom [35]

Answer:I think it could be A or B but I would choose A.

Explanation:

3 0

3 years ago