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Ronch [10]
2 years ago
12

How exactly are synthetic paint brushes made? Nylon?

Chemistry
1 answer:
oksian1 [2.3K]2 years ago
6 0

Answer:

<h2>Synthetic brushes <em>are typically made from nylon bristles. Other times, they're made </em><em>from </em><em>plastic</em></h2><h3 /><h3 /><h3>#CARRYONLEARNING</h3>

BY : <em>EDLYN MORENO</em>

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What is the Oort cloud​
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The Oort cloud sometimes called the Öpik–Oort cloud it was first described in 1950 by Dutch astronomer Jan Oort
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3 years ago
The process that makes atmospheric nitrogen available to plants by mutualistic and free-living bacteria is called
artcher [175]
Nitrogen fixation is the process that makes atmospheric nitrogen available to plants by mutualistic and free-living bacteria. The process is undertaken by the rhizobium bacteria that live in root roots of plants such as legumes. The mutualistic relationship is that the plant supplies the bacteria with a habitat in which to live, water, and nutrients, and the bacteria supply nitrogen for making plant proteins. 
3 0
3 years ago
Complete and balance the following acid-base equations:
Alchen [17]

Answer:

a) HClO_4 (aq) + LiOH \rightarrow LiClO_4 + H_2O

b) H_2SO_4(aq) + 2NaOH(aq) \rightarrow Na_2SO_4 (aq) + 2H_2O (l)

c) Ba(OH)2 + 2HF\rightarrow BaF_2 + 2H_2O

Explanation:

a)

when HClO_4 is added to LiOH, lithium chlorate and water is formed.

HClO_4 (aq) + LiOH \rightarrow LiClO_4 + H_2O

Balancing of above reaction,

It can be seen that all the atoms in both the sides are balanced. so it is a balanced reaction.

(b) When aqueous H_2SO_4 is added to NaOH, Na_2SO_4 and H_2O is formed. It is a neutrilization reaction.

H_2SO_4(aq) + NaOH(aq) \rightarrow Na_2SO_4 (aq) + H_2O (l)

Balancing of above reaction,

First balance all the atoms except O and H

S atom is already balanced in either side

No. of Na atom in left hand side = 1

No. of Na atom in right hand side = 2

So multiply NaOH by 2, now the reaction becomes:

H_2SO_4(aq) + 2NaOH(aq) \rightarrow Na_2SO_4 (aq) + H_2O (l)

No. of O atoms in left hand side = 6

No. of O atoms in right hand side = 5

So multiply H2O by 2, now the reaction becomes:

H_2SO_4(aq) + 2NaOH(aq) \rightarrow Na_2SO_4 (aq) + 2H_2O (l)

Now, it can been seen that all the atoms are balanced.

c) When Ba(OH)2 reacts with HF, baroum fluoride and water is formed.

Ba(OH)2 + HF\rightarrow BaF_2 + H_2O

Balancing of above reaction,

Barium atoms are already balanced on either side.

No. of F atoms on left hand side = 1

No. of F atoms on right hand side = 2

So, multiply HF by 2, now reaction becomes

Ba(OH)2 + 2HF\rightarrow BaF_2 + H_2O

In order to balance H and O on either side, multiply H2O by 2.

Ba(OH)2 + 2HF\rightarrow BaF_2 + 2H_2O

7 0
3 years ago
HELP ME OMG WHY IS NOBODY HELPING ME SJCNEOVKRM
Sholpan [36]

I believe it would be CS and GE (option 3). i hope i helped ya out.

3 0
3 years ago
One of the reactions that occurs in a blast furnace, in which iron ore is converted to cast iron, is Fe2O3 + 3CO → 2Fe + 3CO2 Su
Tpy6a [65]

Answer : The percent purity of Fe_2O_3 in the original sample is 87.94 %

Explanation :

The given balanced chemical reaction is:

Fe_2O3+3CO\rightarrow 2Fe+3CO_2

First we have to calculate the mass of Fe.

\text{Moles of }Fe=\frac{\text{Mass of }Fe}{\text{Molar mass of }Fe}

Molar mass of Fe = 55.8 g/mole

\text{Moles of }Fe=\frac{1.79\times 10^3kg}{55.8g/mole}=\frac{1.79\times 10^3\times 1000g}{55.8g/mole}=3.15\times 10^4mole

Now we have to calculate the moles of Fe_2O_3

From the balanced chemical reaction we conclude that,

As, 2 moles of Fe produced from 1 mole of Fe_2O_3

So, 3.15\times 10^4mole of Fe produced from \frac{3.15\times 10^4}{2}=15750 mole of Fe_2O_3

Now we have to calculate the mass of Fe_2O_3

\text{ Mass of }Fe_2O_3=\text{ Moles of }Fe_2O_3\times \text{ Molar mass of }Fe_2O_3

Molar mass of Fe_2O_3 = 159.69 g/mole

\text{ Mass of }Fe_2O_3=(15750moles)\times (159.69g/mole)=2.515\times 10^6g=2.515\times 10^3kg

Now we have to calculate the percent purity of Fe_2O_3 in the original sample.

Mass of original sample = 2.86\times 10^3kg

\text{Percent purity}=\frac{\text{Mass of }Fe_2O_3}{\text{Mass of sample}}\times 100

\text{Percent purity}=\frac{2.515\times 10^3kg}{2.86\times 10^3kg}\times 100=87.94\%

Therefore, the percent purity of Fe_2O_3 in the original sample is 87.94 %

3 0
2 years ago
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