Question

Determine the empirical formula of kelvar use in making bullet proof vest is 70.6% c, 4.2%h, 11.8%n, ad 13,4%n

Answer 1

Here is your answer ⤵⤵⤵

Assuming a basis of 100 g of Kevlar, therefore the masses

are:

C = 70.6 g

H = 4.2 g

N = 11.8 g

O = 13.4 g

 

We know the molar masses are:

C = 12.01, H = 1.008, N = 14.01, O = 16.00

 

Calculating for the moles per element:

C = 70.6 / 12.01 = 5.88 mol

H = 4.2 / 1.008 = 4.17 mol

N = 11.8 / 14.01 = 0.84 mol

O = 13.4 / 16 = 0.84 mol

 

To get the empirical formula, the next step is to divide

everything by the smallest moles = 0.84. So:

C = 7

H = 5

N = 1

O = 1

 

Hence the empirical formula is:

C7H5NO

HOPE IT HELPS YOU ☺☺ !!!

Answer 2

Answer:

Hii

Step-by-step explanation:

Let the mass of compound be 100g

Let the mass of compound be 100gC=70.6g=

Let the mass of compound be 100gC=70.6g= 12

Let the mass of compound be 100gC=70.6g= 1270.6

Let the mass of compound be 100gC=70.6g= 1270.6

Let the mass of compound be 100gC=70.6g= 1270.6 moles =5.88 moles

Let the mass of compound be 100gC=70.6g= 1270.6 moles =5.88 moles H=4.2g=

Let the mass of compound be 100gC=70.6g= 1270.6 moles =5.88 moles H=4.2g= 1

Let the mass of compound be 100gC=70.6g= 1270.6 moles =5.88 moles H=4.2g= 14.2

Let the mass of compound be 100gC=70.6g= 1270.6 moles =5.88 moles H=4.2g= 14.2

Let the mass of compound be 100gC=70.6g= 1270.6 moles =5.88 moles H=4.2g= 14.2 moles =4.2 moles

Let the mass of compound be 100gC=70.6g= 1270.6 moles =5.88 moles H=4.2g= 14.2 moles =4.2 moles N=11.8g=

Let the mass of compound be 100gC=70.6g= 1270.6 moles =5.88 moles H=4.2g= 14.2 moles =4.2 moles N=11.8g= 14

Let the mass of compound be 100gC=70.6g= 1270.6 moles =5.88 moles H=4.2g= 14.2 moles =4.2 moles N=11.8g= 1411.8

Let the mass of compound be 100gC=70.6g= 1270.6 moles =5.88 moles H=4.2g= 14.2 moles =4.2 moles N=11.8g= 1411.8

Let the mass of compound be 100gC=70.6g= 1270.6 moles =5.88 moles H=4.2g= 14.2 moles =4.2 moles N=11.8g= 1411.8 moles =0.84 moles

Let the mass of compound be 100gC=70.6g= 1270.6 moles =5.88 moles H=4.2g= 14.2 moles =4.2 moles N=11.8g= 1411.8 moles =0.84 moles O=13.4g=

Let the mass of compound be 100gC=70.6g= 1270.6 moles =5.88 moles H=4.2g= 14.2 moles =4.2 moles N=11.8g= 1411.8 moles =0.84 moles O=13.4g= 16

Let the mass of compound be 100gC=70.6g= 1270.6 moles =5.88 moles H=4.2g= 14.2 moles =4.2 moles N=11.8g= 1411.8 moles =0.84 moles O=13.4g= 1613.4

Let the mass of compound be 100gC=70.6g= 1270.6 moles =5.88 moles H=4.2g= 14.2 moles =4.2 moles N=11.8g= 1411.8 moles =0.84 moles O=13.4g= 1613.4

Let the mass of compound be 100gC=70.6g= 1270.6 moles =5.88 moles H=4.2g= 14.2 moles =4.2 moles N=11.8g= 1411.8 moles =0.84 moles O=13.4g= 1613.4 moles =0.84 moles

Let the mass of compound be 100gC=70.6g= 1270.6 moles =5.88 moles H=4.2g= 14.2 moles =4.2 moles N=11.8g= 1411.8 moles =0.84 moles O=13.4g= 1613.4 moles =0.84 moles C:H:N:O=5.88:4.2:0.84:0.84

Let the mass of compound be 100gC=70.6g= 1270.6 moles =5.88 moles H=4.2g= 14.2 moles =4.2 moles N=11.8g= 1411.8 moles =0.84 moles O=13.4g= 1613.4 moles =0.84 moles C:H:N:O=5.88:4.2:0.84:0.84C:H:N:O=7:5:1:1

Let the mass of compound be 100gC=70.6g= 1270.6 moles =5.88 moles H=4.2g= 14.2 moles =4.2 moles N=11.8g= 1411.8 moles =0.84 moles O=13.4g= 1613.4 moles =0.84 moles C:H:N:O=5.88:4.2:0.84:0.84C:H:N:O=7:5:1:1Therefore, empirical formula of Kelvar is C 7H5NO

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