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tatiyna
3 years ago
13

A small hole in the wing of a space shuttle requires a 16.2 cm2 patch. (b) If the patching material costs NASA $3.58/in2, what i

s the cost of the patch to the nearest cent?
Chemistry
1 answer:
Tju [1.3M]3 years ago
6 0

The cost of the patch : = 887.84 cents

<h3>Further explanation</h3>

Given

16.2 cm² patch

cost = $3.58/in²

Required

The cost of the patch

Solution

We must pay attention to the conversion factor when changing units

1 cm² = 0.155 in²

16 cm² = 2.48 in²

Cost :

= $3.58/in² x 2.48 in²

= $8.8784

1 dollar is equal to 100 cents, so :

= $8.8784 x 100 cents/$

= 887.84 cents

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A certain sample of coal contains 1.60 percent sulfur by mass. when the coal is burned, the sulfur is converted to sulfur dioxid
alina1380 [7]
<span>1.40 x 10^5 kilograms of calcium oxide The reaction looks like SO2 + CaO => CaSO3 First, determine the mass of sulfur in the coal 5.00 x 10^6 * 1.60 x 10^-2 = 8.00 x 10^4 Now lookup the atomic weights of Sulfur, Calcium, and Oxygen. Sulfur = 32.065 Calcium = 40.078 Oxygen = 15.999 Calculate the molar mass of CaO CaO = 40.078 + 15.999 = 56.077 Since 1 atom of sulfur makes 1 atom of sulfur dioxide, we don't need the molar mass of sulfur dioxide. We merely need the number of moles of sulfur we're burning. divide the mass of sulfur by the atomic weight. 8.00 x 10^4 / 32.065 = 2.49 x 10^3 moles Since 1 molecule of sulfur dioxide is reacted with 1 molecule of calcium oxide, just multiply the number of moles needed by the molar mass 2.49 x 10^3 * 56.077 = 1.40 x 10^5 So you need to use 1.40 x 10^5 kilograms of calcium oxide per day to treat the sulfur dioxide generated by burning 5.00 x 10^6 kilograms of coal with 1.60% sulfur.</span>
5 0
3 years ago
One of the compounds used to increase the octane rating of gasoline is toluene (pictured). Suppose 43.3 mL of toluene (d = 0.867
Thepotemich [5.8K]

<u>Answer:</u>

(A)

Density = Mass / Volume

So  

Mass = Density × Volume

= 0.867 g/mL \times 43.3mL = 37.5411 g Toluene

1C_6 H_5 CH_3  + 9 O_2  > 7 CO_2  + 4 H_2 O

Mole ratio of toluene : Oxygen is 1 : 9

$37.5411 g \text { Toluene } \times \frac{1 \text {mol} \text {toluene}}{92 g \text { toluene}} \times \frac{9 {mol} O_{2}}{1 \text {mol} \text { toluene }} \times \frac{32 g O_{2}}{1 {mol} O_{2}}=117 g O_{2}(\text {Answer})$

(B)

1 mole of Toluene produces 7 moles of CO_2 gas and 4 moles of H_2 O Vapour

So the mole ratio is 1 : 11

37.5411 g Toluene $\times \frac{1 \text { mol toluene }}{92 g \text { toluene }} \times \frac{11 \mathrm{mol} \text { gas }}{1 \text { mol toluene }} $$\\\\=4.49 \text { mol gaseous products (Answer) } $

(C)

1mole contains 6.022\times10^{23} molecules

37.5411 g Toluene $\times \frac{1 \text { mol toluene }}{92 g \text { toluene}} \times \frac{4 \mathrm{mol} \mathrm{H}_{2} \mathrm{O}}{1 \mathrm{mol} \text { toluene }} \times \frac{6.022 \times 10^{23} \text { molecules } \mathrm{H}_{2} \mathrm{O}}{1 \mathrm{mol} \mathrm{H}_{2} \mathrm{O}} $\\\\$=9.82 \times 10^{23} \text { molecules } \mathrm{H}_{2} \mathrm{O} \text { (Answer) } $

6 0
4 years ago
Comparison between alkyl nucleophilic substitution with acyl nucleophilic substitution
S_A_V [24]

Answer:

in both nucleophil attach the c and leaving group leave but in acyl nu. subsituation c of carbonyl because of double bond with o have bigger positive charge and is better electrophil so do it faster,also alkyl nu. subsituation can have rearangment if going from sn1 and in sn2 sterichemistry of molecule change , acyl nu. subsituation most of time is better

4 0
3 years ago
Is a diamond a homogeneous or heterogeneous mixture?
Ivanshal [37]

Answer:

It's a heterogeneous mixture.

Explanation:

Diamond is made of just one element: carbon. Each carbon atom in diamond is connected to four other carbon atoms, in a crystal that extends on and on. There are other forms of pure carbon where the atoms are bonded differently, notably charcoal and graphite.

7 0
4 years ago
Read 2 more answers
Calculate the molecular weight when a gas at 25.0 ∘C and 752 mmHg has a density of 1.053 g/L . Express your answer using three s
stiks02 [169]

Answer:

26.0 g/mol is the molar mass of the gas

Explanation:

We have to combine density data with the Ideal Gases Law equation to solve this:

P . V = n . R .T

Let's convert the pressure mmHg to atm by a rule of three:

760 mmHg ____ 1 atm

752 mmHg ____ (752 . 1)/760 =  0.989 atm

In density we know that 1 L, occupies 1.053 grams of gas, but we don't know the moles.

Moles = Mass / molar mass.

We can replace density data as this in the equation:

0.989 atm . 1L = (1.053 g / x ) . 0.082 L.atm/mol.K . 298K

(0.989 atm . 1L) / (0.082 L.atm/mol.K . 298K) = 1.053 g / x

0.0405 mol = 1.053 g / x

x =  1.053 g / 0.0405  mol = 26 g/mol

7 0
4 years ago
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