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2 years ago

1. Use the following thermochemical equation.

Chemistry

1 answer:

alekssr [168]2 years ago

7 0

Answer:

That is alot to answer so i am not sure could answer this it is too hard maybe you could break it down for me cause this alot

Explanation:

I wish i knew
Its too much
I don't know this kind of chemistry
if you have another chemistry question i could answer plz let me know
cause this too hard

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If a 1.00 mL sample of the reaction mixture for the equilibrium constant experiment required 32.40 mL of 0.258 M NaOH to titrate

andrey2020 [161]

Answer:

The concentration of acetic acid is 8.36 M

Explanation:

Step 1: Data given

Volume of acetic acid = 1.00 mL = 0.001 L

Volume of NaOH = 32.40 mL = 0.03240 L

Molarity of NaOH = 0.258 M

Step 2: The balanced equation

CH3COOH + NaOH → CH3COONa + H2O

Step 3: Calculate the concentration of the acetic acid

b*Ca*Va = a*Cb*Vb

⇒with b = the coefficient of NaOH = 1

⇒with Ca = the concentration of CH3COOH = TO BE DETERMINED

⇒with Va = the volume of CH3COOH = 1.00 mL = 0.001L

⇒with a = the coefficient of CH3COOH = 1

⇒with Cb = the concentration of NaOH = 0.258 M

⇒with Vb = the volume of NaOH = 32.40 mL = 0.03240 L

Ca * 0.001 L = 0.258 * 0.03240

Ca = 8.36 M

The concentration of acetic acid is 8.36 M

6 0

2 years ago

In Part A, you found the amount of product (1.80 mol P2O5 ) formed from the given amount of phosphorus and excess oxygen. In Par

Finger [1]

First step is to balance the reaction equation. Hence we get P4 + 5 O2 => 2 P2O5

Second, we calculate the amounts we start with

P4: 112 g = 112 g/ 124 g/mol – 0.903 mol

O2: 112 g = 112 g / 32 g/mol = 3.5 mol

Lastly, we calculate the amount of P2O5 produced.

2.5 mol of O2 will react with 0.7 mol of P2O5 to produce 1.4 mol of P2O5.

This is 1.4 * (31*2 + 16*5) = 198.8 g

3 0

2 years ago

A small hole in the wing of a space shuttle requires a 20.7-cm2 patch, (a) What is the patch's area in square kilometers (km2)?

bixtya [17]

Answer:

(a) The area of space shuttle is $2.07\times 10^{-9} km^2$ .

(b) $10.427 is the cost of the patch.

Explanation:

Area of the patch of the space shuttle = $20.7 cm^2$

a) 1 cm = 0.00001 km

$1 cm^2= (0.00001 km)^2=10^{-10}km^2$

$20.7 cm^2=20.7\times 10^{-10}km^2=2.07\times 10^{-9} km^2$

b) 1 cm = 0.393701 inch

$1 cm ^2=0.1550 inch^2$

$20.7 cm^2=20.7\times 0.1550 inch^2=3.2085 inch^2$

Cost of patching area = $\$3.25/inch^2$

Cost of patching $3.2085 inch^2$ are:

$\$3.25/inch^2\times 3.2085 inch^2=\$10.427$

8 0

2 years ago

Synthesis of disubstituted benzenes involves two steps, each introducing one of the functional groups. For the synthesis of p-ch

Setler79 [48]

Answer:

The synthesis of p-chlorobenzenesulfonic acid with the reagents for each step and the structure of the monosubstituted intermediate compound is shown in the drawing below.

Explanation:

Benzenesulfonic acid is obtained from the sulfonation of benzene, using concentrated smoking sulfuric acid.

Chlorination of benzenesulfonic acid with chlorine without catalyst requires a polar solvent, such as acetic acid to obtain the selectivity in the position for and the ortho derivative in a low proportion.

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2 years ago

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2 years ago

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