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3 years ago

How do I set this up?

Chemistry

1 answer:

Allisa [31]3 years ago

4 0

Answer: -

280mg

Explanation: -

The chemical formula is C14H18N2O5

Molar mass of the compound = 12 x 14 + 1 x 18 + 14 x 2 + 16 x 5 = 294 gram

Number of nitrogen atoms present in the formula = 2

Mass of nitrogen present= 14 x 2 = 28 gram

Thus 294 gram of the compound has 28 gram.

Hence 2.95grams of the compound has (28 gram x 2.95 gram / 294 gram)

= 0.28 gram

= 280 mg

Thus 280 mg of nitrogen is present in 2.95 grams of C14H18N2O5

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Consider these chemical equations. N2(g) + 3H2(g) → 2NH3(g) C(s) + 2H2(g) → CH4(g) 4H2(g) + 2C(s) + N2(g) → 2HCN(g) + 3H2(g) Whi

joja [24]

Multiply the second equation by 2

3 0

3 years ago

Read 2 more answers

8. When a 2.5 mol of sugar (C12H22O11) are added to a certain amount of water the boiling point is raised by 1 Celsius degree. I

Julli [10]

Hey there!:

8) ΔTb = i*Kb*m

m is molality

Since same number of mol is added to same amount of water in both cases

m will be same for both

is 1 for glucose since it is covalent compound

is 4 of Al(NO3)3 as it breaks into 1 Al₃⁺ and 3 NO₃⁻

So, ΔTb will be 4 times in aluminum nitrate case

So, boiling point will change by 4ºC

9) use Q = m* L

L = heat of vaporization so:

T1=T2=100ºC

5.40 * 1000 => 5400 cal/g

Q = 5400 / 540

Q = 10 grams

Hope that thlps!

5 0

3 years ago

Start with 100.00 mL of 0.10 M acetic acid, CH3COOH. The solution has a pH of 2.87 at 25 oC. a) Calculate the Ka of acetic acid

jasenka [17]

Answer: a) The $K_a$ of acetic acid at $25^0C$ is $1.82\times 10^{-5}$

b) The percent dissociation for the solution is $4.27\times 10^{-3}$

Explanation:

$CH_3COOH\rightarrow CH_3COO^-H^+$

cM 0 0

$c-c\alpha$ $c\alpha$ $c\alpha$

So dissociation constant will be:

$K_a=\frac{(c\alpha)^{2}}{c-c\alpha}$

Give c= 0.10 M and $\alpha$ = ?

Also $pH=-log[H^+]$

$2.87=-log[H^+]$

$[H^+]=1.35\times 10^{-3}M$

$[CH_3COO^-]=1.35\times 10^{-3}M$

$[CH_3COOH]=(0.10M-1.35\times 10^{-3}=0.09806M$

Putting in the values we get:

$K_a=\frac{(1.35\times 10^{-3})^2}{(0.09806)}$

$K_a=1.82\times 10^{-5}$

b) $\alpha=\sqrt\frac{K_a}{c}$

$\alpha=\sqrt\frac{1.82\times 10^{-5}}{0.10}$

$\alpha=4.27\times 10^{-5}$

$\% \alpha=4.27\times 10^{-5}\times 100=4.27\times 10^{-3}$

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3 years ago

How do the major and minor scale divisions relate the precision of measurements taken with a metric ruler or graduated cylinder?

Nata [24]

Answer: Major scale and minor scale both relate the precision of measurements with more true value of measurement.

Explanation:

Major scale in a measuring device is maximum unit value on the scale that can be measured.

Minor scale in a measuring device is a least unit value on the scale that can be measured.

In a metric ruler major scale is 1 cm and minor scale is 0.1 mm which means one can measure accurate value up-to one decimal point and in 10 ml of graduated cylinder major scale division is 1 ml minor scale division is 0.01.

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3 years ago

A gas sample contained in a cylinder equipped with a moveable piston occupied 300.0 mL at a pressure of 2.00 atm. What would be

yKpoI14uk [10]

Answer:

Explanation

hope this helps!

6 0

3 years ago