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2 years ago

What is the length of the shaded grey area? write your answer to the correct number of significant digits. The unit is cm

Chemistry

1 answer:

alukav5142 [94]2 years ago

8 0

Answer:

1.37cm

Explanation:

It's less than 1.4cm but more than 1.3cm. It's also more than 1.35cm so I guess the best answer would be 1.37cm or round up to 1.4cm

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n the first step of the industrial process for making nitric acid, ammonia reacts with oxygen in the presence of a suitable cata

MariettaO [177]

Answer:

10.54 L

Explanation:

Given that:-

Moles of $O_2$ = 1.00 moles

According to the given reaction:-

$4NH_3+7O_2\rightarrow 4NO_2+6H_2O$

7 moles of oxygen gas reacts with 4 moles of ammonia

1 mole of oxygen gas reacts with $\frac{4}{7}$ moles of ammonia

Moles of ammonia = 0.5714 moles

Also, Given that:

Temperature = 850 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15

So,

T₁ = (850 + 273.15) K = 1123.15 K

n = 0.5714 moles

P = 5.00 atm

Using ideal gas equation as:

$PV=nRT$

where,

P is the pressure

V is the volume

n is the number of moles

T is the temperature

R is Gas constant having value = 0.0821 L atm/ K mol

Applying the equation as:

5.00 atm × V = 0.5714 moles ×0.0821 L atm/ K mol × 1123.15 K

⇒V = 10.54 L

<u>10.54 L of $NH_3$ required.</u>

5 0

3 years ago

Can someone help I have to much stuff to do tonight so any help is appreciated.

Anit [1.1K]

Sorry I'm not in chemistry but do you have any other work

6 0

3 years ago

An organic compound contains , , , and . Combustion of 0.1023 g of the compound in excess oxygen yielded 0.2587 g and 0.0861 g .

Kisachek [45]

The question displayed below shows the missing information which therefore completes the question.

An organic compound contains C, H, N and O. Combustion of 0.1023 g of the compound in excess oxygen yielded 0.2587 g of CO2 and 0.0861 g of H2O. A sample of 0.4831 g of the compound was analyzed for nitrogen by the Dumas method. The compound is first reacted by passage over hot: The product gas is then passed through a concentrated solution of to remove the. After passage through the solution, the gas contains and is saturated with water vapor. At STP, 38.9 mL of dry N2 was obtained. In a third experiment, the density of the compound as a gas was found to be 2.86 g/L at 127°C and 256 torr. What are the empirical and molecular formulas of the compound? (Enter the elements in the order: C, H, N, O.)

Answer:

the empirical formula = $\mathbf {C_3H_6O_{12}N}$

the molecular formula = $\mathbf {C_3H_6O_{12}N}$

Explanation:

From the given information:

$\bigg ( 0.2587 \ g \ of CO_2 \bigg) \times \dfrac{1 \ mol \ of CO_2}{44 \ of \ CO_2} \times \dfrac{1 \ mol \ of \ C}{1 \ mol \ of CO_2}$

$= 0.00588 \ mol \ of \ C \times \dfrac{12.01 \ g \ of \ C}{1 \ mol \ of \ C }$

= 0.0706g of C

$\bigg ( 0.0861\ g \ of H_2O \bigg) \times \dfrac{1 \ mol \ of H_2O}{18.02 \ g \ of \ H_2O} \times \dfrac{2 \ mol \ of \ H}{1 \ mol \ of H_2O}$

$=0.0096 \ mol \times \dfrac{1.008 \ g \ of \ H}{1 mol \ H}$

0.0097g of H

Given that N2 at STP = 1 atm, 273 K and V = 0.0389 L

PV = nRT

n = PV/RT

$n = \dfrac{1 \ atm \times 0.0389 \ of \ H_2}{0.0821 \ L.atm /mol.K \times 273 \ K }$

n = 0.00173 mol of N2

The oxygen in the sample = The total grams in sample - gram in H - gram in C

The oxygen in the sample = 0.1023 g - 0.0097 g - 0.706 g

The oxygen in the sample = 0.022 g of O

The number of moles of $O_2 = \dfrac{0.02}{16}$

= 0.001375 mol of O

$O \ in \ product = (0.00588 \ mol \ of \ C ) \times \dfrac{2 \ mol \ of \ O }{1 \ mol \ of \ C }+ \bigg ( 0.0096 \ mol \ of \ H ) \times \dfrac{1 \ mol \ of \ O }{1 \ mol \ of \ H}$

O in product = 0.02136 mol of O

∴

we are meant to divide the moles of each compound by the smallest number of moles; we have:

$C = \dfrac{0.00588}{0.00173} \simeq 3$

$H = 0.0096 = \dfrac{0.0096}{0.00173} \simeq 6$

$O = 0.0199= \dfrac{0.0199}{0.00173} \simeq 12$

$N = 0.00173= \dfrac{0.00173}{0.00173} \simeq 1$

Thus; the empirical formula = $\mathbf {C_3H_6O_{12}N}$

To estimate the molecular formula; we have:

$MM = \dfrac{dRT}{P}$

$MM = \dfrac{2.80 \ g/ L \times 0.0821 \ L.atm /mol.K \times 400 \ K }{0.337 \ atm}$

MM = 272.86 g/mol

Also; the molar mass of $\mathbf {C_3H_6O_{12}N}$ = 248 g/mol

∴

$= \dfrac{272.86 \ g/mol}{248 \ g/mol}$

$=1$

Thus; we can conclude that empirical formula as well as the molecular formula are the same.

6 0

3 years ago

The second ionization energy of lithium is much greater than the first ionization energy

Hitman42 [59]

Mg: [Ne] 3s2. The second ionization energy of Mg is larger than the first because it always takes more energy to remove an electron from a positively charged ion than from a neutral atom. The third ionization energy of magnesium is enormous, however, because the Mg2+ ion has a filled-shell electron configuration.

7 0

4 years ago

g When water and perchloric acid (HClO4) are mixed, heat is released. The resulting solution is not an ideal solution, in terms

Natalija [7]

Answer:

A) boiling point higher than expected vapor pressure lower than expected.

Explanation:

When water is introduced with heat it vaporize at a temperature of 100 degree centigrade. When hydrochloric acid is added to the water its boiling point of water increases slightly. The difference in the boiling point is due to the presence of heavier molecules in acid.

7 0

3 years ago