What is temperature ?

If CO2 is not removed by boiling the solution after the addition of HCl, how will this affect the amount of NaOH required to rea

zimovet [89]

Answer:

See explanation

Explanation:

CO2 reacts with water to form carbonic acid as follows;

CO2(g) + H2O(l) ---------> H2CO3(aq)

This makes the solution more acidic hence it will require more NaOH to reach end point due to the increase in acidity.

Hence, if CO2 is not removed by boiling the solution, a greater amount of NaOH is required to reach end point.

6 0

3 years ago

What makes the atomic radius change down a column of the periodic table?

LUCKY_DIMON [66]

“When we move down a group then there is increase in the size of elements because there occurs increase in atomic mass of the elements.
It means that there occurs increase in number of electrons due to which there will be increase in number of new shells. As a result, size of atom or element increases due to which there is increase in atomic radius of the element.
Therefore, we can conclude that increase in number of electrons makes the atomic radius change down a column of the periodic table.”

4 0

3 years ago

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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

High-temperature superconducting oxides hold great promise in the utility, transportation, and computer industries.

-Dominant- [34]

The main class of high-temperature superconductors are in the class of copper oxides (only some particular copper oxides) especially the Rare-earth barium copper oxides (REBCOs) such as Yttrium barium copper oxide (YBCO).

<h3>What superconducting material works with the highest temperature?</h3>

As of 2020, the material with the highest accepted superconducting temperature is an extremely pressurized carbonaceous sulfur hydride with a critical transition temperature of +15°C at 267 GPa.

<h3>How do high-temperature superconductors work?</h3>

High-temperature superconductivity, the ability of certain materials to conduct electricity with zero electrical resistance at temperatures above the boiling point of liquid nitrogen, was unexpectedly discovered in copper oxide (cuprate) materials in 1987.

Learn more about high temperature superconductors here:

<h3>brainly.com/question/1657823</h3><h3 /><h3>#SPJ4</h3>

3 0

2 years ago

Consider the reaction 2CuCl2 + 4KI → 2CuI + 4KCl + I2. If 4 moles of CuCl2 react with 4 moles of KI, what is the limiting reacta

Tatiana [17]

Explanation:

here's the answer to your question

4 0

3 years ago