Uranium in the isotope below had 140 nutrons

How many valence electrons are in Tr2-

myrzilka [38]

Answer:

4 valence electrons

Explanation:

8 0

3 years ago

Read 2 more answers

Krypton is named after the Greek word that means “secret.” Which explains why krypton was most likely given this name?

irinina [24]

Answer:

<h2>Krypton is a halogen, so it should be kept secret because it is so reactive and toxic. Krypton is a noble gas, so it was difficult for chemists to find it, as though it was a secret.</h2>

Explanation:

<h2>PLEASE MARK ME BRAINLIEST AND FOLLOW M E LOTS OF LOVE FROM MY HEART AND SOUL DARLING TEJASWI HERE ❤️</h2>

7 0

3 years ago

Read 2 more answers

Which of the following is a scientific question?

Serjik [45]

Answer:

D. How frequently do earthquakes occur in San Francisco?

8 0

3 years ago

What can be said about an Endothermic reaction with a negative entropy change?The reaction isa. spontaneous at all temperatures.

Lyrx [107]

Answer:

1. d. The reaction is spontaneous in the reverse direction at all temperatures.

2. c. The reaction is spontaneous at low temperatures.

Explanation:

The spontaneity of a reaction is associated with the Gibbs free energy (ΔG). When ΔG < 0, the reaction is spontaneous. When ΔG > 0, the reaction is non-spontaneous. ΔG is related to the enthalpy (ΔH) and the entropy (ΔS) through the following expression:

ΔG = ΔH - T. ΔS [1]

where,

T is the absolute temperature (T is always positive)

1. What can be said about an Endothermic reaction with a negative entropy change?

If the reaction is endothermic, ΔH > 0. Let's consider ΔS < 0. According to eq. [1], ΔG is always positive. The reaction is not spontaneous in the forward direction at any temperature. This means that the reaction is spontaneous in the reverse direction at all temperatures.

2. What can be said about an Exothermic reaction with a negative entropy change?

If the reaction is exothermic, ΔH < 0. Let's consider ΔS < 0. According to eq. [1], ΔG will be negative when |ΔH| > |T.ΔS|, that is, at low temperatures.

3 0

3 years ago

Combustion analysis of 0.600 g of an unknown compound containing carbon, hydrogen, and oxygen produced 1.043 g of CO2 and 0.5670

Aliun [14]

Answer : The empirical formula of the compound is $C_3H_8O_2$

Explanation :

The chemical equation for the combustion of hydrocarbon having carbon, hydrogen and oxygen follows:

$C_xH_yO_z+O_2\rightarrow CO_2+H_2O$

where, 'x', 'y' and 'z' are the subscripts of Carbon, hydrogen and oxygen respectively.

We are given:

Mass of $CO_2=1.043g$

Mass of $H_2O=0.5670g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

Molar mass of water = 18 g/mol

For calculating the mass of carbon:

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 1.043 g of carbon dioxide, $\frac{12}{44}\times 1.043=0.284g$ of carbon will be contained.

For calculating the mass of hydrogen:

In 18 g of water, 2 g of hydrogen is contained.

So, in 0.5670 g of water, $\frac{2}{18}\times 0.5670=0.063g$ of hydrogen will be contained.

Mass of oxygen in the compound = (0.600) - (0.284 + 0.063) = 0.253 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{0.284g}{12g/mole}=0.0237moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{0.063g}{1g/mole}=0.063moles$

Moles of Oxygen = $\frac{\text{Given mass of oxygen}}{\text{Molar mass of oxygen}}=\frac{0.253g}{16g/mole}=0.0158moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 0.0158 moles.

For Carbon = $\frac{0.0237}{0.0158}=1.5$

For Hydrogen = $\frac{0.063}{0.0158}=3.98\approx 4$

For Oxygen = $\frac{0.0158}{0.0158}=1$

The ratio of C : H : O = 1.5 : 4 : 1

To make in a whole number we are multiplying the ratio by 2, we get:

The ratio of C : H : O = 3 : 8 : 2

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : O = 3 : 8 : 2

Hence, the empirical formula for the given compound is $C_3H_8O_2$

6 0

3 years ago

Uranium in the isotope below had 140 nutrons​

Uranium in the isotope below had 140 nutrons