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Firdavs [7]
3 years ago
6

What was the time frame?

Chemistry
1 answer:
faltersainse [42]3 years ago
7 0

Answer:

If you are referring to the Precambrian time, the answer is 500 Million Years Ago.

Explanation:

Precambrian time is the time period that started with the formation of Earth and ended about 500 million years ago.

You might be interested in
3. give reasons
soldi70 [24.7K]

Answer:

A.A gas exerts pressure on the walls of the container. Reason: They have high kinetic energy and negligible force of attraction to move constantly moving with high speed in all directions. ... These collisions of the gas particles with the container walls exert pressure on the walls of the container.

B A gas exerts pressure on the walls of the container due to imapact of its gas molecules with hight velocity .

C.A wooden table should be called a solid. Reason: It has a definite shape and volume. It is very rigid and cannot be compressed, i.e., it has a solid characteristic.

D.We can easily move our hand in air but to do the same through a solid block of wood we need a karate expert. It is because the molecules of air has less force of attraction between them and a very small external force can separate them and pass through it.

Explanation:

Plz  mark me as a brainlest

3 0
2 years ago
Net ionic equation of hydrated oxalic acid and sodium hydroxide
blondinia [14]
<span>H2C2O4(aq) + 2OH- --> C2O4^2- + 2H2O(l)</span>
6 0
3 years ago
C-12 and C-13 are naturally-occurring isotopes of the element carbon. C-12 occurs 98.88% of the time and C-13 occurs 1.108% of t
Veseljchak [2.6K]

Answer:

c) (12×0.9889) + (13×0.01108)

Explanation:

Given data:

Percentage of C-12 = 98.89%

Percentage of C-13 = 1.108%

Atomic mass = ?

Solution:

98.89/100 = 0.9889

1.108/ 100 = 0.01108

Atomic mass = (12×0.9889) + (13×0.01108)

Atomic mass = (11.8668 + 0.144034)

Atomic mass = 12.01084

6 0
3 years ago
i am begging anyone to help me with this! (all tutors i've asked said they can't solve it but i need someone to help me out) - i
9966 [12]

First, we need to calculate how much energy we will get from this combustion.

Assuming the combustion is complete, we have the octane reacting with O₂ to form only water and CO₂, so:

C_8H_{18}+O_2\to CO_2+H_2O

We need to balance the reaction. Carbon only appear on two parts, so, we can start by it:

C_8H_{18}+O_2\to8CO_2+H_2O

Now, we balance the hydrogen:

C_8H_{18}+O_2\to8CO_2+9H_2O

And in the end, the oxygen:

C_8H_{18}+\frac{25}{2}O_2\to8CO_2+9H_2O

We can multiply all coefficients by 2 to get integer ones:

2C_8H_{18}+25O_2\to16CO_2+18H_2O

Now, we need to use the enthalpies of formation to get the enthalpy of reaction of this reaction.

The enthalpy of reaction can be calculated by adding the enthalpies of formation of the products multiplied by their stoichiometric coefficients and substracting the sum of enthalpies of formation of the reactants multiplied by their stoichiometric coefficients.

For the reactants, we have (the enthalpy of formation of pure compounds is zero, which is the case for O₂):

\begin{gathered} \Delta H\mleft\lbrace reactants\mright\rbrace=2\cdot\Delta H\mleft\lbrace C_8H_{18}\mright\rbrace+25\cdot\Delta H\mleft\lbrace O_2\mright\rbrace \\ \Delta H\lbrace reactants\rbrace=2\cdot(-250.1kJ)+25\cdot0kJ \\ \Delta H\lbrace reactants\rbrace=-500.2kJ+0kJ \\ \Delta H\lbrace reactants\rbrace=-500.2kJ \end{gathered}

For the products, we have:

\begin{gathered} \Delta H_{}\mleft\lbrace product\mright\rbrace=16\cdot\Delta H\lbrace CO_2\rbrace+18\cdot\Delta H\lbrace H_2O\rbrace \\ \Delta H_{}\lbrace product\rbrace=16\cdot(-393.5kJ)+18\cdot(-285.5kJ) \\ \Delta H_{}\lbrace product\rbrace=-6296kJ-5139kJ \\ \Delta H_{}\lbrace product\rbrace=-11435kJ \end{gathered}

Now, we substract the rectants from the produtcs:

\begin{gathered} \Delta H_r=\Delta H_{}\lbrace product\rbrace-\Delta H\lbrace reactants\rbrace \\ \Delta H_r=-11435kJ-(-500.2kJ) \\ \Delta H_r=-10934.8kJ \end{gathered}

Now, this enthalpy of reaction is for 2 moles of C₈H₁₈, so for 1 mol of C₈H₁₈ we have half this value:

\Delta H_c=\frac{1}{2}\Delta H_r=\frac{1}{2}\cdot(-10934.8kJ)=-5467.4kJ

Now, we have 100 g of C₈H₁₈, and its molar weight is approximately 114.22852 g/mol, so the number of moles in 100 g of C₈H₁₈ is:

\begin{gathered} M_{C_8H_{18}}=\frac{m_{C_8H_{18}}}{n_{C_8H_{18}}} \\ n_{C_8H_{18}}=\frac{m_{C_8H_{18}}}{M_{C_8H_{18}}}=\frac{100g}{114.22852g/mol}\approx0.875438mol \end{gathered}

Since we have approximately 0.875438 mol, and 1 mol releases -5467.4kJ when combusted, we have:

Q=-5467.4kJ/mol\cdot0.875438mol\approx-4786.37kJ

Now, for the other part, we need to calculate how much heat it is necessary to melt a mass, <em>m</em>.

First, we have to heat the ice to 0 °C, so:

\begin{gathered} Q_1=m\cdot2.010J/g.\degree C\cdot(0-(-10))\degree C \\ Q_1=m\cdot2.010J/g\cdot10 \\ Q_1=m\cdot20.10J/g \end{gathered}

Then, we need to melt all this mass, so we use the latent heat now:

Q_2=n\cdot6.03kJ/mol

Converting mass to number of moles of water we have:

\begin{gathered} M=\frac{m}{n} \\ n=\frac{m}{M}=\frac{m}{18.01528g/mol} \end{gathered}

So:

Q_2=\frac{m}{18.01528g/mol}_{}\cdot6.03kJ/mol\approx m\cdot0.334716kJ/g

Adding them, we have a total heat of:

\begin{gathered} Q_T=m\cdot20.10J/g+m\cdot0.334716kJ/g \\ Q_T=m\cdot0.02010kJ/g+m\cdot0.334716kJ/g \\ Q_T=m\cdot0.354816kJ/g \end{gathered}

Since we have a heat of 4786.37 kJ form the combustion, we input that to get the mass (the negative sign is removed because it only means that the heat is released from the reaction, but now it is absorbed by the ice):

\begin{gathered} 4786.37kJ=m\cdot0.354816kJ/g \\ m=\frac{4786.37kJ}{0.354816kJ/g}\approx13489g\approx13.5\operatorname{kg} \end{gathered}

Since we have a total of 20kg of ice, we can clculate the percent using it:

P=\frac{13.5\operatorname{kg}}{20\operatorname{kg}}=0.675=67.5\%

5 0
9 months ago
The mass of a deuterium nucleus 21H is less than its components masses. Calculate the mass defect.________ amu
Inessa [10]

Answer:

The mass defect of a deuterium nucleus is 0.001848 amu.

Explanation:

The deuterium is:

^{A}_{Z}X \rightarrow ^{2}_{1}H  

The mass defect can be calculated by using the following equation:

\Delta m = [Zm_{p} + (A - Z)m_{n}] - m_{a}

Where:

Z: is the number of protons = 1

A: is the mass number = 2      

m_{p}: is the proton's mass = 1.00728 amu  

m_{n}: is the neutron's mass = 1.00867 amu

m_{a}: is the mass of deuterium = 2.01410178 amu

Then, the mass defect is:

\Delta m = [1.00728 amu + (2- 1)1.00867 amu] - 2.01410178 amu = 0.001848 amu

Therefore, the mass defect of a deuterium nucleus is 0.001848 amu.

I hope it helps you!  

5 0
2 years ago
Read 2 more answers
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