What are the elements and number of atoms in the compound c6h12o6?

Conservation of mass was discussed in the background. Describe how conservation of mass (actual, not theoretical) could be check

Sedbober [7]

Answer:

Conservation of mass can be checked in an experiment . There are three steps to do it in a best way:

1. Weigh all the equipment and materials required in the experiment before the experiment.

2. Avoid spillage and evaporation during the experiment.

3. Weigh all the equipment and materials after the experiment.

If the mass is conserved then weight from step 1 is equal to weight from step

3 0

3 years ago

Which of the following combinations of quantum numbers (n, l, ml, ms) do not represent permissible solutions of the schrodinger

Softa [21]

Option c. 6, -5, -1, 1/2

Because the second quantum number, l, cannot be negative.

The values of l are restricted to the range 0 to n -1, which in this case is 0, 1, 2, 3, 4 and 5.

8 0

3 years ago

Write the nuclear equation for the beta decay Ni-60

viktelen [127]

$^6^0_2_7CO\;\rightarrow\;^6^0_2_8Ni\;+\;^0_-1e$ is the nuclear equation for the beta decay Ni-60.

<h3>What is beta decay?</h3>

Beta-decay occurs when, in a nucleus with too many protons or too many neutrons, one of the protons or neutrons is transformed into the other.

The nuclear equation for the beta decay Ni-60

$^6^0_2_7CO\;\rightarrow\;^6^0_2_8Ni\;+\;^0_-1e$

In beta minus (β−) decay, a neutron is converted to a proton, and the process creates an electron and an electron antineutrino; while in beta plus (β+) decay, a proton is converted to a neutron and the process creates a positron and an electron neutrino. β+ decay is also known as positron emission.

Hence, $^6^0_2_7CO\;\rightarrow\;^6^0_2_8Ni\;+\;^0_-1e$ is the nuclear equation for the beta decay Ni-60.

Learn more about the beta decay here:

brainly.com/question/25455333

#SPJ1

8 0

1 year ago

Air at 30.0% relative humidity is cooled isobarically at 1 atm absolute from 75.0°C to 35.0°C. a. Estimate the dew point and the

Dvinal [7]

Answer:

a. Dew point: 48.7°C. Degrees of superheat 26.3°C

b. $3.33mol/m^3$

c. $P=3.34atm$

Explanation:

a. Based on the psychometric chart of air, the specific volume of air at the given conditions is:

$v =\frac{RT}{PM} =\frac{0.082\frac{atm*L}{mol*K}*348.15K}{1atm*28.97g/mol} *\frac{1m^3}{1000L}*\frac{1000g}{1kg} =0.98544m^3/kg$

The dew point at the specific volume and the 30%-humidity has a value of 48.7°C, it means that there are 75°C-48.7°C=26.3°C of superheat.

b. At 75°C the molar fraction of water is 11580Pa/101625Pa=0.114 moles per cubic meter of feed gas are:

$\frac{n}{V}=y_{H_2O}\frac{P}{RT}=0.114\frac{1atm}{0.082\frac{atm*L}{mol*K}*348.15K} }*\frac{1000L}{1m^3} \\\frac{n}{V}=4mol/m^3$

Once the 35°C are reached, the mole fraction of water is 1688Pa/101325Pa=0.017 and remaining moles per cubic meter of feed gas are:

$\frac{n}{V}=y_{H_2O}\frac{P}{RT}=0.017\frac{1atm}{0.082\frac{atm*L}{mol*K}*308.15K} }*\frac{1000L}{1m^3} \\\frac{n}{V}=0.673mol/m^3$

So the condensed moles per cubic meter of feed gas are:

$4mol/m^3-0.673mol/m^3=3.33mol/m^3$

c. Considering the Raoult's law, one computes the pressure as follows:

$P=\frac{P^{sat}_{H_2O}}{y_{H_2O}}$

At 75°C and 30%-humidity, the saturation water vapor pressure has a value of 38599Pa, thus:

$P=\frac{38599Pa}{0.114}*\frac{1atm}{101325Pa} \\P=3.34atm$

Best regards.

6 0

3 years ago

What is the mass percent of sucrose (C12H22O11, Mm = 342 g/mol) in a 0.329-m sucrose solution?

Hitman42 [59]

Answer:

$\% m/m=10.1\%$

Explanation:

Hello,

In this case given the molal solution of sucrose, we can assume there are 0.329 moles of sucrose in 1 kg of solvent, thus, computing both the mass of sucrose and solvent in grams, we obtain:

$m_{sucrose}=0.329mol*\frac{342g}{1mol}=112.5g$

$m_{solvent}=1000g$

In such a way, we proceed to the calculation of the mass percent as follows:

$\% m/m=\frac{112.5g}{112.5g+1000g}*100\%\\ \\\% m/m=10.1\%$

Regards.

8 0

3 years ago