diamond has a density of 3.26 g/cm3. what is the mass of a diamond that has a volume of 0.313 cm cubed

Convert the following to Celsius

Naya [18.7K]

9. (80-32)5/9

48×5/9

240/9

26.66°C

9. (90-32)5/9

58×5/9

32.22°C

10 (212-32)5/9

180×5/9

20×5

100°C

8 0

3 years ago

A scientist plants two rows of corn for experimentation. She puts fertilizer on row 1 but does not put fertilizer on row 2. Both

IrinaVladis [17]

Answer:

Q9. The independent variable in this experiment is the fertilizer. It is independent because she manipulating the variable to compare the growth.

Q10. The dependent variable in this experiment is the amount of growth of the corn. It is this because the growth depends on what the scientist did on the corn.

Q11. The variable controlled in this experiment is the amount of sun and water. These two variables never change so this is why it is the control.

Explanation:

5 0

2 years ago

Read 2 more answers

8. The time period of artificial satellite in a circular orbit of radius R is T. The radius of the orbit in which time period is

Elena-2011 [213]

Explanation:

It is given that,

The time period of artificial satellite in a circular orbit of radius R is T. The relation between the time period and the radius is given by :

$T^2\propto R^3$

The radius of the orbit in which time period is 8T is R'. So, the relation is given by :

$(\dfrac{T}{T'})^2=(\dfrac{R}{R'})^3$

$(\dfrac{T}{8T})^2=(\dfrac{R}{R'})^3$

$\dfrac{1}{64}=(\dfrac{R}{R'})^3$

$R'=4\times R$

So, the radius of the orbit in which time period is 8T is 4R. Hence, this is the required solution.

4 0

3 years ago

How much heat energy is needed to heat 300g of water from 10 degrees Celsius to 50 degrees Celsius

elixir [45]

Answer:

There is 50.2 kJ heat need to heat 300 gram of water from 10° to 50°C

Explanation:

Step 1: Data given

mass of water = 300 grams

initial temperature = 10°C

final temperature = 50°C

Temperature rise = 50 °C - 10 °C = 40 °C

Specific heat capacity of water = 4.184 J/g °C

Step 2: Calculate the heat

Q = m*c*ΔT

Q = 300 grams * 4.184 J/g °C * (50°C - 10 °C)

Q = 50208 Joule = 50.2 kJ

There is 50.2 kJ heat need to heat 300 gram of water from 10° to 50°C

8 0

2 years ago

. Determine the standard free energy change, ɔ(G p for the formation of S2−(aq) given that the ɔ(G p for Ag+(aq) and Ag2S(s) are

olga nikolaevna [1]

Answer: The standard free energy change of formation of $S^{2-}(aq.)$ is 92.094 kJ/mol

Explanation:

We are given:

$K_{sp}\text{ of }Ag_2S=8\times 10^{-51}$

Relation between standard Gibbs free energy and equilibrium constant follows:

$\Delta G^o=-RT\ln K$

where,

$\Delta G^o$ = standard Gibbs free energy = ?

R = Gas constant = $8.314J/K mol$

T = temperature = $25^oC=[273+25]K=298K$

K = equilibrium constant or solubility product = $8\times 10^{-51}$

Putting values in above equation, we get:

$\Delta G^o=-(8.314J/K.mol)\times 298K\times \ln (8\times 10^{-51})\\\\\Delta G^o=285793.9J/mol=285.794kJ$

For the given chemical equation:

$Ag_2S(s)\rightleftharpoons 2Ag^+(aq.)+S^{2-}(aq.)$

The equation used to calculate Gibbs free change is of a reaction is:

$\Delta G^o_{rxn}=\sum [n\times \Delta G^o_f_{(product)}]-\sum [n\times \Delta G^o_f_{(reactant)}]$

The equation for the Gibbs free energy change of the above reaction is:

$\Delta G^o_{rxn}=[(2\times \Delta G^o_f_{(Ag^+(aq.))})+(1\times \Delta G^o_f_{(S^{2-}(aq.))})]-[(1\times \Delta G^o_f_{(Ag_2S(s))})]$

We are given:

$\Delta G^o_f_{(Ag_2S(s))}=-39.5kJ/mol\\\Delta G^o_f_{(Ag^+(aq.))}=77.1kJ/mol\\\Delta G^o=285.794kJ$

Putting values in above equation, we get:

$285.794=[(2\times 77.1)+(1\times \Delta G^o_f_{(S^{2-}(aq.))})]-[(1\times (-39.5))]\\\\\Delta G^o_f_{(S^{2-}(aq.))=92.094J/mol$

Hence, the standard free energy change of formation of $S^{2-}(aq.)$ is 92.094 kJ/mol

8 0

3 years ago