Which one of these is a mixture ?

Type your response in the box.

Wittaler [7]

Answer:

Here is a set of possible answers:

By how much can CO2 emissions be reduced?

How much is it going to cost the city to install the wind farms?

Will families have to pay for part of the wind farms?

What is the out-of-pocket cost of installing solar panels on a house after the assistance has been granted?

Are there any yearly operating costs to use solar panels?

How reliable is solar energy in this area? In an average year, how many months can solar energy be used?

Do we have a backup source of energy if the wind farm or solar panels don’t generate enough electricity?

Explanation:

This is the actucal answer so you might want to change it up

8 0

2 years ago

What did J. J. Thomson observe when he applied electric voltage to a cathode ray tube in his famous experiment?

irinina [24]

QUICK ANSWER

J.J. Thomson's cathode ray experiment was a set of three experiments that assisted in discovering electrons. He did this using a cathode ray tube or CRT. It is a vacuum sealed tube with a cathode and anode on one side.

7 0

2 years ago

Read 2 more answers

2C2H6(g) + 7O2(g) → 4CO2(g) + 6H2O(g)

Ket [755]

Answer:- 10 L of ethane.

Solution:- The given balanced equation is:

$2C_2H_6(g)+7O_2(g)\rightarrow 4CO_2(g)+6H_2O(g)$

From this equation, ethane and oxygen react in 2:7 mol ratio, the ratio of volumes would also be same if they are at same temperature and pressure.

Since 14 L of each gas are taken, the oxygen will be the limiting reactant and ethane will be the excess reactant. Let's calculate the volume of ethane used:

$14LO_2(\frac{2LC_2H_6}{7LO_2})$

= $4LC_2H_6$

From above calculations, 4 L of ethane are used. So, excess volume of ethane left after the completion of reaction = 14 L - 4 L = 10 L

Hence, 10 L of ethane will be remaining.

5 0

2 years ago

Calculate the unit cell edge length for an 85 wt% fe-15 wt% v alloy. All of the vanadium is in solid solution, and, at room temp

Lady bird [3.3K]

Answer is 0.289nm.

Explanation: The wt % of Fe and wt % of V is given for a Fe-V alloy.

wt % of Fe in Fe-V alloy = 85%

wt % of V in Fe-V alloy = 15%

We need to calculate edge length of the unit cell having bcc structure.

Using density formula,

$\rho_{ave}=\frac{Z\times M_{ave}}{a^3\times N_A}$

For calculating edge length,

$a=(\frac{Z\times M_{ave}}{\rho_{ave}\times N_A})^{1/3}$

For calculating $M_{ave}$ , we use the formula

$M_{ave}= \frac{100}{\frac{(wt\%)_{Fe}}{M_{Fe}}+\frac{(wt\%)_{V}}{M_V}}$

Similarly for calculating $(\rho)_{ave}$ , we use the formula

$\rho_{ave}= \frac{100}{\frac{(wt\%)_{Fe}}{\rho_{Fe}}+\frac{(wt\%)_{V}}{\rho_V}}$

From the periodic table, masses of the two elements can be written

$M_{Fe}= 55.85g/mol$

$M_{V}=50.941g/mol$

Specific density of both the elements are

$(\rho)_{Fe}=7.874g/cm^3\\(\rho)_{V}=6.10g/cm^3$

Putting $M_{ave}$ and $\rho_{ave}$ formula's in edge length formula, we get

$a=\left [\frac{Z\left (\frac{100}{\frac{(wt\%)_{Fe}}{M_{Fe}}+\frac{(wt\%)_{Fe}}{M_{Fe}}} \right )}{N_A\left (\frac{100}{\frac{(wt\%)_V}{\rho_V}+\frac{(wt\%)_V}{\rho_V}} \right )} \right ]^{1/3}$

$a=\left [\frac{2atoms/\text{unit cell}\left (\frac{100}{\frac{85\%}{55.85g/mol}+\frac{15\%}{50.941g/mol}} \right )}{(6.023\times10^{23}atoms/mol)\left (\frac{100}{\frac{85\%}{7.874g/cm^3}+\frac{15\%}{6.10g/cm^3}} \right )} \right ]^{1/3}$

By calculating, we get

$a=2.89\times10^{-8}cm=0.289nm$

7 0

2 years ago

Wich scientist is credited with developing the first scientific atomic theory

Fantom [35]

Answer:

john dalton

Explanation:

8 0

2 years ago