How many moles are in 50.0 grams of H3PO4?

I need help with 17,18,19,& 21

valentinak56 [21]

Answer:

17. tie hair up

wear goggles

18. i don't know

19. you light it on the side of the box

20. it should be orange because that is the safest.

21. you turn down the electricity thing

3 0

3 years ago

A gas exerts a force of 1000 N on a surface with an area of 5.0 m squared. What is the pressure on the area?

TEA [102]

The answer is 200 pascal

6 0

3 years ago

In boyle's law, the initial volume of a gas is 120 ml at 256 kpa. what will be the volume if the pressure increases to 385 kpa?

Vinil7 [7]

<h3>Answer:</h3>

79.8 ml

<h3>Explanation:</h3>

Boyle's law describes the relationship between pressure and volume when the temperature is constant. For this question, I will be rounding to significant figures.

Boyle's Formula

When describing the relationship between pressure and volume, the formula $\displaystyle P_{1} V_{1}= P_{2} V_{2}$ . In this formula, $P_{1}V_{1}$ is the initial pressure and volume. On the other side, $P_{2} V_{2}$ is final pressure and volume. So, to find a missing variable you must plug in the values you are given.

Final Volume

Remember when solving this question to remain constant in your units. When we plug in the values we know we are given:

$256kPa*120mL=385kPa*V_{2} mL$

Now, we can multiply the left side

$30720kPa*mL=385kPa*V_{2} mL$

Finally, we can divide by 385 to isolate the final volume

$79.8mL=V_{2}$

This gives the final volume of 79.8mL.

7 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

3 years ago

13. To refine aluminum from its ore, aluminum oxide is electrolyzed to form aluminum and oxygen. At which electrode does oxygen

victus00 [196]

Answer:

I think B

Explanation:

7 0

3 years ago