What kinds of materials can laboratories test?

Which is and element <br> A) Salad<br> B) Sugar<br> C) Water <br> D) Potassium

Semenov [28]

Answer:

the only element above is potassium

3 0

3 years ago

Consider 10.0 g of helium gas (He) in a rigid steel container. If you add 10.0 g of neon gas (Ne) to this container, which of th

Norma-Jean [14]

Answer: (e) The pressure in the container increases but does not double.

Explanation:

To solve this, we need to first remember our gas law, Boyle's law states that the pressure and volume of a gas have an inverse relationship. That is, If volume increases, then pressure decreases and vice versa, when temperature is held constant. Therefore, increasing the volume in this case does not double the pressure owning to out gas law, but an increase in pressure would be noticed if temperature is constant

6 0

2 years ago

You are working the grill at a restaurant and are having trouble getting the grill lit through electrical means. You light a mat

never [62]

Answer:

in this situation I would a little bold

Explanation:

first I don't know what extinguisher I would use pretty much any that helps with fires. I'll back people up, take the hood and put it on the small fire that way it will light out more and if I open the hood and there still a little fire I would use the extinguisher and no one gets hurt :)

6 0

2 years ago

At its critical point, ammonia has a density of 0.235 g cm23. You have a special thickwalled glass tube that has a 10.0mm outs

salantis [7]

Answer:

$\large \boxed{\text{69.3 mg}}$

Explanation:

1. Volume of sealed tube

Assume the sealed tube is a right circular cylinder in which the cap and the base are also 4.20 mm thick.

Its outside dimensions are 155 mm long × 10.0 mm diameter.

Its inside dimensions are

h = 155 mm - 2 × 4.20 mm = 146.6 mm

r = 5.0 mm - 4.20 mm = 0.8 mm

V = πr²h = π(0.8)²× 146.6 mm³ = 294.8 mm³ = 0.2948 cm³

2. Calculate the mass of NH₃

$\begin{array}{rcl}\text{Density} & = & \dfrac{\text{Mass}}{\text{Volume}}\\\\\rho & = &\dfrac{m}{V}\\\\0.235 \text{ g$\cdot$ cm}^{-3} & = & \dfrac{m}{\text{0.2948 cm}^{3}}\\\\m & = & \text{0.0693 g}\\& = & \textbf{69.3 mg}\\\end{array}\\\text{You must seal $\large \boxed{\textbf{69.3 mg}}$ of ammonia in the tube.}$

4 0

3 years ago

Information-

tekilochka [14]

Answer:

So first thing to do in these types of problems is write out your chemical reaction and balance it:

Mg + O2 --> MgO

Then you need to start thinking about moles of Magnesium for moles of Magnesium Oxide. Based on the above equation 1 mole of Magnesium is needed to make one mole of Magnesium Oxide.

To get moles of magnesium you need to take the grams you started with (.418) and convert to moles by dividing by molecular weight of Mg (24.305), this gives you .0172 moles of Mg.

The theoretical yield would be the assumption that 100% of the magnesium will be converted into Magnesium Oxide, so you would get, based on the first equation, .0172 mol of MgO. Multiplying this by the molecular weight of MgO (24.305+16) gives us .693 g of MgO.

The percent yield is what you actually got in the experiment, and for this you subtract off the total mass from the crucible mass, or 27.374 - 26.687, which gives .66 g of MgO obtained.

Percent yield is acutal/theoretical, .66/.693, or 95.24%.

I'll let you do the same for the second trial, and average percent yield is just an average of the two trials percent yield.

Hope this helps.

6 0

2 years ago