What we're looking for here is the gas sample's molar mass given its mass, pressure, volume, and temperature. Recalling the gas law, we have

or

where R is <span>0.08206 L atm / mol K, P is the given pressure, T is the temperature, and V is the volume.
Before applying the values given, it is important to make sure that they are to be converted to have consistent units with that of R.
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Thus, we have
P = 736/ 729 = 0.968 atm
T = 28 + 273.15 = 301.15 K
V = 250/1000 = 0.250 L
Now, applying these converted values into the gas law, we have


Given that the mass of the sample is 0.430 g, we have

Thus, the gas sample has a molar mass of 43.9 g/mol.
Answer:
Answer is A
Explanation:
Now we know that an atom wants to complete its outer shell while keeping electrons in pairs of two now in A there are four electrons which which can be ejected while in B will want to accept 3 electrons to complete its shell as ejecting five will take lot of energy similar case will be for C,D and E which would want to accept 2,1,0 electrons respectively
Depends on how the sword is made, what materials are used and temperature used but yes they can shatter.
When molecules cool down they stop vibrating and moving as much and so they "shrink" and the metal of the sword becomes brittle. sometimes they shrink at different phases which cause tension in the sword if this tension is strong enough it can cause the metallic bonds to break causing the sword to shatter.
hope that helps
Answer:
Determining the Slope on a p-t Graph. It was learned earlier in Lesson 3 that the slope of the line on a position versus time graph is equal to the velocity of the object. ... If the object has a velocity of 0 m/s, then the slope of the line will be 0 m/s. The slope of the line on a position versus time graph tells it all.
Explanation:
#<em>c</em><em>a</em><em>r</em><em>r</em><em>y</em><em>o</em><em>n</em><em>l</em><em>e</em><em>a</em><em>r</em><em>n</em><em>i</em><em>n</em><em>g</em><em> </em>
One atom of carbon weighs exactly 12/6.022x10^23 = 1.9927x10^-23 grams<span>.</span>