| Agency overview | |
|---|---|
| Dissolved | July 1, 2018 |
| Parent agency | Manitoba Sport, Culture and Heritage |
| Website | www |
The Manitoba Film Classification Board (MFCB) was a provincial government organization responsible for rating films and video games rented, sold, or shown in the province of Manitoba. In mid 2018, the Board was dissolved, with its duties being outsourced to British Columbia for film classifications, and transferred to the Entertainment Software Rating Board (ESRB) for video games.[1][2][3]
The MFCB consisted of a minimum of 16 community members, and was tasked with providing ratings information about film, videos, DVDs, computer and video games distributed in Manitoba.
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Balancing Chemical Equations
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Transcription
Like the concept of the mole, balancing equations is one of those ideas that you learn in first-year chemistry class. It tends to give a lot of students a hard time, even though it is a fairly straightforward concept. I think what makes it difficult is that there's a bit of an art to it. So before we talk about balancing chemical equations, what is a chemical equation? Well here's some examples right here, and I have some more in the rest of this video. But it essentially just describes a chemical reaction. You've got some aluminum. You have some oxygen gas or a diatomic oxygen molecule. And then you end up with aluminum oxide. And you would say, ok, fine, that's an equation. It looks nice. I have my reactants, or the things that react. These are the reactants. And then I have the products of this reaction. What's left there to do? Well you have a problem here. The way I've written it right now, I have one atom of aluminum plus two atoms of oxygen, right? They're bonded to each other, but there's two atoms of oxygen here. One molecule of diatomic oxygen, or one molecule of oxygen, but they have two oxygen atoms here. And when you add them together, I have two atoms of aluminum and three atoms of oxygen. So I have a different number of aluminums on both sides of this equation. On this side, I have one aluminum. On this side, I have two aluminums. And then I have a different number of oxygens. On this side, I have two oxygens. And on that side, I have three oxygens. So balancing equations is all about fixing that problem, so that I have the same number of aluminum on both sides of the equation, and the same number of oxygens on both sides. So let's try to do that. I'll do it in orange. So I said I have one aluminum here and I have two aluminums there. So maybe a simple thing is just to put a two out here. So now I have two aluminums on this side and I have two aluminums on this side. The aluminums look happy. Now let's look at the oxygen. Here I have two oxygens on the left-hand side of the equation. And on the right-hand side of the equation I have three oxygens. What can I do here? Well if I could kind of have half atoms, I could just multiply this by one and a half. 1.5. 1.5 times 2 is 3. So now I have three oxygens on both sides of this equation and I have two aluminums on both sides of the equation. Am I done? Well, no, you can't have half an atom, or one and a half of an atom. That's not cool. So what you do is you just multiply this so that you end up with whole numbers. So let's just multiply both sides of this whole equation by two. And we have four aluminums plus three oxygens yields-- we multiplied everything by two-- two molecules of aluminum oxide. Now, you might have been tempted at some point in this exercise to say, oh, well why don't I just tweak part of the aluminum oxide? Why don't I put a 2/3 in front of this oxygen. You cannot do that. The equation is as it is. The molecule aluminum oxide is aluminum oxide. You can't change the relative ratios of the aluminum and the oxygen within the aluminum oxide molecule. You can just change the number of aluminum oxide molecules as a whole that you have, in this case, two. So what did we do? We looked at the aluminum. We said, OK, we need two aluminums to have both sides of that be two. And then when we looked at oxygen, we said, well, if I multiply this by one and a half, then that becomes three oxygens here, because one and a half times two oxygens. And three oxygens there. And then all we said is, oh, I can't have a one half there, so let me multiply both sides by two. And I ended up with four aluminums plus three oxygen molecules, or six oxygen atoms, yields two molecules of aluminum oxide. Let's see if we can do some more of these. So here I have methane. And that g in parentheses, I just wanted to expose you to that. That just means it's a gas. So I have methane gas plus oxygen gas yields carbon dioxide gas plus water. That's an l there, so liquid water. So what can we do here? So the general thing is, do the complicated molecules first, and then at the end you can worry about the single atom molecules. Because those are very easy to play with. And the reason why you do that, whenever you change a number here-- let's say we're trying to engineer how many carbons we have on both sides of this equation-- if I set a number here, I'm also changing the number of hydrogens. Then I'll have to play with the hydrogens there. And at the end, I'll have some number of oxygens. And then I have to tweak this number right there. So let's just start with the carbons. It seems complicated, but when you go step by step and you kind of play with things a little bit, it should proceed fairly smoothly. So here I have two carbons. And here I have, on the right-hand side, only one carbon. So ideally I'd want two carbons on both sides of this equation. So let me put a two out here. So there you go, my carbons are happy. I now have two carbons and two carbons. Now let me move to the hydrogens. Remember, I wanted to do the oxygens last, because I can just set this to whatever I want it to be without messing up any of the other atoms. So on this side of the equation I have four hydrogen atoms. How many hydrogen atoms do I have on this side of the equation? So I have four hydrogen atoms here. How many do I have on this side? Well, I have two hydrogen atoms right there. So I want to have four on both sides, so let me put a two in front of the water. So now I have four hydrogen atoms. Cool. Finally, oxygen. On the left-hand side I have two oxygen atoms. And on the right-hand side, what do I have? I have two times this one oxygen, right here, so right now that's two, right? Two waters. And in each water I have one oxygen atom, but I have two water molecules, so I have two oxygens. And then in the carbon dioxide I have two oxygens in each carbon dioxide. And I have two carbon dioxides, right? When I put that magenta two out front. So how many oxygens do I have? Two times two. I have four oxygens. So on the left-hand side I have two oxygens. On the right-hand side I have six oxygens, two in the two molecules of water and four in the two molecules of carbon dioxide. So how do I make this two into six? I want to have six oxygens on this side. I want to them to have six oxygens on the left-hand side as well. Well, if I put a three out here, now I have six oxygens. And our equation has been balanced. I have two carbons on this side, two carbons on that side, four hydrogens on this side, four hydrogens on this side, six oxygens on this side, and then-- four plus two-- six oxygens on that side. Next equation to balance. This actually becomes quite fun once you get the knack of it. So I have ethane plus oxygen gas yielding carbon dioxide and water. So this is a combustion process. Let's look at the carbons first. I have two carbons here. I have one carbon there. So let me put a two here. So now I have two carbons. Fair enough. Remember, I'm going to worry about the oxygen last. This one's actually not too different than the last problem. I have six hydrogens here. I only have two hydrogens in the water. So then we have three water molecules. And now I've balanced out the hydrogens. I have six hydrogens on both sides of the equation. And now let's deal with the water. Here on the right-hand side-- I'll do it in this orange color-- I have two oxygens in each carbon dioxide molecule. And I have two molecules, so I have four oxygens. And I have one oxygen in each water molecule. And I have three molecules, so I have three oxygens here. Is that right? Three oxygens on the right-hand side. Yep, three water molecules. And then I have four oxygens in the carbon dioxide. Right. So I have seven oxygens. I have seven oxygens on this side and I only have two on this side. So how can I make this two into seven. Well I could multiply it by three and a half. Right? Remember, I just want to have seven on both sides. If I have three and a half of these diatomic oxygen molecules-- three and a half times two is seven-- so now I have seven oxygens on both sides of the equation. Four plus three and seven here. Two carbons. And then six hydrogens. And I'm almost done, except for the fact that you can't have fractions of molecules. So what you do is just multiply both sides of this equation by two or three, or whatever you need to multiply it to get rid of the fractions. So if I multiply everything by two, I end up with two molecules of ethane plus seven molecules of diatomic oxygen, seven O2's, yielding two molecules of carbon dioxide-- oh, sorry, I'm multiplying everything by two, so four molecules of carbon dioxide, plus six molecules of water. And just to make sure that all still works, if you want to, you can check. How much carbon do we have? I have four carbons here. I have four carbons here. How much hydrogen do I have? I have two times six. I have 12 hydrogens here. I have 12 hydrogens here. How much oxygen do I have? Let me do a different color. I have 14 oxygens here. And here I have eight oxygens. And here I have-- six times one-- six oxygens. So six plus eight is 14. So my equation has been balanced. So that one, a lot of people might find that to be a hard problem, because you have three and a half, and just going straight to this might seem non-intuitive. But if you just work from the more complicated molecules and you go atom by atom, and if you end up with any fractions you just multiply both sides by some number to get rid of the fraction, of and then you're done. All right. Let's do another one. So this one looks all hairy. I have this iron oxide plus sulfuric acid yields all this hairy stuff. But the key here to realize is that this group right here, this sulfate group, stays together. Right? You have this SO4 there and you have this SO4 there. So to really simplify things for our head, you can kind of treat that like an atom. So let's make a substitution. We'll substitute that for x. So if we rewrite this as-- I'll do it in a vibrant color-- iron oxide plus H2-- there's only one sulfate group here-- x. And then that yields two irons, this molecule with two irons. And it has three of these sulfate groups. And then plus water. All I did is replace the sulfate with an x. And now we can treat that x like an atom, and we can just balance the equation. Let's see, on the left-hand side, how many irons do we have? We have two irons here. We have two irons there. So the irons look balanced, at first glance. Let's deal with the oxygens. So if we have three oxygens here-- let me do it in a different color-- and we only have one oxygen here. Remember, there were some oxygens in this x group, but the sulfate group stayed together, so we can just treat those separately. We want to have three oxygens on the right-hand side, as well, so let's stick a three here. Oops. So let me put that three there. So now we have three oxygens, as well. And then finally, let's look at the hydrogens. Let's see, how many hydrogens do we have here. We have six hydrogens now, on the right-hand side. Six hydrogens here. Three times two. And we want to have six hydrogens here, so we have to have three of these molecules. And then finally, let's look at the sulfate group, that x. We have three x's here. And lucky for us, we have three x's over there. So our equation has been balanced. And if we want to write it back in terms of the sulfate terms, we can just un-substitute the x, and we're left with one molecule of iron oxide plus three molecules of sulfuric acid, H2SO4 -- I just un-substituted the x with SO4 -- yields one of these molecules, SO4 3. I just un-substituted the x plus three molecules of water. Let's do one more. And then I think we'll be all balanced out. Carbon dioxide plus hydrogen gas yields methane plus water. So let's deal with the carbon first. I have one carbon here, one carbon there. The carbons look happy. Let's look at the oxygen. I have two oxygens here. I have one oxygen there. So I want two oxygens here so let me stick a two there. And then let's deal with the hydrogens last, because this is the easiest one to play with because it doesn't affect any of the other atoms. So if I have two here and I have four, plus four here, right? I have four hydrogens there. And I have four hydrogens there. So I need eight hydrogens on the left-hand side. So I just put a four there. We're all done balancing. Anyway, hopefully you found that useful.
History
Film censorship in Manitoba began in 1911 with An Act to Regulate Moving Picture Exhibitions. As Winnipeg was the only place in the province showing films, such censorship would the responsibility of civic government.
In 1916, as films began showing in other centres, the Manitoba Censor Board (MCB) was created under the Public Amusements Act (assented 10 March 1916), with regulation by the Amusements Act soon after. Manitoba would not be the only jurisdiction to establish a film censor board in the wake of cinema:[3]
The proliferation of film censor boards in Canada in the quarter century following the birth of the cinema mirrored, to a degree, the situation in the United States where, from 1907 onwards, state and city bodies sought to control the content of motion pictures. Each Canadian province had its own jurisdiction by the 1930s.
— James M. Skinner, deputy presiding member of the MFCB (1981-1987)
Only two classifications—general and adult—were used by MCB; all films were permissible to everyone since any undesirable material was removed by film editing. (In cases where editing was not possible, films were banned.) By 1933, "permissiveness on the screen had reached a peak with the phenomenal popularity of Mae West, whose dialogue was peppered with innuendo and double entendres."[3]
Until 1934, the Board fell under the Treasury Department, forming part of the section devoted to collecting the amusement taxes. However, public concern would develop over the MCB's dual and conflicting role as the body responsible for both censoring film and taxing film.
As result, the Board would be moved in 1935 to the Department of the Municipal Commissioner; at this time, the person who was appointed Commissioner was also appointed to the chairmanship of the MCB on a part-time basis. In 1948, the MCB was transferred to the newly-created Department of Public Utilities. The board would be composed of the chair, two censors, an inspector, a couple projectionists, and a secretary.[3]
In 1959, the Board rejected the internationally-acclaimed film Laronde, due to the theme of adultery that could not be edited out. The ensuing controversy caused a new film classification to be developed: the Restricted classification, which would be applicable to films deemed suitable for strictly adult audiences, and any one under the age of 17 was inadmissible.[3]
The Board's ability to censor made it the subject of several court cases in 1972, wherein defendants would successfully argue that if a government board had approved a film for public exhibition, another arm of government (in this case, the Department of Justice) should not charge them with the exhibition of that film. In response, the Amusements Act was amended to revoke the Board's authority to censor films. The MCB thus evolved into a classification board, the Manitoba Film Classification Board (MFCB), with the power to classify, but not edit or ban, making Manitoba the only province to not provide its board with the authority to censor.[3]
In 1991, the Home-Use Video Classification and Licensing Regulation of Manitoba came into effect, allowing MFCB the power to classify films that were in video format, as well as to license and regulate the home-use video industry.[3]
In 2003, MFCB classifications were changed its classification system from four tiers to five (G, PG, 14A, 18A, R), using the same symbols as the Canadian Home Video Rating System and the boards of B.C., Alberta, Saskatchewan, Ontario, and the Maritimes. The name for adult video/DVD products for home use was changed from 18+ to Adult, whereby all such products rented and sold in Manitoba must now display the Manitoba bar code.[3]
On 1 June 2005, through the Amusement Amendment Act, the MFCB would gain the power to regulate and enforce age restrictions on video and computer games sold in Manitoba. Under these regulations, video and computer games classified by the Entertainment Software Rating Board (ESRB) were adopted and some age restrictions were now enforced in Manitoba. Thereafter, it would be against the law to sell or rent video and computer games classified as M (mature) 17+ to anyone under the age of 17, or as AO (Adults Only) 18+ to people under 18.[3]
In 2017, the Government of Manitoba announced that it would dissolve the Manitoba Film Classification Board in order to "create efficiency" and streamline the content-classification system in Canada. On 1 July 2018, the MFCB was eliminated and its duties were outsourced to Consumer Protection British Columbia, following the footsteps of Saskatchewan.[2] While the government would maintain the right to inspect or review all film and video games under the new legislation, the new framework replacing the MFCB would nonetheless provide film festivals the freedom to classify movies on their own accord.[2]
Ratings
The following categories were the film and video ratings used by the Manitoba Film Classification Board:[4]
- General (G): Suitable for viewing by all ages.
- Parental Guidance (PG): Parental guidance is advised. Theme or content may not be suitable for all children.
- 14A: Suitable for viewing by persons 14 years of age or older. Persons under 14 must be accompanied by an adult. May contain violence, coarse language, and/or sexually suggestive scenes.
- 18A: Suitable for viewing by persons 18 years of age or older. Persons aged 14 to 17 must be accompanied by an adult. In the province of Manitoba, Children under the age of 14 are banned from viewing the film. May contain explicit violence, frequent coarse language, sexual activity and/or horror.
- Restricted (R): Admittance restricted to persons 18 years of age and older. Persons under 18 years of age are not permitted to attend under any circumstances. Contains frequent sexual activity, brutal/graphic violence, intense horror and/or other disturbing content that is inappropriate and therefore off-limits for viewing by persons under 18.
- Exempt (E): Films not reviewed or rated. "Any film not subject to, or exempt from classification."
References
- ^ "Province of Manitoba | Sport, Culture, and Heritage | Manitoba Film Classification Board". www.gov.mb.ca. Retrieved 2021-01-13.
- ^ a b c "Another board bites the dust: Tories axe board in charge of rating movies, video games | CBC News". CBC. Retrieved 2021-01-13.
- ^ a b c d e f g h i Manitoba Film Classification Board. "Historic Overview of the MFCB" and "Historic Highlights."
- ^ "Province of Manitoba | Sport, Culture, and Heritage | Manitoba Film Classification Board". www.gov.mb.ca. Retrieved 2021-01-13.
External links
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This page was last edited on 8 January 2024, at 09:04