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3 years ago

Who goes to K12 and if so what grade

Chemistry

2 answers:

olchik [2.2K]3 years ago

8 0

Answer:

I might next year with my cuz and nex year ima be in 8th

Explanation:

grandymaker [24]3 years ago

5 0

Answer:

Hey! I`m In K12. Currently in 7th, going into 8th next year. I have Mrs. Keefe and one of my teachers, what about y`all?

You might be interested in

Which of the following is one way to prevent the corrosion of iron?

aalyn [17]

The best way to prevent the corrosion of iron is to paint the exposed iron with protective paint, because that prevents water from rusting the iron.

8 0

3 years ago

Why do some objects such as windows and doors close in the winter but can be harder to close in the summer?

Elenna [48]

When the temperature rises some materials experience thermal expansion which means they get bigger, so it may make them harder to close because they don't fit as well.

4 0

3 years ago

Read 2 more answers

Se sabe que 10 g de calcio reaccionan con 4 g de oxígeno para obtener 14 g de óxido de calcio. Indica la cantidad de óxido de ca

egoroff_w [7]

Answer:

Si se usan 50 gramos de calcio y óxigeno, se obtienen 70 gramos de óxido de calcio.

Explanation:

Hola,

En este caso, la reacción llevada a cabo es:

$2Ca+O_2\rightarrow 2CaO$

De este modo si asumimos el ejemplo dado, 50 gramos de calcio, cuya masa atómica es 40 g/mol y 50 g de oxígeno, cuya masa atómica como gas diatómico es 32 g/mol, antes de calcular los gramos de óxido de calcio producidos, debemos identificar el reactivo límite. Así, calculamos las moles de calcio disponibles en 50 g:

$mol_{Ca}^{disponible}=50gCa*\frac{1molCa}{40gCa} =1.25molCa$

Y también las moles de calcio consumidas por los 50 g de oxígeno, utilizando su relación molar 2:1:

$mol_{Ca}^{consumidas\ por\ O_2}=50gO_2*\frac{1molO_2}{32gO_2} *\frac{2molCa}{1molO_2} =3.125molCa$

Por lo tanto, hay menos calcio disponible que el que consume el oxígeno, por lo que el calcio esel reactivo límite. Ahora, con este, calculamos los gramos de óxido de calcio, cuya masa molar es 56 g/mol, que se producen:

$m_{CaO}=1.25molCa*\frac{2molCaO}{2molCa}* \frac{56gCaO}{1molCaO}\\ \\m_{CaO}=70gCaO$

Esto quiere decir que de 50 gramos de oxígeno, solo 20 gramos reaccionan para formar 70 gramos de óxido de calcio.

Saludos!

4 0

3 years ago

Read 2 more answers

Sulfur reacts with fluorine to produce three different compounds. The mass ratio of fluorine to sulfur for each compound is give

NARA [144]

Answer:

Explanation:

Mass of F / Mass of S = 2.962/1 =2.962 X 32 / 32 = 94.78/32

Mass of F / Mass of S = 2.370 /1 = 2.370 X 32 / 32 = 75.84 /32

Mass of F /Mass of S = 3.555/1 = 3.555 x 32 / 32 = 113.76 / 32 .

Now constant mass of S that is 32 g reacts with different mass of F. They are as follows :

94.78 g , 75.84 g , and 113.76 g

Their ratio = 94.78 : 75.84 : 113.76

divide them by 19

their ratio = 5 : 4 : 6

So these data are consistent with law of multiple proportion.

8 0

3 years ago

Hello,

nikitadnepr [17]

Answer:

$a)\ 2FeSO4 \xrightarrow{\text{Heat}} SO_3+SO_2+Fe_2O_3\\b)\ Decomposition\ Reaction$

Explanation:

Ferrous Sulphate $(FeSO4)$ is generally found as Lime-Green Crystals. On heating, these crystals almost immediately turn white-yellow. They then, break down to produce an anhydrous mixture of Sulphur Trioxide $(SO_3)$ , Sulphur Dioxide $(SO_2)$ as well as Ferric Oxide $(Fe_2O_3)$ .

We can hence, frame a skeletal equation of this reaction and try to balance it.

Hence,

$FeSO4 \xrightarrow{\text{Heat}} SO_3+SO_2+Fe_2O_3$

Now,

a)In order to balance it through the 'Hit &Trial Method', we'll follow a series of steps:

1. First, lets compare the number of Fe (Iron) atoms on the RHS and LHS. We find that, the no. of Fe Atoms on the RHS is twice the number of Fe Atoms on the LHS. We hence, add a co-effecient 2 beside $FeSO_4$ .

2. Now, Iron atoms, Sulphur Atoms and Oxygen atoms occur 2, 2, 8 respectively on both the sides:

Hence, As all the other elements as well as iron, balance, we've arrived upon our Balanced Equation :

$2FeSO4 \xrightarrow{\text{Heat}} SO_3+SO_2+Fe_2O_3$

b) We know that, decomposition reactions are [generally] endothermic reactions in which Large Compounds decompose into smaller elements and compounds. Here, as Ferrous Sulphate decomposes into Sulphur Dioxide, Sulphur Trioxide and Ferric Oxide, the reaction that occurs here is Decomposition Reaction.

7 0

2 years ago