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3 years ago

5

Many manufacturing processes involve chemical reactions that reach equilibrium. Why would chemists adjust conditions to favor th

e forward reaction in a process

2 answers:

ivann1987 [24]3 years ago

7 0

To increase the amount of product

storchak [24]3 years ago

4 0

The answer is to increase the amount of product -gradpoint

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1) If calcium were to combine with the compound barium nitrate, which member of the barium nitrate would calcium replace?

snow_tiger [21]

Answer:

Calcium would displace barium.

Explanation:

Ba(NO₃)₂ + Ca --> Ca(NO₃)₂ + Ba

There are two types of compounds: molecular/covalent and ionic.

Molecular/covalent compounds are non-metal + non-metal.

Ionic compounds are metal + non-metal.

Looking at the periodic table, barium is a metal. Calcium is also a metal.

Checking a polyatomic ions chart would tell you NO₃⁻ is a non-metal because it has a negative charge.

Since there is no metal + metal compound, the calcium metal would displace barium. The compound remains ionic.

7 0

3 years ago

Calculate the mass of magnesium carbonate ( MgCO3), in grams, required to produce 110.0 g of carbon dioxide using the following

bearhunter [10]

Answer:

$210.7~g~MgCO_3$

Explanation:

We have to start with the <u>reaction</u>:

$MgCO_3~->~MgO~+~CO_2$

We have the same amount of atoms on both sides, so, we can continue. The next step is to find the <u>number of moles</u> that we have in the 110.0 g of carbon dioxide, to this, we have to know the <u>atomic mass of each atom</u>:

C: 12 g/mol

O: 16 g/mol

Mg: 23.3 g/mol

If we take into account the number of atoms in the formula, we can calculate the <u>molar mass</u> of carbon dioxide:

$(12*1)+(16*2)=44~g/mol$

In other words: $1~mol~CO_2=~44~g~CO_2$ . With this in mind, we can calculate the moles:

$110~g~CO_2\frac{1~mol~CO_2}{44~g~CO_2}=25~mol~CO_2$

Now, the <u>molar ratio</u> between carbon dioxide and magnesium carbonate is 1:1, so:

$2.5~mol~CO_2=2.5~mol~MgCO_3$

With the molar mass of $MgCO_3$ ( $(23.3*1)+(12*1)+(16*3)=84.3~g/mol$ . With this in mind, we can calculate the <u>grams of magnesium carbonate</u>:

$2.5~mol~MgCO_3\frac{84.3~g~MgCO_3}{1~mol~MgCO_3}=210.7~g~MgCO_3$

I hope it helps!

8 0

3 years ago

If a star is 33.11 trillion kilometers away how much light years would that be

abruzzese [7]

33.11 trillion kilometers is equivalent to 3.5 light years

hope this helps :)<span />

8 0

3 years ago

AC current is produced by which of the following things?

pogonyaev

Answer:

B

Explanation:

5 0

2 years ago

Read 2 more answers

Why is the answer C for this problem?

pshichka [43]

Answer:

$\boxed{\text{(C) X}$_{3}$P$_{2}}$

Explanation:

Step 1. Identify the Group that contains X

We look at the consecutive ionization energies and hunt for a big jump between them

$\begin{array}{crc}n & IE_{n} & IE_{n} - IE_{n-1}\\1 & 730 & \\2 & 1450 & 720\\3 & 7700 & 6250\\4 & 10500 & 2800\\\end{array}$

We see a big jump between n = 2 and n = 3. This indicates that X has two valence electrons.

We can easily remove two electrons, but the third electron requires much more energy. That electron must be in the stable, filled, inner core.

So, X is in Group 2 and P is in Group 15.

Step 2. Identify the Compound

X can lose two valence electrons to reach a stable octet, and P can do the same by gaining three electrons.

We must have 3 X atoms for every 2 P atoms.

The formula of the compound is $\boxed{\text{X}$_{3}$P$_{2}}$}$ .

4 0

3 years ago