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Nina [5.8K]
3 years ago
10

What is the product of the unbalanced equation below?

Chemistry
1 answer:
Stolb23 [73]3 years ago
3 0

Answer:

D

Explanation:

The answer is D. I'm not sure that it is a solid. I don't think it is a ppte, which is the only way it can be a true solid. It is ionic if the reaction is taking place in water and there is someway to start the reaction. Be that as it may, the internal balace numbers of the chemical produced is the only possible answer. The balanced eq;uatioon is

2Al + 3Br2 ==> 2AlBr3

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If the heat capacity of an object is known, what other information will need to be known to calculate its specific heat capacity
zimovet [89]

Answer:

B. its temperature

Explanation:

4 0
1 year ago
Read 2 more answers
Consider the following Reaction.
jeyben [28]

Hey there!:

Molar mass of Mg(OH)2 = 58.33 g/mol


number of moles Mg(OH)2 :

moles of Mg(OH)2 = 30.6 / 58.33 =>  0.5246 moles

Molar mass of H3PO4 =  97.99 g/mol

number of moles H3PO4:

moles of Mg(OH)2 = 63.6 / 97.99 => 0.649 moles

Balanced chemical equation is:


3 Mg(OH)2 + 2 H3PO4 --->  Mg3(PO4)2 + 6 H2O


3 mol of Mg(OH)2 reacts with 2 mol of H3PO4  ,for 0.5246 moles of Mg(OH)2, 0.3498 moles of H3PO4 is required , but we have 0.649 moles of H3PO4, so, Mg(OH)2 is limiting reagent !

Now , we will use Mg(OH)2 in further calculation  .

Molar mass of Mg3(PO4)2 = 262.87 g/mol

According to balanced equation  :

mol of Mg3(PO4)2 formed = (1/3)* moles of Mg(OH)2


= (1/3)*0.5246


= 0.1749 moles of Mg3(PO4)2

use :

mass of Mg3(PO4)2 = number of mol * molar mass


= 0.1749 *  262.87

= 46 g of Mg3(PO4)2

Therefore:

% yield = actual mass * 100 / theoretical mass

% = 34.7 * 100 / 46

% = 3470 / 46

= 75.5%


Hope that helps!




3 0
3 years ago
How many grams of F2 gas are there in a 5.00-L cylinder at 4.00 × 10^3 mm Hg and 23°C?
GuDViN [60]

Answer:

41.17g

Explanation:

We are given the following parameters for Flourine gas(F2).

Volume = 5.00L

Pressure = 4.00× 10³mmHG

Temperature =23°c

The formula we would be applying is Ideal gas law

PV = nRT

Step 1

We find the number of moles of Flourine gas present.

T = 23°C

Converting to Kelvin

= °C + 273k

= 23°C + 273k

= 296k

V = Volume = 5.00L

R = 0.08206L.atm/mol.K

P = Pressure (in atm)

In the question, the pressure is given as 4.00 × 10³mmHg

Converting to atm(atmosphere)

1 mmHg = 0.00131579atm

4.00 × 10³ =

Cross Multiply

4.00 × 10³ × 0.00131579atm

= 5.263159 atm

The formula for number of moles =

n = PV/RT

n = 5.263159 atm × 5.00L/0.08206L.atm/mol.K × 296K

n = 1.0834112811moles

Step 2

We calculate the mass of Flourine gas

The molar mass of Flourine gas =

F2 = 19 × 2

= 38 g/mol

Mass of Flourine gas = Molar mass of Flourine gas × No of moles

Mass = 38g/mol × 1.0834112811moles

41.169628682grams

Approximately = 41.17 grams.

3 0
3 years ago
What is the concentration of the unknown h3po4 solution? the neutralization reaction ish3po4(aq)+3naoh(aq)→3h2o(l)+na3po4(aq) -g
hjlf
First, we need to get moles of NaOH:

when moles NaOH = volume * molarity 

                                  = 0.02573L * 0.11 M

                                 = 0.0028 moles 

from the reaction equation:

H3PO4(aq) + 3NaOH → 3 H2O(l) + Na3PO4(aq)

we can see that when 1 mol H3PO4 reacts with→ 3 mol NaOH

 ∴ X mol H3PO4 reacts with → 0.0028 moles NaOH

∴ moles H3PO4 = 0.0028 mol / 3 = 9.4 x 10^-4 mol

now we can get the concentration of H3PO4:

∴[H3PO4] = moles H2PO4 / volume

               = 9.4 x 10^-4 / 0.034 L

               = 0.028 M
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What are alternative periodic tables
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Tabulations of chemical elements differing in their organization from the traditional seen periodic system
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