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ikadub [295]
3 years ago
12

Balance the equations by inserting coefficients as needed.

Chemistry
1 answer:
drek231 [11]3 years ago
4 0

Answer:

first one is already balanced

2.   MgN2+6HCl->3MgCl2+2NH3

Explanation:

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Barium nitrate is combined with sodium phosphate. When the balanced net ionic equation is constructed for this reaction, the sto
alexandr1967 [171]
The complete equation for this reaction is,

Ba(NO3)2 + Na2PO4 = 2NaNO3 + BaPO4

Among the compounds present in the reaction, Barium Nitrate, Sodium Phosphate and Sodium Nitrate are soluble ionic compounds. Hence, they will completely ionize into ions. Only BaSO4 is insoluble which becomes the precipitate. Ionic equation is:

Ba2+ + 2NO3- + 2Na+ + PO42- = 2Na+ +NO3- + BaPO4

Cancel like ions,

Ba2+ + PO42- = BaPO4.

Thus, the stoichiometric coefficient of Barium ion is 1.
3 0
3 years ago
Complete combustion of 4.20 g of a hydrocarbon produced 12.9 g of CO2 and 6.15 g of H2O. What is the empirical formula for the h
umka21 [38]
We calculate first the number of moles of CO2 and H2O produced by dividing the given masses by the molar masses of CO2 and H2O.
 moles CO2 = (12.9 g CO2) x (1 mole CO2 / 12 g CO2) = 1.075 moles.
 moles H2O = (6.15 g H2O) x (1 mole H2O / 18 g H2O) = 0.36 moles
Then, we count the number of C, H, and O moles. This gives us 1.075 moles C, 2.5 moles O and 0.72 moles H. The empirical formula is,
                             C1.075H0.72O2.5
Simplifying, 
                             C4H3O10
4 0
3 years ago
SiS2 is a _____ compound
Lerok [7]

Answer:

SiS2, silicone disulfide, is a linear, nonpolar compound.

7 0
3 years ago
A sample of methane collected when the temp was 30 C mmHg measures 398 mL. What would be the volume of the sample at -5 C and 61
Levart [38]
<h2>Question </h2>

A sample of methane collected when the temp was 30 C and 760mmHg measures 398 mL. What would be the volume of the sample at -5 C and 616 mmHg pressure

<h2>Answer:</h2>

434.32mL

<h2>Explanation:</h2>

Using the combined gas law:

\frac{PV}{T} = k

Where;

P = Pressure

V = Volume

T = Temperature

k = constant.

It can be deduced that:

\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2} = k           ---------------------(i)

Where:

P₁ and P₂ are the initial and final pressures of the given gas

V₁ and V₂ are the initial and final volumes of the given gas

T₁ and T₂ are the initial and final temperatures of the gas.

<em>From the question:</em>

the gas is methane

P₁  = 760mmHg

P₂ = 616mmHg

V₁ = 398mL

V₂ = ?

T₁ = 30°C = (30 +273)K = 303K

T₂ = -5°C = (-5 +273)K = 268K

Substitute these values into equation (i) as follows;

\frac{760*398}{303} = \frac{616*V_2}{268}

Solve for V₂

V₂ = \frac{760*398*268}{616*303}

V₂ = 434.32mL

Therefore, the volume of the sample at -5C and 616mmHg pressure is 434.32mL

5 0
3 years ago
What is the formula for Disulfur Pentafluoride?
dolphi86 [110]

Answer:

please mark me as brainliest so difficulty to find me and edit

7 0
3 years ago
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