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AVprozaik [17]
3 years ago
7

Need help balancing equations please.

Chemistry
1 answer:
densk [106]3 years ago
3 0

Answer:

A, 4,3,2

B, 1,1,2

C, 1,1,1,2

You might be interested in
What is the volume of 9.5 g fluorine gas, F2, at STP?
Tju [1.3M]

Answer:

5.6L

Explanation:

At STP, the pressure and temperature of an ideal gas is

P = 1 atm

T = 273.15k

Volume =?

Mass = 9.5g

From ideal gas equation,

PV = nRT

P = pressure

V = volume

n = number of moles

R = ideal gas constant =0.082J/mol.K

T = temperature of the ideal gas

Number of moles = mass / molar mass

Molar mass of F2 = 37.99g/mol

Number of moles = mass / molar mass

Number of moles = 9.5 / 37.99

Number of moles = 0.25moles

PV = nRT

V = nRT/ P

V = (0.25 × 0.082 × 273.15) / 1

V = 5.599L = 5.6L

The volume of the gas is 5.6L

5 0
3 years ago
What would be the composition and ph of an ideal buffer prepared from lactic acid (ch3chohco2h), where the hydrogen atom highlig
Mashutka [201]

Answer:

P_H =2.86

c=1.4\times 10^{-4}

Explanation:

first write the equilibrium equaion ,

C_3H_6O_3  ⇄ C_3H_5O_3^{-}  +H^{+}

assuming degree of dissociation \alpha =1/10;

and initial concentraion of C_3H_6O_3 =c;

At equlibrium ;

concentration of C_3H_6O_3 = c-c\alpha

[C_3H_5O_3^{-}  ]= c\alpha

[H^{+}] = c\alpha

K_a = \frac{c\alpha \times c\alpha}{c-c\alpha}

\alpha is very small so 1-\alpha can be neglected

and equation is;

K_a = {c\alpha \times \alpha}

[H^{+}] = c\alpha = \frac{K_a}{\alpha}

P_H =- log[H^{+} ]

P_H =-logK_a + log\alpha

K_a =1.38\times10^{-4}

\alpha = \frac{1}{10}

P_H= 3.86-1

P_H =2.86

composiion ;

c=\frac{1}{\alpha} \times [H^{+}]

[H^{+}] =antilog(-P_H)

[H^{+} ] =0.0014

c=0.0014\times \frac{1}{10}

c=1.4\times 10^{-4}

6 0
3 years ago
Assume that your empty crucible weighs 15.98 g, and the crucible plus the sodium bicarbonate sample weighs 18.56 g. After the fi
Savatey [412]

The question is incomplete, the complete question is;

Assume that your empty crucible weighs 15.98 g, and the crucible plus the sodium bicarbonate sample weighs 18.56 g. After the first heating, your crucible and contents weighs 17.51 g. After the second heating, your crucible and contents weighs 17.50 g.

What is the theoretical yield of sodium carbonate?

What is the experimental yield of sodium carbonate?

What is the percent yield for sodium carbonate?

Which errors could cause your percent yield to be falsely high, or even over 100%?

Answer:

See Explanation

Explanation:

We have to note that water is driven away after the second heating hence we are concerned with the weight of the pure dry product.

Hence;

From the reaction;

2 NaHCO3 → Na2CO3(s) + H2O(l) + CO2(g)

Number of moles of  sodium bicarbonate = 18.56 - 15.98 = 2.58 g/87 g/mol

= 0.0297 moles

2 moles of sodium bicarbonate yields 1 mole of sodium carbonate

0.0297 moles of 0.015 moles  sodium bicarbonate yields 0.0297 * 1/2 = 0.015 moles

Theoretical yield of sodium carbonate = 0.015 moles * 106 g/mol = 1.59 g

Experimental yield of sodium bicarbonate = 17.50 g - 15.98 g = 1.52 g

% yield = experimental yield/Theoretical yield * 100

% yield = 1.52/1.59 * 100

% yield = 96%

The percent yield may exceed 100% if the water and CO2 are not removed from the system by heating the solid product to a constant mass.

5 0
2 years ago
Complex molecules are formed by the bond formation between monomers . Which process is being referred to in this statement? A.)
Vesnalui [34]
I think the correct answer from the choices listed above is option B. Polymerization is the process of forming c<span>omplex molecules by the bond formation between monomers. There are two types of this process which are the addition and condensation polymerization.</span>
6 0
3 years ago
Read 2 more answers
How many Ne atoms are contained in 32.0 g of the element?
Svetlanka [38]
Mass atomic of Ne=20.18 u
Therefore:
molar mass=20.18 g/1 mol

1 mole=6.022*10²³ particles (atoms or molecules)

Then: 6.022*10²³ atoms are contained in 20.18g

Now, We can solve this problem by the three rule.

6.022*10²³ atoms-------------------20.18 g
x------------------------------------------32 g

x=(6.022*10²³ atoms * 32 g)/20.18 g=9.55*10²³ atoms.

Answer: 9.55*10²³ Ne atoms are contained in 32 g of the element.
3 0
3 years ago
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