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nekit [7.7K]
2 years ago
7

Consider the following reaction: 2HI(aq)+Ba(OH)2(aq) → 2H2O(l)+BaI2(aq) Enter the complete ionic equation for this reaction. Exp

ress your answer as a complete ionic equation. Identify all of the phases in your answer
Chemistry
1 answer:
babunello [35]2 years ago
3 0

Answer:

2 HI(aq) + Ba²⁺(aq) + 2 OH⁻(aq) → 2 H₂O(l) + Ba²⁺(aq) + 2I⁻(aq)

Explanation:

Let's consider the following balanced molecular equation.

2 HI(aq) + Ba(OH)₂(aq) → 2 H₂O(l) + BaI₂(aq)

In the complete ionic equation we include all the ions and the species that do not dissociate in water. HI is a weak electrolyte so it exists mostly in the molecular form. H₂O has a very low equilibrium constant (Kw = 10⁻¹⁴) si it exists mainly in the molecular form.

2 HI(aq) + Ba²⁺(aq) + 2 OH⁻(aq) → 2 H₂O(l) + Ba²⁺(aq) + 2I⁻(aq)

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Consider four sealed, rigid containers with the following volumes: 50 mL, 100 mL, 250 mL, and 500 mL. If each of these contains
dsp73
First, we assume that helium behaves as an ideal gas such that the ideal gas law is applicable.
                                     PV = nRT
where P is pressure, V is volume, n is number of moles, R is universal gas constant, and T is temperature. From the equation, if n, R, and T are constant, there is an inverse relationship between P and V. From the given choices, the container with the greatest pressure would be the 50 mL. 
5 0
2 years ago
A student dissolved 4.00 g of Co(NO3)2 in enough water to make 100. mL of stock solution. He took 4.00 mL of the stock solution
mihalych1998 [28]

Answer:

0.08097 grams of nitrate ions are there in the final solution.

Explanation:

Moles of cobalt(II) nitrate ,n= \frac{4.00 g}{245 g/mol}=0.01633 mol

Volume of the cobalt(II) nitrate solution, V = 100.0 mL = 0.1 L

Molarity=\frac{n}{V(L)}

Let the molarity of the solution be M_1

M_1=\frac{0.01633 mol}{0.1 L}=0.1633 M

A students then takes 4 .00 mL of M_1 solution and dilute it to 275 ml.

M_1=0.1633 M

V_1=4.00 mL

M_2=? (molarity after dilution)

V_2=275 mL (after dilution)

M_1V1=M-2V_2

M_2=\frac{M_1V_1}{V_2}=\frac{0.1633 M\times 4.00 mL}{275 mL}=0.002375 M

Molarity of the of solution after dilution is 0.002375 M.

Co(NO_3)_2(aq)\rightarrow Co^{2+}(aq)+2NO_3^{-}(aq)

1 mol of cobalt(II) nitrate gives 2 moles of nitrate ions. Then 0.002375 M solution of cobalt (II) nitrate will give:

[NO_3^{-}]=\frac{2}{1}\times 0.002375 M=0.004750 M

Moles of nitrate ions = n

Volume of the solution = 275 mL = 0.275 L

Molarity of the nitrate ions = [NO_3^{-}]=0.004750 M

[NO_3^{-}]=\frac{n}{0.275 L}

n = 0.001306 mol

Mass of 0.001306 moles of nitrate ions:

0.001306 mol × 62 g/mol= 0.08097 g

0.08097  grams of nitrate ions are there in the final solution.

4 0
2 years ago
7. You are about to perform some intricate electrical studies on single skeletal muscle fibers from a gastronemius muscle. But f
Vilka [71]

Answer:

58.61 grams

Explanation:

Taking The molecular weight of NaCl = 58.44 grams/mole

<u>Determine how many grams of NaCl to prepare the bath solution </u>

first we will calculate the moles of NaCl that is contained in 6L of 170 mM of NaCI solution

= ( 6 * 170 ) / 1000

= 1020 / 1000 = 1.020 moles

next

determine how many grams of NaCl

= moles of NaCl * molar mass of NaCl

= 1.020 * 58.44

= 58.61 grams

4 0
2 years ago
Element X is a radioactive isotope such that every 82 years, its mass decreases by half. Given that the initial mass of a sample
lesantik [10]

Answer: 17 years

Explanation:

Expression for rate law for first order kinetics  for radioactive substance is given by:

t=\frac{2.303}{k}\log\frac{a}{a-x}

where,

k = rate constant

t = age of sample

a = let initial amount of the reactant

a - x = amount left after decay process  

a) for completion of half life:

Half life is the amount of time taken by a radioactive material to decay to half of its original value.

t_{\frac{1}{2}}=\frac{0.693}{k}

k=\frac{0.693}{82years}=8.4\times 10^{-3}years^{-1}

b) for 8900 g of the mass of the sample to reach  7700 grams

t=\frac{2.303}{8.4\times 10^{-3}}\log\frac{8900}{7700}

t=17years

Thus it will take 17 years

4 0
2 years ago
Convert 1.72 moles of magnesium carbonate to formula units
snow_tiger [21]
One mole any substance contains 6.022 ₓ 10²³ particles called Avogadro's Number.

The relation between moles and number of particles is given as,

                        # of particles  =  moles ₓ Avogadro's number

In our case the particles are formula units of MgCO₃. So, 1 mole of MgCO₃ contain 6.022 ₓ 10²³ formula units, then the number of formula units in 1.72 moles are calculated as,

              # of formula units  =  1.72 mol ₓ 6.022 ₓ 10²³ formula units / mol
            
             # of formula units   =  1.035 ₓ 10²⁴ Formula Units
8 0
2 years ago
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