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alexdok [17]
3 years ago
10

Calculate the amount in grams in 0.52600 moles of vanadium.

Chemistry
1 answer:
mrs_skeptik [129]3 years ago
4 0

Answer:

26.795229

Explanation:

This is how much it is

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5.00 ml of commercial bleach was diluted to 100.0 ml. 25.0 ml of the diluted sample was titrated with 4.56 ml of 0.100 m s2o3 2-
aev [14]

The solution would be like this for this specific problem:

 

4 NaOCl + S2O3{2-} + 2 OH{-} → 2 SO4{2-} + H2O + 4 NaCl

<span>(0.00456 L) x (0.100 mol/L S2O3{2-}) x (4 mol NaOCl / 1 mol S2O3{2-}) x (100.0 mL / 25 mL) x </span><span>
<span>(74.4422 g NaClO/mol) = 0.54313 g </span></span>

<span>(5.00 mL) x (1.08 g/mL) = 5.40 g solution </span>

(0.54313 g) / (5.40 g) = 0.101 = 10.1%

 

So, the average percent by mass of NaClO in the commercial bleach is 10.1%.

3 0
3 years ago
A certain first-order reaction has a rate constant of 2.15×10−2 s−1 at 20 ∘C. What is the value of k at 55 ∘C if Ea = 72.0 kJ/mo
adell [148]

Answer:

k_2=0.504s^{-1}

Explanation:

Hello there!

In this case, according to the given information, it turns out possible for us to calculate the rate constant at 55 °C by using the temperature-variable version of the Arrhenius equation:

ln(\frac{k_2}{k_1} )=-\frac{Ea}{R}(\frac{1}{T_2} -\frac{1}{T_1} )

Thus, we plug in the temperatures, activation energy and universal constant of gases in consistent units to obtain:

ln(\frac{k_2}{0.0215s^{-1}} )=-\frac{72000\frac{J}{mol}}{8.3145\frac{J}{mol*K}}(\frac{1}{55+273} -\frac{1}{20+273} ) \\\\ln(\frac{k_2}{0.0215s^{-1}} )=3.154\\\\k_2=0.0215s^{-1}exp(3.154)\\\\k_2=0.504s^{-1}

Regards!

3 0
3 years ago
Barium reacts with a polyatomic ion to form a compound with the general formula Ba3(X)2. What would be the most likely formula f
AfilCa [17]

Answer:

D. Na₃X

Explanation:

We have the neutral compound Ba₃(X)₂. <em>The total charge (zero) is equal to the sum of the charges of the ions times the number of ions in the molecule</em>.

3 × qBa + 2 × qX = 0

3 × (+2) + 2 × qX = 0

2 × qX = -6

qX = -3

If we have the cation Na⁺ and X³⁻, a neutral molecule would require 3 Na⁺ and 1 X³⁻. The resulting compound is Na₃X.

3 0
3 years ago
Read 2 more answers
Explain how you would calculate the q for warming 100.0 grams of liquid water from 0°C to 100 °C.
mojhsa [17]

Answer:

mass = 100 g

T1 = 0°C

T2 = 100 °C

C = 1 cal/g°C

Q = mC(T2 -T1)

Q = 100(1)(100 - 0)

Q = 100(100)

Q = 10000 cal

Explanation:

6 0
3 years ago
In addition, they are poor conductors of
MA_775_DIABLO [31]

Answer: electricity

Explanation:

6 0
3 years ago
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