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Vinvika [58]
3 years ago
6

Mass of NaOH = 0.2 Volume of solution 500 ml

Chemistry
1 answer:
Yuliya22 [10]3 years ago
6 0
Density will be mass/volume= 0.2/500 = 0.0004
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If some salt were accidently spilled as it was transferred from the balance to the cup would your calculated enthalpy of solutio
morpeh [17]

Answer:

It would be to low

Explanation:

7 0
2 years ago
Write balanced equations for the following reactions
Arte-miy333 [17]

Answer:

See explanation

Explanation:

The shorthand nuclear reaction equations have been given; the first particle in the parentheses is a reactant particle while the second particle is a product particle. These can now be rewritten as the longhand equations as follows;

238/92U + 4/2 He -------> 241/94Pu + 1/0 n

238/92U + 4/2 He ------> 241/94Pu + 1/0 n

14/7N + 4/2 He------> 17/8O + 1/1 p

56/26Fe + 2 4/2 He----> 60/29Cu + 4/2 He

4 0
3 years ago
Carbon-14 has a half-life of 5,730 years. How long will it take for 112. 5 g of a 120. 0-g sample to decay radioactively? 5,730
vichka [17]

The time taken by Carbon-14 to decay radioactively from 120g to 112.5g is 22,920 years.

<h3>How do we calculate the total time of decay?</h3>

Time required for the whole radioactive decay of any substance will be calculated by using the below link:

T = (n)(t), where

  • t = half life time = 5730 years
  • n = number of half life required for the decay

Initial mass of Carbon-14 = 120g

Final mass of Carbon-14 = 112.5g

Left mass = 120 - 112 = 7.5g

Number of required half life for this will be:

  • 1: 120 → 60
  • 2: 60 → 30
  • 3: 30 → 15
  • 4: 15 → 7.5

4 half lives are required, now on putting values we get

T = (4)(5730) = 22,920 years

Hence required time for the decay is 22,920 years.

To know more about radioactive decay, visit the below link:

brainly.com/question/24115447

#SPJ1

3 0
2 years ago
Nitric acid is a strong acid, sodium hydroxide is a strong base, and sodium nitrate is a soluble salt. Which of the following is
Evgesh-ka [11]
Can’t help with this but good luck
4 0
3 years ago
When 5.00 g of Al2S3 and 2.50 g of H2O are reacted according to the following reaction: Al2S3(s) + 6 H2O(l) → 2 Al(OH)3(s) + 3 H
Debora [2.8K]

Answer:

Y=58.15\%

Explanation:

Hello,

For the given chemical reaction:

Al_2S_3(s) + 6 H_2O(l) \rightarrow 2 Al(OH)_3(s) + 3 H_2S(g)

We first must identify the limiting reactant by computing the reacting moles of Al2S3:

n_{Al_2S_3}=5.00gAl_2S_3*\frac{1molAl_2S_3}{150.158 gAl_2S_3} =0.0333molAl_2S_3

Next, we compute the moles of Al2S3 that are consumed by 2.50 of H2O via the 1:6 mole ratio between them:

n_{Al_2S_3}^{consumed}=2.50gH_2O*\frac{1molH_2O}{18gH_2O}*\frac{1molAl_2S_3}{6molH_2O}=0.0231mol  Al_2S_3

Thus, we notice that there are more available Al2S3 than consumed, for that reason it is in excess and water is the limiting, therefore, we can compute the theoretical yield of Al(OH)3 via the 2:1 molar ratio between it and Al2S3 with the limiting amount:

m_{Al(OH)_3}=0.0231molAl_2S_3*\frac{2molAl(OH)_3}{1molAl_2S_3}*\frac{78gAl(OH)_3}{1molAl(OH)_3} =3.61gAl(OH)_3

Finally, we compute the percent yield with the obtained 2.10 g:

Y=\frac{2.10g}{3.61g} *100\%\\\\Y=58.15\%

Best regards.

7 0
3 years ago
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