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d1i1m1o1n [39]
4 years ago
6

You have seven closed containers, each with equal masses of chlorine gas (cl2). you add 10.0 g of sodium to the first sample, 20

.0 g of sodium to the second sample, and so on (adding 70.0g of sodium to the seventh sample). sodium and chlorine react to form sodium chloride according to the equation: 2na(s) + cl2(g)  2nacl(s) after the reaction is complete, you collect and measure the amount of sodium chloride
Chemistry
1 answer:
Phoenix [80]4 years ago
7 0

The reaction between Na and Cl2 is

2Na + Cl2 ---> 2NaCl

So two moles of Na will react with one mole of Cl2

However here in the given question the moles of Cl2 is constant as mass is constant. Say we have 71 of Cl2 in each container, the moles = 71 / 35.5 = 2 moles of Cl2

Moles of Na = Mass / atomic mass

The container will have following moles of Na

1 ) Moles = 10 / 23 = 0.435  (these will react with 0.2175 moles of Cl2 to give 0.435 moles of NaCl]

2) moles = 20 / 23 = 0.87 ( (these will react with 0.435 moles of Cl2 to give 0.87 moles of NaCl]

3) moles = 30 / 23 = 1.3 ( (these will react with 0.65 moles of Cl2 to give 1.3 moles of NaCl]

4) moles = 40 / 23 = 1.74 (these will react with 0.87 moles of Cl2 to give 1.7 moles of NaCl]

5) moles = 50 / 23 = 2.18 (these will react with 1.09 moles of Cl2 to give 2.18 moles of NaCl]

6) moles = 60 / 23 = 2.61  (these will react with 1.3 moles of Cl2 to give 2.61 moles of NaCl]

7) moles = 70 / 23 =  3 moles  (these will react with 1.5 moles of Cl2 to give 3 moles of NaCl]

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Find the empirical formula of the compound ribose, a simple sugar often used as a nutritional supplement. A 14.229 g sample of r
MakcuM [25]

Answer:

CH2O

Explanation:

Firstly, we need to convert the masses of the elements to percentage compositions. This can be done by placing the mass of each element over the total mass multiplied by 100% . We can start with carbon.

C = 5.692/14.229 * 100 = 40%

O = 7.582/14.229 * 100 = 53.29%

H = 0.955/14.229 * 100 = 6.71%

We then proceed to divide each percentage composition by their atomic mass of 12, 16 and 1 respectively.

C = 40/12 = 3.333

O = 53.29/16 = 3.33

H = 6.71/2 = 6.71

Dividing by the smaller value which is 3.33

C = 3.33/3.33 = 1

O = 3.33/3.33= 1

H = 6.71/3.33 = 2

The empirical formula of the compound ribose is CH2O

6 0
4 years ago
Read 2 more answers
A container has the dimensions of 10cm x 80mm x 0.15m. The density of
vesna_86 [32]

Answer:

2.52kg

Explanation:

80mm = 8 cm

0.15m = 15 cm

volume of the container = 10m × 8cm × 15cm

= 1200 cubic centimetres

mass = density × volume

= 2.1 × 1200

= 2520g

2520/ 1000

= 2.52kilograms

8 0
3 years ago
How many structures are possible for a octahedral molecule with a formula of ax4y2?
vekshin1

There are only two possible structures for an octahedral molecule with a formula of ax4y2. One is when the two y’s are next to each other and the other one is when the two y’s are of the opposite side of the molecule. Different structures can be drawn but only two types of molecule can be formed.

3 0
4 years ago
How many atoms of iodine are in 12.75g of CaI2? Hint. How many Iodines are there in one CaI2 particle?
xenn [34]

Answer:

5.225x10^{22}atoms\ I

Explanation:

Hello!

In this case, since 12.75 g of calcium iodide has the following number of moles (molar mass = 293.89 g/mol):

n_{CaI_2}=12.75gCaI_2*\frac{1molCaI_2}{293.89gCaI_2}=0.0434molCaI_2

In such a way, since 1 mole of calcium iodide contains 2 moles of atoms of iodine, and one mole of atoms of iodine contains 6.022x10²³ atoms (Avogadro's number), we compute the resulting atoms as shown below:

atoms\ I=0.0434molCaI_2*\frac{2molI}{1molCaI_2} *\frac{6.022x10^{23}atoms\ I}{1molI} \\\\atoms\ I = 5.225x10^{22}atoms\ I

Best regards!

4 0
3 years ago
PLEASEEEE HELP MEEE
Nookie1986 [14]
The answer is i don’t know because i’m just doing this for points sorry!!! hope this helps
4 0
3 years ago
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