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solniwko [45]
2 years ago
13

B. Calculate how many atoms of hydrogen are in 1. 75 moles of glucose (c6h12o6).

Chemistry
1 answer:
ycow [4]2 years ago
4 0

Answer:

1.26x10^25 atoms of hydrogen

Explanation:

because there are 12 atoms of hydrogen in a molecule of glucose, multiply 12 by Avogadro's number (6.02x10^23) to get how many molecules of hydrogen there are in a mole of glucose. Then multiply that number by 1.75, which is the number of moles of glucose there is in this problem.

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100x100 wsbntxrdfdmxnjhdtnvj
CaHeK987 [17]

Answer:

10000

Explanation:

just add the zeros

hope dis helps ^-^

6 0
3 years ago
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Doctors sometimes use computer models to track the spread of an infectious disease . How might this help them better understand
kotegsom [21]
It could them know how it start when it started wats in the infectious disease and why
7 0
3 years ago
Gold has a molar mass of 197 g/mol. (a) how many moles of gold are in a 3.98 g sample of pure gold? (b) how many atoms are in th
Natali5045456 [20]
A.)49.4974874 moles or 49.5 moles
B.)2.980808730172671e+25 or 3e+25
6 0
3 years ago
A compound is 2.00% H by mass, 32.7% S by mass, and 65.3% O by mass. What is its empirical
Katarina [22]

Answer:

empirical formula: H_2SO_4

2 g H

32.7 g S

65.3 g O

Explanation:

Like the problem said, the first thing we can do is calculate the mass of each of the 3 elements in a 100-gram sample:

- 2.00% * 100g = 2 g H

- 32.7% * 100g = 32.7 g S

- 65.3% * 100g = 65.3 g O

Now we need to find the empirical formula from these. To do so, convert all of those masses into moles by using the molar mass for each element:

- the molar mass of H is 1.01 g/mol

- the molar mass of S is 32.06 g/mol

- the molar mass of O is 16 g/mol

2 g H ÷ 1.01 g/mol = 1.98 mol H

32.7 g S ÷ 32.06 g/mol = 1.02 mol S

65.3 g O ÷ 16 g/mol = 4.08 mol O

Our ratio of H : S : O is now:

1.98 mol : 1.02 mol : 4.08 mol

Divide them all by the smallest number, which is 1.02:

1.98/1.02  :  1.02/1.02  :  4.08/1.02

1.94 : 1 : 4

1.94 ≈ 2

So:

2 : 1 : 4

Thus, the empirical formula is: H_2SO_4.

7 0
3 years ago
In the important industrial process for producing ammonia (the Haber Process), the overall reaction is:
Kisachek [45]

Answer:

Explanation:

Here we have to use stoichiometry.

First of all, we have to calculate the mass of 100% of yield:

1.7 g ------- 98%

X -------- 100%

X = 1.73 g (approximately)

Second, we have to calculate the mass of N2 that is necessary to react to produce the mass of 1.73g of NH3. To do that, we have to use the Molar mass of N2 and NH3 and don't forget the stoichiometric relationship between them.

Molar Mass N2 : 14x2 = 28 g/mol

Molar Mass NH3: 14 + 3 = 17 g/mol

28g (N2) ------- 17x2 (NH3)

X ------------ 1.73 g

X = 1.42 g (approximately)

5 0
3 years ago
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