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aleksandrvk [35]
3 years ago
6

What is beta radiation?

Chemistry
1 answer:
Nutka1998 [239]3 years ago
6 0

Answer:

emission at a high speed or energy which often forms electrons

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Which formula represents an ionic compound?<br> a. NaCl<br> b. N2O<br> c. HCl<br> d.H2O
icang [17]
I think the correct answer from the choices listed above is option A. It is NaCl the formula that represents an ionic compound. An ionic compound is held my oppositely charged atoms. Therefore, this would mean that one is a positive and another is a negative held together. NaCl is held by these forces.
3 0
3 years ago
3. The alcohol in "gasohol" burns according to the following equation. C2H6O + 3 O2 --------&gt; 2 CO2 + 3 H2O (a) If 25 moles o
Vanyuwa [196]
For all question, all you need to use is the mole-mole ratio. 

a) 25 moles C2H6O (3 moles O2/ 1 mol C2H6O)= 75 moles O2 

b) 30 moles O2 (1 moles C2H6O/ 3 moles O2)= 10 moles C2H6O

c) 23 moles CO2 (3 moles O2/ 2 moles CO2) = 34.5 moles O2

d)  41 moles H2O ( 1 moles C2H6O/ 3 moles H2O= 13.7 moles C2H6O
7 0
3 years ago
Hi no Link please
d1i1m1o1n [39]

Answer:

Tell him u like him and be respect his idea of what he say it’s a 50/50 chance he likes u. I really believe u can do it.

Explanation:

5 0
3 years ago
How many moles are in 234.4 grams of oxygen
uysha [10]

Answer:

7.325 mol.

Explanation:

use the formula n=m/mr

so that makes:

234.4/16*2

which is 7.325 mol.

4 0
3 years ago
Sulfuric acid dissolves aluminum metal according to the following reaction:
Fiesta28 [93]

Answer:

m_{H_2SO_4}=81.7gH_2SO_4

m_{H_2}=1.67gH_2

Explanation:

Hello,

Based on the given undergoing chemical reaction is is rewritten below:

2Al (s) + 3H_2SO_4 (aq)\rightarrow  Al _2(SO4)_3 (aq) + 3H_2 (g)

By stoichiometry we find the minimum mass of H2SO4 (in g) as shown below:

m_{H_2SO_4}=15.0gAl*\frac{1molAl}{27gAl}*\frac{3molH_2SO_4}{2molAl}*\frac{98gH_2SO_4}{1molH_2SO_4} \\m_{H_2SO_4}=81.7gH_2SO_4

Moreover, mass of H2 gas (in g) would be produced by the complete reaction of the aluminum block turns out:

m_{H_2}=15.0gAl*\frac{1molAl}{27gAl}*\frac{3molH_2}{2molAl}*\frac{2gH_2}{1molH_2} \\m_{H_2}=1.67gH_2

Best regards.

3 0
3 years ago
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