Answer:
Will group 2 elements lose electrons to bond with nonmetals in group 17 in a
1:2 ratio?
Explanation:
Metals are electropositive in nature. This means that they loose electrons. Thus, metals form ionic bonds by loosing electrons to non metals.
Elements of group 2 have a valency of 2 while those of group 17 has a valency of 1 so the ratio in which group 2 elements bond with elements of group 17 is 1:2. Hence the answer.
The answer is 2 15 12 6
hope this would help you
Answer:
2.60 moles of A remaining.
Explanation:
According to Le Chatelier's principle, the equilibrium would shift if the volume, concentration, pressure, or temperature changes.
In this question, we were told that the volume doubles, that implies that we would have to double the molarity of B/ C (since B=C.)
However, it is obvious and clear from the given equation of the reaction that A is solid in it's activity = 1. Hence, it is then ignored.
So doubling B would be 1.30 M × 2 = 2.60 M
i.e 2.60 M moles of A was consumed.
Now; the number of moles of A remaining is 5.20 - 2.60 = 2.60 moles of A remaining.
Answer:
Hope this helps:)
Explanation:
Your exercise: Step by step:
1*225/88*r*a=1*(0/-1)*e+1*(__/88)*r*a | If the numerator equals zero, the fraction will be zero
1*225/88*a*r=1*0*e+1*(__/88)*r*a | Multiply 0 by e. Any number multiplied by 0 is 0.
1*225/88*a*r=0+1*__/88*r*a | Multiply __/88 by a. You can multiply a fraction by a number by multiplying the number and the numerator.
1*225/88*a*r=0+1*__*a*a*r | Multiply (__*a)/88 by r. You can multiply a fraction by a number by multiplying the number and the numerator.
1*225/88*a*r=0+1*(__*a*r)/88 | · 88
1*225*a*r=88*(__*a*r)/88 | Multiply 88 by (__*a*r)/88. You can multiply a fraction by a number by multiplying the number and the numerator.
225*a*r=88*(__*a*r)/88 | Cancel (__*88*a*r)/88 with 11*2^3
225*a*r=(__*a*r)/1 | Numerator 1 of (__*a*r)/1 can be omitted .
225*a*r=1*__*a*r | : 225
1*a*r=1*__*0.004*a*r | Divide both sides by a .
1*r=1*(__*0.004*a*r)/a
