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kherson [118]
2 years ago
10

If an atom has 35 protons in the nucleus, how many electrons will it have orbiting the nucleus?

Chemistry
1 answer:
stepan [7]2 years ago
5 0
It will have 35 electrons
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reativity of alkali metals increases down the group while reativity decreases down the group in helogens ​
BigorU [14]

Complete question is;

Chemical reactivity of alkali metals increases down the group while reactivity of halogens decreases down the group. Give reasons

Answer:

Explained below

Explanation:

Alkali metals exhibit reactivity due to their electropositivity. Now, for alkalis, their electro-positivity increases down their group. Since their reactivity increases with increase in electropositivity, it means their reactivity also increases down the group.

Whereas, the reactivity of halogens occurs as a result of their electronegativity. Now, electronegativity for halogens decreases down the group. Since their reactivity decreases with decrease in electronegativity, it means that their reactivity will also decrease down the group.

4 0
3 years ago
3. Which of the following sentences apply to neutrons? a. have a negative charge. b. circle around the atom nucleus. c. are the
DerKrebs [107]

Answer:

d.have a mass of 1 amu..

Explanation:

Does not match a as the answer for a is electrons B.electron circle around the nucleus because neutrons are inside the nucleus.C.proton and neutrons make the nucleus so the only answer left is .d .which is correct

3 0
2 years ago
Which element would be the most reactive with K? Br, Na, Ar, or O? And how do you know?
kykrilka [37]
Ar I did this I think i got it right on edunuty
5 0
2 years ago
A tank contains 90 kg of salt and 2000 L of water. Pure water enters a tank at the rate 6 L/min. The solution is mixed and drain
sashaice [31]

Answer:

a) 90 kg

b) 68.4 kg

c) 0 kg/L

Explanation:

Mass balance:

-w=\frac{dm}{dt}

w is the mass flow

m is the mass of salt

-v*C=\frac{dm}{dt}

v is the volume flow

C is the concentration

C=\frac{m}{V+(6-3)*L/min*t}

-v*\frac{m}{V+(6-3)*L/min*t}=\frac{dm}{dt}

-3*L/min*\frac{m}{2000L+(3)*L/min*t}=\frac{dm}{dt}

-3*L/min*\frac{dt}{2000L+(3)*L/min*t}=\frac{dm}{m}

-3*L/min*\int_{0}^{t}\frac{dt}{2000L+(3)*L/min*t}=\int_{90kg}^{m}\frac{dm}{m}

-[ln(2000L+3*L/min*t)-ln(2000L)]=ln(m)-ln(90kg)

-ln[(2000L+3*L/min*t)/2000L]=ln(m/90kg)

m=90kg*[2000L/(2000L+3*L/min*t)]

a) Initially: t=0

m=90kg*[2000L/(2000L+3*L/min*0)]=90kg

b) t=210 min (3.5 hr)

m=90kg*[2000L/(2000L+3*L/min*210min)]=68.4kg

c) If time trends to infinity the  division trends to 0 and, therefore, m trends to 0. So, the concentration at infinit time is 0 kg/L.

6 0
3 years ago
For an ideal gas, evaluate the volume occupied by 0.3 mole of gas.
Serhud [2]

Answer:

V=0.3×22.4=6.72 liters hope this helps

8 0
2 years ago
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