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viva [34]
3 years ago
9

An atom has a mass number of 24 and 13 neutrons. What is the atomic number of this atom? Express

Chemistry
1 answer:
Temka [501]3 years ago
8 0
A=mass number   (protons + neutrons)
Z=atomic number  (protons)
N=number of neutrons.

A=Z+N

Data:
A=24
N=13

Therefore:
24=Z+13
Z=24-13=11

Answer: The atomic number of this atom is 11. (The element is sodium)
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A 1.5 L sample of gas was collected at a pressure of 1.8 atm. What volume will the gas have if the pressure is reduced to 1.0 at
luda_lava [24]
P1V1=P2V2
(1.8)(1.5)=(1.0)V2
2.7=1.0 V2
2.7/1.0= 1.0/1.0V2
2.7=V2
6 0
2 years ago
The net ionic equation for formation of an aqueous solution of nii2 accompanied by evolution of co2 gas via mixing solid nico3 a
Ann [662]

Answer:

NiCO3 (s) + 2H+ (aq) → H2O (l) + CO2 (g) + Ni2+ (aq)

Explanation:

To write the complete ionic equation:

1. Start with a balanced molecular equation.

2. Break all soluble strong electrolytes (compounds with (aq) beside them) into their ions

3. indicate the correct formula and charge of each ion

4. indicate the correct number of each ion

5. write (aq) after each ion

6. Bring down all compounds with (s), (l), or (g) unchanged.

8 0
3 years ago
Which has a larger molar mass, salt (NaCl) or sugar (C12H22011) – how can you tell?
uysha [10]

Answer:

Explanation:

Molar mass of NaCl = (23+35.5)

Molar mass of NaCl = 58.5g/mol

Molar mass of C12H22011

= 12(12) + 22(1) + 16(11)

= 144 +22 + 176

= 342g/mol

4 0
3 years ago
Which of these are formed when
makvit [3.9K]
Option d lo siento si es incorrecto
7 0
2 years ago
The standard enthalpy of formation for glucose [c6h12o6(s)] is −1273.3 kj/mol. what is the correct formation equation correspond
balu736 [363]
The standard formation equation for glucose C6H12O6(s) that corresponds to the standard enthalpy of formation or enthalpy change ΔH°f = -1273.3 kJ/mol is 
     C(s) + H2(g) + O2(g) → C6H12O6(s)
and the balanced chemical equation is 
     6C(s) + 6H2(g) + 3O2(g) → C6H12O6(s)

Using the equation for the standard enthalpy change of formation 
     ΔHoreaction = ∑ΔHof(products)−∑ΔHof(Reactants)
     ΔHoreaction = ΔHfo[C6H12O6(s)] - {ΔHfo[C(s, graphite) + ΔHfo[H2(g)] + ΔHfo[O2(g)]}

C(s), H2(g), and O2(g) each have a standard enthalpy of formation equal to 0 since they are in their most stable forms:
     ΔHoreaction = [1*-1273.3] - [(6*0) + (6*0) + (3*0)]
                           = -1273.3 - (0 + 0 + 0)
                           = -1273.3
8 0
3 years ago
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