Answer:
what is the net ionic equation
H2SO4(aq) + Cal2(aq) → CaSO4(s) + 2Hl(aq)?
A. H++ SO42- + Ca2+ + 21 → CaSO4 + H+ +1-
B. 2H+ + S042- + Ca2+ + 21° → Ca2+ + SO42- + 2H+ + 21
C. S042- + Ca2+ → CaSO4,
D. 2H+ + SO42- + Ca2+ + 2I- → CaSO4 + 2H+ + 2I-
cancel the spectator ion that is the ions which does not take place in the reaction
for this case is 2 H^+ and 2 i^-
Answer:
3.336.
Explanation:
<em>Herein, the no. of millimoles of the acid (HCOOH) is more than that of the base (NaOH).</em>
<em />
So, <em>concentration of excess acid = [(NV)acid - (NV)base]/V total</em> = [(30.0 mL)(0.1 M) - (29.3 mL)(0.1 M)]/(59.3 mL) = <em>1.18 x 10⁻³ M.</em>
<em></em>
<em> For weak acids; [H⁺] = √Ka.C</em> = √(1.8 x 10⁻⁴)(1.18 x 10⁻³ M) = <em>4.61 x 10⁻⁴ M.</em>
∵ pH = - log[H⁺].
<em>∴ pH = - log(4.61 x 10⁻⁴) = 3.336.</em>
Molar mass Na = 23.0 g/mol
1 mol ---- 23.0 g
n mol ---- 69 g
n = 69 / 23.0
n = 3.0 moles
1 mole -------- 6.02x10²³ molecules
3.0 moles ---- ?
3.0 * 6.02x10²³ / 1
= 1.806x10²⁴ molecules
hope this helps!
nuclear power--used to turn turbines...
fossil fuels--burned to provide energy that is....
renewable energy--energy that with come back after use
outlet--a device....
steam--nuclear reactors....
I'm not sure but I tried lol,lemme know if I'm wrong :D
The 7160 cal energy is required to melt 10. 0 g of ice at 0. 0°C, warm it to 100. 0°C and completely vaporize the sample.
Calculation,
Given data,
Mass of the ice = 10 g
Temperature of ice = 0. 0°C
- The ice at 0. 0°C is to be converted into water at 0. 0°C
Heat required at this stage = mas of the ice ×latent heat of fusion of ice
Heat required at this stage = 10 g×80 = 800 cal
- The temperature of the water is to be increased from 0. 0°C to 100. 0°C
Heat required for this = mass of the ice×rise in temperature×specific heat of water
Heat required for this = 10 g×100× 1 = 1000 cal
- This water at 100. 0°C is to be converted into vapor.
Heat required for this = Mass of water× latent heat
Heat required for this = 10g ×536 =5360 cal
Total energy or heat required = sum of all heat = 800 +1000+ 5360 = 7160 cal
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