C is the answer I used to work at McDonald’s I know military time
PH is logarithmically related to the concentration of H+ ions. This will be the formula that will be used for this problem. We calculate as follows:
HCN = H+ + CN-
1.85 g HCN ( 1 mol / 27.02 ) ( 1 mol H+ / 1 mol HCN ) = 0.068 mol H+
Concentration = 0.068 / 3.50 = 0.02 M
pH = -log 0.02
pH = 1.71
Hello!
We have the following data:
Area (
A) = 50 square feet
Mass (
m) = 8.5 ounces
Density (
d) = 2.70 g/cm³
Volume (
V) = ?
Thickness (
T) =? (in mm)
To move on, we must transform the area of 50 ft² in cm², let's see:
1 ft² ------- 929,0304 cm²
50 ft² -----
A![A = 50*929,0304](https://tex.z-dn.net/?f=A%20%3D%2050%2A929%2C0304)
![\boxed{A = 46451,52\:cm^2}\Longleftarrow(Area)](https://tex.z-dn.net/?f=%5Cboxed%7BA%20%3D%2046451%2C52%5C%3Acm%5E2%7D%5CLongleftarrow%28Area%29)
In the same way, we will convert the mass of 8.5 oz in grams, see:
1 oz -------- 28,3495 g
8,5 oz -------
m![m = 8,5*28,3495](https://tex.z-dn.net/?f=m%20%3D%208%2C5%2A28%2C3495%20)
![\boxed{m = 240,97075\:g}\Longleftarrow(mass)](https://tex.z-dn.net/?f=%5Cboxed%7Bm%20%3D%20240%2C97075%5C%3Ag%7D%5CLongleftarrow%28mass%29)
Knowing that the density is 2.70 g/cm³ and the mass is 240.97075 g, we will find the volume, applying the data in the density formula we have:
![d = \dfrac{m}{V} \to V = \dfrac{m}{d}](https://tex.z-dn.net/?f=d%20%3D%20%20%5Cdfrac%7Bm%7D%7BV%7D%20%5Cto%20V%20%3D%20%20%5Cdfrac%7Bm%7D%7Bd%7D%20)
![V = \dfrac{240,97075\:\diagup\!\!\!\!\!g}{2,70\:\diagup\!\!\!\!\!g/cm^3}](https://tex.z-dn.net/?f=V%20%3D%20%20%5Cdfrac%7B240%2C97075%5C%3A%5Cdiagup%5C%21%5C%21%5C%21%5C%21%5C%21g%7D%7B2%2C70%5C%3A%5Cdiagup%5C%21%5C%21%5C%21%5C%21%5C%21g%2Fcm%5E3%7D%20)
![V = 89,24842593... \to \boxed{V \approx 89,25\:cm^3}](https://tex.z-dn.net/?f=V%20%3D%2089%2C24842593...%20%5Cto%20%5Cboxed%7BV%20%5Capprox%2089%2C25%5C%3Acm%5E3%7D)
The statement wants to find the thickness of the packaging, for this we have some important data, such as: V (volume) = 89,25 cm³ and Area (A) = 46451,52 cm² and T (thickness) =? (in mm)
In the calculations of Costs in Surface Treatment of a part within the flat geometry, we will use the following formula:
![V (volume) = A (Area) * T (Thickness)](https://tex.z-dn.net/?f=V%20%28volume%29%20%3D%20A%20%28Area%29%20%2A%20T%20%28Thickness%29)
![89,25\:cm^3 = 46451,52\:cm^2\:*\:T](https://tex.z-dn.net/?f=89%2C25%5C%3Acm%5E3%20%3D%2046451%2C52%5C%3Acm%5E2%5C%3A%2A%5C%3AT)
![46451,52\:cm^2*T = 89,25\:cm^3](https://tex.z-dn.net/?f=46451%2C52%5C%3Acm%5E2%2AT%20%3D%2089%2C25%5C%3Acm%5E3)
![T = \dfrac{89,25\:cm^3}{46451,52\:cm^2}](https://tex.z-dn.net/?f=T%20%3D%20%20%5Cdfrac%7B89%2C25%5C%3Acm%5E3%7D%7B46451%2C52%5C%3Acm%5E2%7D%20)
![T = 0,001921358009...\:cm](https://tex.z-dn.net/?f=T%20%3D%200%2C001921358009...%5C%3Acm)
We will convert to millimeters, going through a decimal place on the right
![T = 0,01921358009..\:mm](https://tex.z-dn.net/?f=T%20%3D%200%2C01921358009..%5C%3Amm%20)
Hope this helps! :))
Answer:
55.5g of N₂
Explanation:
It is possible to find ΔH of a reaction by the sum of half-reaction (Hess's law), thus:
<em>(1) </em>4 NH₃ + 3 O₂ → 2 N₂ + 6 H₂O ΔH= -1556KJ
<em>(2) </em>N₂O + H₂ → N₂ + H₂O ΔH= -389.4 KJ
<em>(3) </em>H₂+ ½ O₂ → H₂O ΔH= -223.9 KJ
(4) = 1/2 (1) + 3 (2):
<em>(4) </em>2 NH₃ + 3 N₂O +3 H₂ + ³/₂ O₂ → 4 N₂ + 6 H₂O
ΔH = 1/2 (-1556kJ) + 3 (-389.4kJ) = -1946.2kJ
(4) - 3(3):
2 NH₃ + 3 N₂O → 4 N₂ + 3 H₂O ΔH = -1946.2kJ - 3 (-223.9 KJ)
ΔH = -1274.5 kJ
As when 5 moles of nitrogen are produced there are released 1274.5kJ, when are released 505kJ:
505kJ × (5 moles N₂ / 1274.5kJ) =<em> 1.98 moles of N₂. </em>In mass:
1.98moles N₂ ₓ (28g / 1mol) = <em>55.5g of N₂</em>
Answer:
The final volume should be 22 mL
Explanation:
For this problem, we will use the dilution equation:
C1*V1 = C2*V2
<u>Step 1</u>: Data given
with C1 = the initial concentration C1 = 0.220 mg/L
with V1 = the initial volume = 10 mL = 10 * 10^-3 L
with C2 = the final concentration = 0.100 mg/L
with V2 = the final volume = TO BE DETERMINED
<u>Step 2</u>: Calculating the final volume
C1*V1 = C2*V2
0.220 mg/L * 10*10^-3 L = 0.100 mg/L * V2
V2 = (0.220 mg/L * 10*10^-3 L) / 0.100 mg/L
V2 =0.022 L = 22 mL
The final volume should be 22 mL