Answer:
plant cells and eukroyatic algae
Answer:
The amount of sodium is 32 mg per cracker, 49 mg per pretzel and 68 mg per cookie.
Explanation:
Let's assume amount of sodium is x mg per cracker, y mg per pretzel and z mg per cookie.
So, the following three equations can be written as per given information:
x+y+z = 149 ........(1)
8y+8z = 936 ........(2)
6x+7y = 535 .........(3)
From equation- (2), y+z =
= 117
By substituting the value of (y+z) in equation- (1) we get,
x = 149-(y+z) = 149-117 = 32
By substituting the value of x into equation- (3) we get,
y =
= 49
By substituting the value of y into equation- (2) we get,
z = (117-49) = 68
So, the amount of sodium is 32 mg per cracker, 49 mg per pretzel and 68 mg per cookie.
Answer:
190 °C
Step-by-step explanation:
The pressure is constant, so this looks like a case where we can use <em>Charles’ Law</em>:
V₁/T₁ = V₂/T₂ Invert both sides of the equation.
T₁/V₁ = T₂/V₂ Multiply each side by V₂
T₂ = T₁ × V₂/V₁
=====
V₁ = 3.75 L; T₁ = (37 + 273.15) K = 310.15 K
V₂ = 5.6 L; T₂ = ?
=====
T₂ = 310.15 × 5.6/3.75
T₂ = 310.15 × 1.49
T₂ = 463 K
t₂ = 463 – 273.15
t₂ = 190 °C
C(HClO) = 0,3 M.
<span>V(HClO) = 200 mL = 0,2 L.
n(HClO) = </span>c(HClO) · V(HClO).
n(HClO) = 0,06 mol.<span>
c(KClO</span>) =
0,2 M.
<span>V(KClO) = 0,3 L.
n(KClO) = 0,06 mol.
V(buffer solution) = 0,2 L + 0,3 L = 0,5 L.
ck</span>(HClO) = 0,06 mol ÷ 0,5 L = 0,12 M.
cs(KClO) = 0,06 mol ÷ 0,5 L = 0,12 M.<span>
Ka(HClO</span>) =
2,9·10⁻⁸.<span>
This is buffer solution, so use Henderson–Hasselbalch
equation:
pH = pKa + log(cs</span> ÷ ck).<span>
pH = -log(</span>2,9·10⁻⁸) + log(0,12 M ÷ 0,12 M).<span>
pH = 7,54 + 0.
pH = 7,54</span>