Answer: Marcia made a more concentrated salt solution.
Explanation:
Marcia used 1 kg of salt, which the exercise tells us is 2.2 lb.
She dissolved it in 3L of water, so we have to find out how much that is in pints. We do that using cross multiplication.
If 1L equals 2.11 pints, then
xpints / 3L = 2.11pints / 1L, so
xpints = (3L . 2.11pints) / 1L = 6.33 pints
Now we know she dissolved 2.2lb in 6.33 pints of water. We use cross multiplication again to find out how much salt she would have in 10 pints, so we can know how concentrated it is compared to Bobby's solution, which is 10 pints.
xlb / 10pints = 2.2lb / 6.33pints
xlb = (10pints . 2.2lb) / 6.33pints = 3.47lb
So Marcia has a concentration of 3.47lb per 10 pints, whereas Bobby only has 3lb per 10 pints.
Answer:
a) [NH2]-,
b) [O]2-
c) [Cl]-
Explanation:
The conjugate base is the base member (X⁻) of a weak acid (HX). In other words, the conjugate base is the remaining substance due to the proton loss in the acid HX.
a) For NH₃, (HX; X: NH₂⁻), conjugate base is NH₂⁻. In the format, <em>[NH2]-</em>.
b) For OH⁻, (HX; X: O²⁻), conjugate base is O²⁻. In the format,<em> [O]2-</em>.
c) For HCl, (HX; X: Cl⁻), conjugate base is Cl⁻. In the format, <em>[Cl]-</em>.
Answer:
Final temperature = T₂ = 328.815 K
Explanation:
Given data:
Given energy = 980 KJ = 980×1000= 980000 J
Volume = 6.2 L
Initial temperature =T₁= 291 K
Specific heat of water = 4.18 j /g .K
Final temperature = T₂ = ?
Formula:
Q = m. c . ΔT
ΔT = T₂ - T₁
we will first convert the litter into milliliter
6.2 × 1000 = 6200 mL
It is given in question that
1 mL = 1 g
6200 mL = 6200 g
Now we will put the values in formula,
Q = m. c . (T₂ - T₁)
980000 j = 6200 g . 4.18 j /g .K . (T₂ - 291 K)
980000 j = 25916 j/ k . (T₂ - 291 K)
980000 j / 25916 j/ k = T₂ - 291 K
37.8145 K + 291 K =T₂
T₂ = 328.815 K
