Answer:
First, we can test Solution 1. We know that Sodium Hydroxide is a strong base. If we test acids on blue litmus paper, they will turn red. If we test bases on red litmus paper, they will turn blue. So, you can test all the of the solutions- water, sodium hydroxide and hydrochloric acid with blue and red litmus paper. HCl, Hydrochloric acid is an acid, so it will turn blue litmus paper red. It will not turn red litmus blue. The acids will turn blue litmus paper red. The bases will turn red litmus paper blue. Only water is a neutral liquid, which will not turn blue litmus paper red or red litmus paper blue. It will not change the colour of it. Thus, if you test all the solutions with blue and red litmus paper, you will know which solution is water. Water is the only one which is neutral. It is the only solution which cannot change the colour of any litmus paper. Thus, you can identify it very easily.
B. 
will decrease.
<h3>Explanation</h3>
The increase in temperature disturbs the equilibrium. The Le Chatelier's Principle suggests that the system would respond in a way that minimizes the degree of the impact.
There's an increase in temperature. The system will try to reduce its temperature. It will favor the backward reaction, which is <em>endothermic </em>and removes heat from the system. Doing so will convert some of its <em>products</em> to <em>reactants</em>.
The of an equilibrium is a fraction of multiple concentrations. For the reversible reaction a A + b B ⇄ c C + d D, ![K_{eq} = \frac{[C]^{c}\cdot [D]^d}{[A]^{a} \cdot [B]^{b}}](https://tex.z-dn.net/?f=K_%7Beq%7D%20%3D%20%5Cfrac%7B%5BC%5D%5E%7Bc%7D%5Ccdot%20%5BD%5D%5Ed%7D%7B%5BA%5D%5E%7Ba%7D%20%5Ccdot%20%5BB%5D%5E%7Bb%7D%7D)
, where [A], [B], [C], and [D] are equilibrium concentrations of the four species. Converting some products back to reactants increases [A] and [B] while reducing [C] and [D]. As a result, will decrease.
Answer:
57,120 J.
Explanation:
This = mass * specific * increase in temperature
= 680 * 0.50 * (190 - 22)
= 340 * 168
= 57,120 J.
Answer:
I think the answer is A. The distance is to great.
Answer:
Q=1170J/s = 279.516cal/s
Explanation:
<em>q=-k.(T₀-T₁)/(x₀- x₁)</em>
q= heat flow density
T= temperature
x= thickness
q= -(1.04 W/m. ºC) . (50ºC-500ºC)/(0.1m-0m)
q=4680W/m²=4,68KW/m²
<u>1KW=1000J/s</u>
∴ q= 4680 J/s.m²
q= Q/A
Q=heat fluw
A= surface= 6 . (0.5m . 0.5m)= 0.25m²
Q= (4680J/s.m²) . (0.25m²)
Q=1170J/s = 279.516cal/s
<u>1J=0.2389cal</u>
