Answer:

Explanation:
Hello,
In this case, the undergoing chemical reaction is:

In such a way, the mercury II sulfate (molar mass 296.65g/mol) is in a 1:1 molar ratio with the mercury II chloride (molar mass 271.52g/mol), for that reason the stoichiometry to find mass in grams of mercury II chloride turns out:

Best regards.
Answer:
NaOH(aq) + HNO3(aq)------>NaNO3(aq) + H2O(l)
Explanation:
A thing to note is that an acid and a base will react to form a metal salt + H2O.
~Hope it helps:).
Answer:
pH = -log₁₀ [H⁺]
Explanation:
pH is a value in chemistry used in to measure solution trying to determine each quality, purity, risks for health of some products, etc.
As you write in the question, [H⁺] = 10^(-pH)
Using logarithm law (log (m^(p) = p log(m):
log₁₀ [H⁺] = -pH
And
<h3>pH = -log₁₀ [H⁺]</h3>
Answer:
a) 90 kg
b) 68.4 kg
c) 0 kg/L
Explanation:
Mass balance:

w is the mass flow
m is the mass of salt

v is the volume flow
C is the concentration





![-[ln(2000L+3*L/min*t)-ln(2000L)]=ln(m)-ln(90kg)](https://tex.z-dn.net/?f=-%5Bln%282000L%2B3%2AL%2Fmin%2At%29-ln%282000L%29%5D%3Dln%28m%29-ln%2890kg%29)
![-ln[(2000L+3*L/min*t)/2000L]=ln(m/90kg)](https://tex.z-dn.net/?f=-ln%5B%282000L%2B3%2AL%2Fmin%2At%29%2F2000L%5D%3Dln%28m%2F90kg%29)
![m=90kg*[2000L/(2000L+3*L/min*t)]](https://tex.z-dn.net/?f=m%3D90kg%2A%5B2000L%2F%282000L%2B3%2AL%2Fmin%2At%29%5D)
a) Initially: t=0
![m=90kg*[2000L/(2000L+3*L/min*0)]=90kg](https://tex.z-dn.net/?f=m%3D90kg%2A%5B2000L%2F%282000L%2B3%2AL%2Fmin%2A0%29%5D%3D90kg)
b) t=210 min (3.5 hr)
![m=90kg*[2000L/(2000L+3*L/min*210min)]=68.4kg](https://tex.z-dn.net/?f=m%3D90kg%2A%5B2000L%2F%282000L%2B3%2AL%2Fmin%2A210min%29%5D%3D68.4kg)
c) If time trends to infinity the division trends to 0 and, therefore, m trends to 0. So, the concentration at infinit time is 0 kg/L.
1. D)
2. I think the correct answer from the choices listed above is option C. The tools that <span>should
be used to record the most complete data about a gas are a manometer
and a thermometer. Pressure and temperature are important measurements
for a gas since from these data we can calculate any other properties of
the gas.</span>