Answer:
ions are deflected and the lighter the ions the more the deflection.
Placing magnetic field in the path of the ions cause them to move in a curved path.
Explanation:
This question is all about the way in which mass spectroscopy works. Mass spectroscopy Is one of the techniques in spectroscopy which is used in the identification of chemical compounds.
Mass spectroscopy works based on the principle of ionization. For a mass spectroscopy to start ionization must first occur that is to say this is the first step in the identification of compound in mass spectroscopy.
The following steps are involve in mass spectroscopy;
=> Ionization: the molecules of the sample are first ionized. The ions formed here are positive ions.
=> Acceleration: the ions in step one are accelerated.
=> Deflection: the smaller ions get deflected more than the bigger ions. The magnetic field is used in the deflection of this ions.
=> Detection: the ions are then detected.
Therefore, to answer the question, the ions are deflected and the lighter the ions the more the deflection.
Placing magnetic field in the path of the ions cause them to move in a curved path.
A silver nitrate solution contains 14.77 g of primary standard AGNO3 ( Molecular weight 169.87) in 1.00 L. What volume of this solution will be needed to react with 0.2631 g of NaCl ( Molecular weight 58.44) ?
Answer:
[tex]V=5.2 mL=0.052L[/tex]
Explanation:
Hello!
In this case, since the chemical reaction between silver nitrate and sodium chloride is:
[tex]AgNO_3(aq)+NaCl(aq)\rightarrow AgCl(s)+NaNO_3(aq)[/tex]
We can see there is a 1:1 mole ratio between each solution; thus, we first compute the moles of each reactant considering their molar masses:
[tex]n_{AgNO_3}=14.77g*\frac{1mol}{169.87g}=0.087molAgNO_3\\\\ n_{NaCl}=0.2631g*\frac{1mol}{58.44}=0.0045molNaCl[/tex]
Now, since the concentration of the silver chloride solution is 0.087 M, we may assume that the concentration of the NaCl solution is the same, so we can compute the volume as shown below:
[tex]V=\frac{n_{NaCl}}{M}=\frac{0.0045mol}{0.087mol/L}\\\\V=0.052L[/tex]
Or:
[tex]V=5.2 mL[/tex]
Best regards!
The volume of solution needed to react with 0.2631 g of NaCl is 0.052 L.
How we calculate the volume?Volume of the solution will be calculated by using the below formula:
M = n/V, where
M = concentration in terms of molarity
n = no. of moles
V = volume
Given chemical reaction is:
AgNO₃(aq) + NaCl(aq) → AgCl(s) + NaNO₃(aq)
First we calculate the moles of given reactants by using the formula:
n = W/M , where
W = given mass
M = molar mass
Moles of AgNO₃ = 14.77g / 169.87g/mole = 0.087 mole
Moles of NaCl = 0.2631g / 58.44g/mole = 0.0045 mole
Concentration of AgNO₃ = 0.087 mole / 1L = 0.087M
From the stoichiometry of the reaction it is clear that mole ration of AgNO₃ & NaCl is 1:1. So, we take the concentration of NaCl is equal to the concentration of AgNO₃ and calculate the volume by using the above formula as:
Volume of NaCl = 0.0045mole / 0.087M = 0.052 L
Hence, 0.052 L is the required volume of NaCl.
To know more about moles, visit the below link:
https://brainly.com/question/17199947
Given the following balanced chemical equation:
N2(g) + 3H2(g) + 2NH3(8)
What is the maximum amount of NH3(g) that can be produced from 2.0 mol H2(g)? Assume that N2(g) is the excess reactant.
Answer:
22 g
Explanation:
Step 1: Write the balanced equation
N₂(g) + 3 H₂(g) ⇒ 2 NH₃(g)
Step 2: Calculate the moles of NH₃ produced from 2.0 moles of H₂
The molar ratio of H₂ to NH₃ is 3:2.
2.0 mol H₂ × 2 mol NH₃/3 mol H₂ = 1.3 mol NH₃
Step 3: Calculate the mass corresponding to 1.3 moles of NH₃
The molar mass of NH₃ is 17.03 g/mol.
1.3 mol × 17.03 g/mol = 22 g
After going through a guided tutorial by selecting Run Grams Demonstration, you can create your own experiment by clicking the Run Experiment button at the end or by clicking the Overview tab and returning to the Experiment tab to select Run Experiment. There are nine reactions you can explore on your own. Sulfur dioxide gas (SO2) and oxygen gas (O2) react to form the liquid product of sulfur trioxide (SO3). How much SO2 would you need to completely react with 6.00 g of O2 such that all reactants could be consumed
Answer: Thus 24.0 g of [tex]SO_2[/tex] would be needed.
Explanation:
To calculate the moles :
[tex]\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}[/tex]
[tex]\text{Moles of} O_2=\frac{6.00g}{32g/mol}=0.1875moles[/tex]
[tex]2SO_2(g)+O_2(g)\rightarrow 2SO_3(l)[/tex]
According to stoichiometry :
1 mole of [tex]O_2[/tex] require = 2 moles of [tex]SO_2[/tex]
Thus 0.1875 moles of [tex]O_2[/tex] will require=[tex]\frac{2}{1}\times 0.1875=0.375moles[/tex] of [tex]SO_2[/tex]
Mass of [tex]SO_2=moles\times {\text {Molar mass}}=0.375moles\times 64g/mol=24.0g[/tex]
Thus 24.0 g of [tex]SO_2[/tex] would be needed to completely react with 6.00 g of [tex]O_2[/tex] such that all reactants could be consumed.
How many grams are in 2.49 x 10^24 atoms of Hg?
Answer:
[tex]m_{Hg}=829.4gHg[/tex]
Explanation:
Hello there!
In this case, considering the Avogadro's number, which helps us to realize that 1 mole of mercury atoms contains 6.022x10²³ atoms and at the same time 1 mole of mercury weights 200.59 g, we obtain:
[tex]m_{Hg}=2.49x10^{24}atoms*\frac{1mol}{6.022x10^{23}atoms} *\frac{200.59g}{1molHg}\\\\m_{Hg}=829.4gHg[/tex]
Best regards!
(trying this again because this test is due at 3 and paimon has to get this done or she will fail!! paimon will try to give brainlist if brainly lets her!!)
The _________________molecules in the food we eat are_____________.
A. Enzyme, water
B. Small, insoluble
C. Large, insoluble
D. Enzyme, insoluble
Gold's natural state has a definite shape and a definite volume. What is gold's natural state(s)?
Answer:
If your asking what golds natural state of matter is it's solid.
Explanation:
Answer:
the answer is soild
Explanation:
i did it on edge :)
A continuously stirred tank reactor is designed to remove the hazardous waste compound trichloroethylene (TCE) from industrial wastewater. The tank is 3250 liters. The wastewater flows into the tank at 200 L/min with a TCE concentration of 25 mg/L. The reactor decays TCE at a reaction rate of 0.20 min-1 . What is the steady-state concentration of TCE in the treated water leaving the reactor
Answer:
the steady-state concentration of TCE in the treated water leaving the reactor is 5.88 mg/L
Explanation:
Given that;
Tank volume v = 3250 liters
wastewater flows into the tank Q = 200 L/min
TCE concentration Co= 25 mg/L
reactor decays TCE at a reaction rate K = 0.20 min-1
mass balance
we know that;
Accumulation = inflow - outflow ± generation
⇒dc/dt = QCo - Qc ± rc.V
now at a steady state; dc/dt = 0
so
0 = QCo - Qc + rcV
where rc = -kc
0 = QCo - Qc - kcV
Qc + kcV = QCo
c(Q + kV) = QCo
c = QCo / (Q + kV)
so we substitute
c = (200 × 25) / (200 + (0.2×3250))
c = 5000 / 850
c = 5.88 mg/L
Therefore, the steady-state concentration of TCE in the treated water leaving the reactor is 5.88 mg/L
In general chemicals enter Ecosystems through which two spears
Answer:
biosphere and lithosphere
Explanation:
The biosphere is described as the zone of life on Earth. It is a sum of all ecosystems. The biosphere is composed of living organisms and non-living factors.
The lithosphere is the outer part of the Earth such that this part is rocky. The lithosphere is made up of the brittle crust.
In general, chemicals enter Ecosystems through the biosphere and lithosphere.