Which of the following metals (M) will form an ionic compound with nitrogen with the general formula M3N2?
Answer:
There are no options provided dude,
But i guess the answer will be a metal with valency 2 for sure as the subscript given for N in 'M3N2' is 2 so the valency of the metal u need to select will be 2 for sure
It can be magnesium or some other if the provided options in real question has Mg then its the answer
Which organism would most likely belong to the plant kingdom?
Answer:
tree
Explanation:
so easy and obvious
Some antacid tables contain aluminum hydroxide. The aluminum hydroxide reacts with stomach acid according to the equation: Al(OH)3 + 3HCl →AlCl3 + 3H2O. Determine the moles of stomach acid (HCl) required if a tablet contains 8.89 moles of Al(OH)3.
Answer:
26.67 mol HCl
Explanation:
Al(OH)₃ + 3HCl → AlCl₃ + 3H₂O
In order to solve this problem, we need to convert Al(OH)₃ moles to HCl moles.
To do so we use the stoichiometric ratios of the balanced reaction:
8.89 mol Al(OH)₃ * [tex]\frac{3molHCl}{1molAl(OH)_{3}}[/tex] = 26.67 mol HClThus 26.67 moles of HCl would react completely with 8.89 moles of Al(OH)₃.
The following aqueous solutions of are mixed: 100.0 mL of 1.00 M lithium bromide and 100.0 mL of 1.00 M lead (II) nitrate. In an organized and clear manner, show all of your work and answers for this problem on the uploaded work. (a) Write the molecular equation for this reaction. (b) Write the total ionic chemical equation for this reaction. (c) Write the net ionic chemical equation for this reaction. (d) Identify the spectator ions in this reaction. (e) Identify the limiting reactant. (f) Determine the mass of solid product that is formed?
Answer and Explanation:
(a) When lithium bromide (LiBr) solution is mixed with a solution of lead (II) nitrate (Pb(NO₃)₂), lithium nitrate (LiNO₃) and lead (II) bromide (PbBr₂) are formed, according to the following molecular equation:
2LiBr(aq) + Pb(NO₃)₂(aq) → 2LiNO₃(aq) + PbBr₂(s) ↓
As the product PbBr₂ is an insoluble solid, it precipitates (↓).
(b) The total ionic equation is written with all ions of the reaction - no matter if they participate in the precipitate formation or not:
2Li⁺(aq) + 2Br⁻(aq) + Pb²⁺(aq) + 2NO₃⁻(aq) → 2Li⁺(aq) + 2NO₃⁻(aq) + PbBr₂(s)
(c) The net ionic equation is written including only the ions which participate in the precipitate formation. In this case, the precipitate is PbBr₂, and it is formed by Pb²⁺ and Br⁻ ions:
Pb²⁺(aq) + 2Br⁻(aq) → PbBr₂(s)
(d) The spectator ions are those which do not participate in the formation of the precipitate. From the total ionic equation, we can see that Li⁺ and NO₃⁻ ions are repeated on both sides of the equation, so they are redundant. Thus, the spectator ions are Li⁺ and NO₃⁻ ions.
(e) To identify the limiting reactant, we first calculate the moles of each compound as the product of the solution concentration and volume:
For LiBr:
C = 1.00 M = 1 mol/L
V = 100.0 mL x 1 L/1000 mL = 0.1 L
moles of LiBr = 0.1 L x 1.00 mol/L = 0.1 mol
The same for Pb(NO₃)₂:
C = 1.00 M = 1 mol/L
V = 100.0 mL x 1 L/1000 mL = 0.1 L
moles of Pb(NO₃)₂ = 0.1 L x 1.00 mol/L = 0.1 mol
From the total ionic equation, we can see that 2 mol of LiBr reacts with 1 mol of Pb(NO₃)₂ to give 1 mol of PbBr₂ (solid product). The stoichiometric molar ratio is 2 mol LiBr/1 mol Pb(NO₃)₂ and we have 0.1 mol of each reactant (0.1 mol LiBr/0.1 mol Pb(NO₃)₂= 1). As 2 mol LiBr/mol Pb(NO₃)₂ > 1 mol LiBr/mol Pb(NO₃)₂, LiBr is the limiting reactant.
(f) From the total ionic equation, we know that 2 moles of LiBr produce 1 mol of PbBr₂. To determine the mass of solid product (PbBr₂) formed, we first multiply the stoichiometric ratio (1 mol PbBr₂/2 mol LiBr) by the actual number of moles of limiting reactant we have (0.1 mol):
moles of PbBr₂ = 0.1 mol LiBr x (1 mol PbBr₂/2 mol LiBr) = 0.05 mol PbBr₂
Finally, we convert the moles of PbBr₂ to gram by using the molar mass of the compound:
Molar mass PbBr₂ = 207.2 g/mol + (2 x 79.9 g/mol) = 367 g/mol
grams of PbBr₂ = 0.05 mol x 367 g/mol = 18.35 g
If 1000g of water from Luboc River contains 0.0075g of Ca2+ ion, what is the concentration in ppm by mass of Ca2+ present in Luboc River?
7.5 ppm
Further explanationGiven
1000 g of water
0.0075g of Ca²⁺ ion
Required
the concentration in ppm by mass of Ca²⁺
Solution
ppm = part per million
solvent = water ⇒ ppm = 1 mg/L(water density is 1 kg / L) or mg/kg
Convert g to mg of Ca²⁺ ion :
0.0075 g = 7.5 x 10⁻³ g = 7.5 mg
Convert g to kg of water :
1000 g = 1 kg water
So the concentration of Ca²⁺ ion :
= 7.5 mg / 1 kg
= 7.5 ppm
What is the ratio
amount (mol) Fatoms
amount (mol) Xe atoms
Enter your answer as an integer.
Pls help
Answer:
Empirical formula
Explanation:
The empirical formula of a compound is the simplest whole number ratio of atoms of each element in the compound. It is determined using data from experiments and therefore empirical.
if you want to know the identity of an unknown substance, what type of formula would you most likely want to know?
Answer:
Smell: Most chemists can identify solvents by their distinctive smells (though this is a pretty bad idea). ¹
Melting point: If you've got very pure crystals, you can use their melting point to figure out which of several possible chemicals you've got.
Explanation:
I hope this helps and pls mark me brainliest :)
Answer:
the answer to the question is smell
please help. im freaking out rn. i have like 40 missing assignments please
Answer:
I'm pretty sure its the one that says very little at the beginning but if I get it wrong I'm sorry
Please explain to me!!!
Answer:
nice handwrtting
Explanation:
Can anyone help me? Plsss
18. What is one of the three things that cause the surface currents of the oceans?
A.differences in salinity
B.temperature differences
C. density differences
D. Coriolis effect
Answer:
b. temperature difference
Pls give a detailed explanation about what are enzyme mutations
Answer:
Enzyme mutations can lead to serious or fatal human disorders and are the consequence of inherited abnormalities in the affected individual's DNA. The mutation may be at a specific position in an enzyme encoded by a mutated gene, just like a single abnormal amino acid residue.
Explanation:
Determine the chemical equation for the reaction between Fe3+ and Cu2+ with NH3
Answer:
Cu2 + 4 NH3 = Cu (NH3) 42+
so - - >
[Cu (NH3) 6] ^ 2+
Explanation:
I need help with this!!!
Answer:
0.73g/cm^3
Explanation:
d=m/v
d=11/15
d=0.73
1. Which of the following is an inherited trait?
19. Which wind water interaction is responsible for cycling nutrient-rich waters
from the ocean floor to the sea surface?
A. upwelling
B.storm surge
C.downwelling
D.surface currents
Explanation:
ksiiaisisjjsjajajhahajajjalañLzlzl
Explanation: A . upwelling .
The formation of ethyl alcohol (C2H5OH) by the fermentation of glucose (C6H12O6) may be represent by the following: C6H12O6 --> 2 C2H5OH 2 CO2 If a particular glucose fermentation process is 70.0% efficient, how many grams of glucose would be required for the production of 51.0 g of ethyl alcohol (C2H5OH)
Answer:
142.5 g
Explanation:
According to the chemical reaction:
C₆H₁₂O₆ --> 2 C₂H₅OH + 2 CO₂
1 mol of glucose (C₆H₁₂O₆) forms 2 moles of ethyl alcohol (C₂H₅OH) and 2 moles of carbon dioxide (CO₂).
We first convert the moles to grams by using the molecular weight (Mw) of each compound:
Mw (C₆H₁₂O₆) = (12 g/mol x 6) + (1 g/mol x 12) + (16 g/mol x 6)= 180 g/mol
1 mol C₆H₁₂O₆ = 180 g/mol x 1 mol = 180 g
Mw(C₂H₅OH) = (12 g/mol x 2) + (1 g/mol x 5) + 16 g/mol + 1 g/mol= 46 g/mol
2 mol C₂H₅OH = 2 mol x 46 g/mol = 92 g
Thus, when the process is 100% efficient, 180 grams of glucose produce 92 grams of ethyl alcohol. To form 51.0 grams of ethyl alcohol, we will need:
51.0 g C₂H₅OH x (180 g C₆H₁₂O₆/92 g C₂H₅OH) = 99.8 g C₆H₁₂O₆
As the process has a lower efficiency (70.0%), we will need more glucose to obtain the required yield. So, we divide the mass of glucose required for a process 100% efficient by the actual efficiency:
mass of glucose required = 99.8 g C₆H₁₂O₆/(70%) = 99.8 g C₆H₁₂O₆ x 100/70 = 142.5 g
Therefore, it would be required 142.5 grams of glucose to obtain 51.0 grams of ethyl alcohol.
Decide whether the element is a metal or nonmetal, if you can.
Element is a hard silvery-gray solid. Wires are fastened to each side of a 2 cm slab of it, and an ordinary household 9 V battery is hooked up so that it can feed electricity through the slab to an LED. But the LED stays dark
Answer:
The element is a nonmetal
Explanation:
Elements are broadly classified into metals and non metals. Metals conduct electricity while non metals do not conduct electricity.
If we look at this scenario described in the question, we can easily decipher that the element is a nonmetal because the LED stays dark. The LED should have been lit if electricity was passed through the element in question.
Hence, the element is a non metal.
Can someone help me with this
Answer:
wow!
5. C
6. B
7. B
8. A
Explanation:
Hi do you know this?
Answer:
2
Explanation
It seems logical
A 9.725-g gaseous mixture contains ethane () and propane (). Complete combustion to form carbon dioxide and water requires 1.115 moles of oxygen gas. Calculate the mass percent of ethane in the original mixture.
Answer:
% = 33.83%
Explanation:
To do a better understanding of this, we can treat the mixture of the combustion as two separate reactions, in that way, we can have an idea of what is happening and how to calculate the mass percent.
So the combustion reactions in this mixture are:
2C₂H₆ + 7O₂ ---------> 4CO₂ + 6H₂O
C₃H₈ + 5O₂ ----------> 3CO₂ + 4H₂O
Now that we have both reactions (And balanced) we can hace an idea of the mole ratio between every compound in the mix.
For practical purposes, let's call "a" the mass of ethane, and "b" the mass of propane. The innitial mix have 9.725 g, so this means that:
a + b = 9.725 g (1)
Now that we have this, we can write a relation between the moles of oxygen and the moles of the gases. If we have 1.115 moles of oxygen, and also know the mole ratio of oxygen to "a" and "b", so:
moles O₂ = moles a (moles O₂/moles a) + moles b (moles O₂/moles b) (2)
And we know that moles a and moles b are:
moles a = a / MW
moles b = b / MW
The MW of a is 30 g/mol and the MW of b is 44 g/mol
Replacing the given data we have:
1.115 moles O₂ = (a/30)(7 moles O₂/2 moles a) + (b/44)(5 moles O₂/1 mole b)
1.115 moles O₂ = (0.1167a) moles O₂ + (0.1136b) moles O₂
To keep solving this, we can use expression (1) to solve for b, and then, replace here and have only one equation with 1 incognite:
a + b = 9.725 g
b = 9.725 - a (3)
Replacing above we have:
1.115 = 0.1167a + 0.1136(9.725 - a)
1.115 = 0.1167a + (1.1048 - 0.1136a)
1.115 - 1.1048 = 0.1167a - 0.1136a
0.0102 = 0.0031a
a = 3.29 g
Now, that we have the mass of the ethane, we can calculate the mass percent:
% = (3.29 / 9.725) * 100
% = 33.83%Hope this helps
How many grams of KCl can be dissolved in 100 g of
water at 80°C?
Answer:
50
Explanation:
5000/100 = 50
Also what happens if you decrease NOCL
Answer:
el que paso del one jhaulio
Explanation:
How many atoms of Kr (Krypton) are in a balloon that contains 2.00 mol of Kr? (4)
Answer:
[tex]atoms= 1.204x10^{24}atoms[/tex]
Explanation:
Hello!
In this case, according to the Avogadro's number, it is possible to compute the atoms of Kr in 2.00 moles as shown below:
[tex]atoms=2.00mol*\frac{6.022x10^{23}atoms}{1mol} \\\\atoms= 1.204x10^{24}atoms[/tex]
Best regards!
which temperature do they need to perform the experiment for perovskite to conduct electric current?
A: T=1980°C
B: T>1980°
C: T>_ 1980°C
D:T< 1980°C
For C it's suppose to be an less than equal sign .
Answer:
Correct answer is D : T < 1980°C
Explanation:
Correct answer is D:T< 1980°C
Perovskite proton conductors belong to the class of high temperature proton conductors (HTPCs) , solid that conduct electricity by transporting H⁺ ions (protons ) at temperatures above ambient , typically 400 - 1000°C
So, we get T < 1980°C
If expending 3500 kcal is equal to a loss of 1.0 lb, how many days will it take Charles to lose 5.0 lb? Express your answer to two significant figures. Charles consumes 1800 kcal per day.
Answer:31 days
Explanation:
3500kcal/lb x 5lb =17500
1800-1230=570
17500/570=30.70
Round up =31
This question involves the concepts of energy, weight, and time.
It will take Charles "9.7 days" to lose 5.0 lb.
TIME TO LOSE WEIGHTSince the loss of 1.0 lb weight requires expending 3500 kcal energy. Therefore, by using the unitary method,
1.0 lb loss = 3500 kcal
(5)(1.0) lb loss = (5)(3500 kcal)
5.0 lb loss = 17500 kcal ----- eqn(1)
Now, the time required to consume 1800 kcal is 1 day. Hence, by the unitary method,
1800 kcal = 1 day
1 kcal = [tex]\frac{1}{1800}\ day[/tex]
17500 kcal = [tex]\frac{17500}{1800}[/tex] day
17500 kcal = 9.7 days ------ eqn(2)
Comparing eqn (1) and eqn (2)
5.0 lb loss = 9.7 days
Learn more about energy here:
https://brainly.com/question/1932868
Convert −319 ∘F to kelvins.
Answer:
it is 78.15 Kelvins
Explanation:
-319 Fahrenheit to 78.15 Kelvins
Write the equilibrium expression of each chemical equation.
2H2S(g) 2H2(g) + S2(g)
Answer:
[H2]2[S2][H2S]2Kc=[H2]2[S2][H2S]2
Explanation:
2H2S(g)⇋2H2(g)+S2(g)2H2S(g)⇋2H2(g)+S2(g)
The equilibrium constant expression in terms of concentrations is:
Kc=[H2]2[S2][H2S]2Kc=[H2]2[S2][H2S]2.
The equilibrium expression for the given reaction can be written in terms of equilibrium constant which is the ratio of power of molar concentration of the product to the product of power of molar concentration of the reactants.
What is equilibrium?Equilibrium is a state for a reversible reaction where, the rate of forward reaction is equal to the rate of backward reaction. The rate of a reaction is the rate of decrease in the concentration of reactants or the rate of increase in the concentration of the products.
The given reaction at equilibrium state is written as:
[tex]\rm 2H_{2}S (g)\leftrightharpoons 2H_{2} (g)+ S_{2}(g)[/tex]
The equilibrium constant Kb is ratio of power of molar concentration of the product to the product of power of molar concentration of the reactants.
[tex]Kb = \rm \frac{[H_{2}S]^{2}}{[H_{2}]^{2} [S_{2}]}[/tex]
The rate of the reaction will be r = Kb [H₂]² [S₂].
To find more on equilibrium constant, refer here:
https://brainly.com/question/15118952
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Scoring Scheme: 3-3-2-1 Part II. You considered the properties of two acid-base indicators, phenolphthalein and methyl orange. Many indicators are weak acids in water and establish the equilibrium: HIn(aq)(Color 1) H2O(l) H3O (aq) In-(aq)(Color 2). Indicators change color depending on whether they are in a protonated (HIn) or unprotonated (In-) form. What is the equilibrium expression for the phenolphthalein indicator in water and what colors are the protonated and unprotonated forms of the indicator
Answer:
Explanation:
Phenolphthalein is a protonated indicator and methyl orange is a basic indicator having hydroxyl ionisable part .
Phenolphthalein can be represented by the following formula
HPh which ionizes in water as follows
HPh + H₂O ⇄ H₃O⁺ + Ph⁻
( colourless ) ( pink )
In acidic solution it is in the form of protonated Hph form which is colourless
In basic medium , it ionises to give H₃O⁺ and unprotonated Ph⁻ whose colour is pink .
Please answer, this is due in 30 minutes
Answer:
0.591 g of magnesium phosphate is the theoretical yield.
Magnesium nitrate is the limiting reactant.
Explanation:
Hello!
In this case, since the balanced reaction turns out:
[tex]3Mg(NO_3)_2+2Na_3PO_4\rightarrow Mg_3(PO_4)_2+6NaNO_3[/tex]
Next, we compute the grams of magnesium phosphate yielded by each reactant, considering the present mole ratios and molar masses:
[tex]m_{Mg_3(PO_4)_2}^{by\ Mg(NO_3)_2}=1.00gMg(NO_3)_2*\frac{1molMg(NO_3)_2}{148.31gMg(NO_3)_2}*\frac{1molMg_3(PO_4)_2}{3molMg(NO_3)_2} *\frac{262.86gMg_3(PO_4)_2}{1molMg_3(PO_4)_2} \\\\m_{Mg_3(PO_4)_2}^{by\ Mg(NO_3)_2}= 0.591gMg_3(PO_4)_2\\\\m_{Mg_3(PO_4)_2}^{by\ Na_3PO_4}=1.00gNa_3PO_4*\frac{1molNa_3PO_4}{163.94gNa_3PO_4}*\frac{1molMg_3(PO_4)_2}{2molNa_3PO_4} *\frac{262.86gMg_3(PO_4)_2}{1molMg_3(PO_4)_2} \\\\m_{Mg_3(PO_4)_2}^{by\ Na_3PO_4} = 0.802gMg_3(PO_4)_2[/tex]
Thus, we infer that the correct theoretical yielded mass is 0.591 g as magnesium nitrate is the limiting reactant for which it produces the fewest grams of product.
However, is not possible to compute the percent yield since no actual yield is given, and must be provided or indicated by the problem or an experiment and it not here, nevertheless, you may compute the percent yield by dividing the actual yield by the theoretical and then multiplying by 100:
[tex]Y=\frac{actual}{0.591g}*100\%[/tex]
Best regards!