Answer:
Where the products are H2O and Ba(NO3)2
Explanation:
A base, as, barium hydroxide (Ba(OH)2) reacts with an acid (HNO3), producing water (H2O), and the related salt (Ba(NO3)2) in a reaction called neutralization reaction.
The balanced reaction is:
Ba(OH)2 + 2 HNO3 → 2 H2O + Ba(NO3)2
Where the products are H2O and Ba(NO3)2
There are _______ alkanes with molecular formula C10H22
a. 74
b. 75
c. 76
d. 77
How long do spent fuel rods remain dangerously radioactive?
Answers
A.
The rods are no longer radioactive because the radioisotopes are used up.
B.
Spent fuel rods remain radioactive for several years after the fuel is exhausted.
C.
It takes tens of thousands of years for the radioisotopes in the rods to decay to safe levels.
D.
It is impossible to determine how long it will take for the radioisotopes to decay because they last too long.
Answer:
c
Explanation:
it takes 10,000 years to just reduce down the decay
Which of the following metals will liberate hydrogen from dilute HCL? A. Ag B.Au C.Hg D.Sn
Answer:
ag and au are sure not to react. but hg and sn might or might not
Calculate the no. of moles in 15g of CaCl2
Answer:
[tex]\boxed {\boxed {\sf 0.14 \ mol \ CaCl_2}}[/tex]
Explanation:
We are asked to calculate the number of moles of 15 grams of calcium chloride (CaCl₂).
To convert from grams to moles, we use the molar mass, or the mass of 1 mole of a substance. Molar masses are found on the Periodic Table because they are equivalent to the atomic masses, but the units are grams per mole instead of atomic mass units.
Look up the individual elements in the compound: calcium and chloride.
Ca: 40.08 g/mol Cl: 35.45 g/molNotice the chemical formula has a subscript of 2 after Cl or chlorine. There are 2 moles of chlorine in every 1 mole of calcium chloride. We must multiply chlorine's molar mass by 2 before adding calcium's molar mass.
Cl₂: 35.45 * 2 = 70.9 g/mol CaCl₂= 40.08 + 70.9 = 110.98 g/molWe will convert using dimensional analysis, so we must create a ratio using the molar mass.
[tex]\frac {110.98 \ g \ CaCl_2}{ 1 \ mol \ CaCl_2}[/tex]
We are converting 15 grams of calcium chloride to moles, so we must multiply the ratio by this value.
[tex]15 \ g \ CaCl_2 *\frac {110.98 \ g \ CaCl_2}{ 1 \ mol \ CaCl_2}[/tex]
Flip the ratio so the units of grams of calcium chloride cancel.
[tex]15 \ g \ CaCl_2 *\frac { 1 \ mol \ CaCl_2}{110.98 \ g \ CaCl_2}[/tex]
[tex]15 *\frac { 1 \ mol \ CaCl_2}{110.98}[/tex]
[tex]\frac { 15}{110.98} \ mol \ CaCl_2[/tex]
[tex]0.1351594882\ mol \ CaCl_2[/tex]
The original measurement of grams (15) has 2 significant figures, so our answer must have the same. For the number we calculated, that is the hundredth place. The 5 in the thousandth place tells us to round the 3 up to a 4.
[tex]0.14 \ mol \ CaCl_2[/tex]
15 grams of calcium chloride is approximately 0.14 moles of calcium chloride.
Based on the reaction below:
[tex]N_2 + 3H_2[/tex] ↔ [tex]2NH_3 + heat[/tex]
If we decrease the temperature, equilibrium will shift towards the...
Please explain!
N₂ + 3H₂ ⇄ 2NH₃ + heat
In the given equilibrium, we notice that the heat is on the right. which means that if the heat requirements don't meet, the reactants on the right will no longer react due to the lack of heat
but because the reactants on the left don't have such weaknesses, they will keep reacting hence producing more and more ammonia until a new equilibrium is reached
where there will be more ammonia and less nitrogen and hydrogen as compared to the equilibrium we had initially
Answer:
Explanation:
heat is given out as 1 of the products, along w/ NH3 in the forward reaction. so its an exothermic reaction
decreasing temperature favors exothermic reaction as more heat can be absorbed by the environment
so equilibrium will shift towards the products
draw styrene
draw the structure of cyrene
In the reaction A + B + C + D, what are the reactants?
O A. Just B
B. Cand D
O c. A and B
O D. A and C
Answer:
C.
Explanation:
I believe that it should be A and B.
Please comment the chart!!!!!
thanks!!
Answer:
Exothermihic chart
Explanation:
a leaking tap drops water at the rate of 3 drops every second.each drop is approximately 1 ml. how many liters of water will leak from the tap during a day?
a. 5 liters
b. 50 liters
c.500 liters
d.15 liters
A leaking tap that drops water at the rate of 3 drops every second, will leak 259.2 L in a day.
We know that a leaking tap drops water at the rate of 3 drops every second and that each drop is approximately 1 ml. The milliliters of water dropped every second are:
[tex]\frac{3drop}{1s} \times \frac{1mL}{1drop} = \frac{3mL}{1s}[/tex]
We want to know the number of seconds in 1 day. We will use the following conversion factors:
1 day = 24 h1 h = 60 min1 min = 60 s[tex]1day \times \frac{24h}{1day} \times \frac{60min}{1h} \times \frac{60s}{1min} = 86400 s[/tex]
3 mL of water are dropped every second. The mL of water dropped in 86400 s are:
[tex]86400 s \times \frac{3mL}{1s} = 259200 mL[/tex]
Finally, we will convert mL to L using the conversion factor 1 L = 1000 mL.
[tex]259200 mL \times \frac{1L}{1000 mL} = 259.2 L[/tex]
Approximately 259.2 L of water will be dropped in 1 day.
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A tree is an example
of a vascular plant that
is
because it
has deep roots.
A. tall
B. tiny
C. small
Dyshort
Question 1 of 10
What is the correct orientation of the bar magnet, based on the magnetic field
lines shown?
A. North pole on the left end and south pole on the right end
B. South pole on the top edge and north pole on the bottom edge.
C. South pole on the left end and north pole on the right end
D. North pole on the top edge and south pole on the bottom edge
Answer:
d. north pole on the top edge and south pole on the bottom edge
Explanation:
What volume of 6.9 M NaOH is needed to completely titrate 0.42 L of 2.39 M phosphoric acid according to
the equation:
H3PO4(aq) + 3NaOH(aq) + Na3PO4(aq) + 3H2O(aq)
A) O 0.05 L
B) O6.93 L
C) O0.44 L
D) 03.01 L
E) 436.43 L
Taking into account the definition of molarity and the stoichiometry of the reaction, the correct option is option C) 0.44 L of 6.9 M NaOH is needed to completely titrate 0.42 L of 2.39 M phosphoric acid.
The balanced reaction is:
H₃PO₄ (aq) + 3 NaOH (aq) → Na₃PO₄ (aq) + 3 H₂O(aq)
Then, by stoichiometry of the reaction (that is, the relationship between the amount of reagents and products in a chemical reaction), the following amounts of moles of each compound participate in the reaction:
H₃PO₄: 1 mole NaOH: 3 moles Na₃PO₄: 1 mole H₂O: 3 molesMolarity is the number of moles of solute that are dissolved in a given volume.
Molarity is determined by:
[tex]Molarity=\frac{number of moles of solute}{volume}[/tex]
Molarity is expressed in units [tex]\frac{moles}{liter}[/tex].
In this case, 0.42 L of 2.39 M phosphoric acid reacts. So, by definition of molarity, the number of moles that participate in the reaction is calculated as:
[tex]2.39 \frac{moles}{liter}=\frac{number of moles of phosphiric acid}{0.42 liters}[/tex]
Solving:
number of moles of phosphiric acid= 2.39 [tex]\frac{moles}{liter}[/tex]* 0.42 liters
number of moles of phosphiric acid= 1.0038 moles ≅ 1 mole
Approaching 1 mole of the amount of phosphoric acid required, then by stoichiometry of the reaction, 3 moles of NaOH are necessary to react with 1 mole of the acid.
Then by definition of molarity and knowing that 6.9 M NaOH is needed, you can calculate the necessary volume amount of NaOH by:
[tex]6.9 \frac{moles}{liter} =\frac{3 moles}{volume}[/tex]
Solving:
6.9 [tex]\frac{moles}{liter}[/tex]* volume= 3 moles
[tex]volume=\frac{3 moles}{6.9\frac{moles}{liter} }[/tex]
volume= 0.44 L
The correct option is option C) 0.44 L of 6.9 M NaOH is needed to completely titrate 0.42 L of 2.39 M phosphoric acid.
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What is the molecule shown below?
A. Pentane
B. Trimethylethane
C. 2,2-dimethylpropane
D. 3-dipropane
Q2
Answer:
C
Explanation:
if we were to followw the IUPAC
how many moles of KF are present in 46.5 grams of KF
Explanation:
here's the answer to your question
Answer:
0.8017
Explanation:
Find the molar Mass of KF
K = 39
F = 19
Total = 58
Note: these numbers are approximate. Use your periodic table to get the exact numbers.
mols = given mass / molar mass
given mass = 46.5
molar mass = 58
mols = 46.5 / 58
mols = 0.8017
Carbonic acid (H₂CO₃) is a polyprotic acid. When carbonic acid dissolves in water, which is higher, the concentration of HCO₃- ions or the concentration of CO₃²- ions?
Please explain!
The concentration of CO₃²⁻ ions will be higher
To explain, I want you to imagine H₂CO₃ in water.
we know that it will lose 2 of it's protons, and form 2 ions
The ion which is more stable will have a higher concentration because that ion will refuse to react with anything else, so once something turns into that specific ion, it's not going back... unless there's a more stabler ion possible
In this case, the 2 ions formed are: HCO₃⁻ and CO₃⁽²⁻⁾, drawing the structures of both the ions tells us that both of them have resonance, but the CO₃⁽²⁻⁾ ion has more resonance structures and hence is more stable
Titanium is a metal often used as an alloying agent to provide materials that are strong, lightweight, and temperature-resistant Which of the following represents the correct ground-state configuration for a neutral atom of titanium?
A) 1s 2s 2p 3s 3p 48°30°
B) 1s 2s 2p 3s 3p 4s3d
C) 15*2s2p 3s 3p 4s
D) 15°2s 2p 3s 3p 3d
Answer:B) 1s 2s 2p 3s 3p 4s 3d
Explanation:
The ground state electron configuration shows how the electrons in the atomic orbitals of an atom are in their lowest , most stable energy arrangements and since Electrons must be filled following the Aufbau's principle(electrons fill lowest energy shells first)
Now, Titanium lies in period IV and group 4 of the periodic table with 22 as its atomic number
Thus, the ground-state electron configuration of a neutral atom of titanium is 1s²2s²2p⁶3s²3p⁶4s²3d².
The standard entropy change of a reaction has a positive value. This reaction results in: Select the correct answer below: a decrease in entropy. an increase in entropy. no entropy change. neither an entropy increase nor decrease.
Explanation:
The standard entropy change of a reaction has a positive value. This reaction results in an increase in entropy.
Positive entropy means the system has increased its degree of disorderness.
You find a clean 100-ml beaker, label it "#1", and place it on a tared electronic balance. You add small amount of unknown solid and place the
beaker with its contents on the balance. The recorded data is:
mass of the empty, clean beaker #1: 74.605 g
mass of the beaker #1 with the white solid: 74.896 g
Using the Law of Conservation of Mass, what is the mass of the unknown solid you placed in beaker #1?
Answer:
the mas is .291 g
Explanation:
the mass of a object does not change. so when added the substance the beaker. you had the mass of both objects together. you know the mass of the beaker and you know the mass of both. since mass does not change. the beakers mass is still 74.605g. the mass of both objects is 74.896. all you have to do is subtract the mass of the beaker from the total mass. 74.896 - 74.605 equals .291g. so the mass of the unknown substance Is .291g
According to Arrhenius, NH4+ an acid or a base? Write an equation to support
According to Arrhenius definition of acids, [tex]NH4^+[/tex] is an acid.
According to Arrhenius definition of acids and bases, acid is any substance that produces hydrogen ion in solution as its only positive ion.
Following this definition, let us now consider what happens when [tex]NH4^+[/tex] is introduced into a water;
[tex]NH4^+[/tex](aq)-------> NH3(aq) + [tex]H^+[/tex](aq)
Hence, according to Arrhenius definition of acids, [tex]NH4^+[/tex] is an acid.
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What mass of oxygen is needed for the complete combustion of 1.60-10^-3
g
of methane?
Express your answer with the appropriate units.
Answer:
6.4×10¯³ g of O₂.
Explanation:
We'll begin by writing the balanced equation for the reaction. This is given below:
CH₄ + 2O₂ —> CO₂ + 2H₂O
Next, we shall determine the masses of CH₄ and O₂ that reacted from the balanced equation. This can be obtained as follow:
Molar mass of CH₄ = 12 + (4×1)
= 12 + 4
= 16 g/mol
Mass of CH₄ from the balanced equation = 1 × 16 = 16 g
Molar mass of O₂ = 2 × 16 = 32 g/mol
Mass of O₂ from the balanced equation = 2 × 32 = 64 g
SUMMARY:
From the balanced equation above,
16 g of CH₄ reacted with 64 g of O₂.
Finally, we shall determine the mass of O₂ needed to react with 1.6×10¯³ g of CH₄. This can be obtained as illustrated below:
From the balanced equation above,
16 g of CH₄ reacted with 64 g of O₂.
Therefore, 1.6×10¯³ g of CH₄ will react with = (1.6×10¯³ × 64) / 16 = 6.4×10¯³ g of O₂
Thus, 6.4×10¯³ g of O₂ is needed for the reaction.
How many chromosomes do we not understand?
Answer:
we don't understand why humans have only 46 chromosomes
Answer:
46 chromosomes is what we don't understand
When determining the amount of oxidant present by titration, you can use iodine/starch as an indicator. First, the oxidant, like hypochlorite, oxidizes Choose... When starch and iodine are both present, the solution is Choose... During the titration, a titrant like thiosulfate reduces the
The question is incomplete, the complete question is;
When determining the amount of an oxidant present by titration, you can use iodine and starch as an indicator.
First, the oxidant, like hypochlorite, oxidizes
Choose...
neutral iodine into iodide ion
iodide ion into neutral iodine
iodate polyatomic ion into iodide ion
When starch and iodine are both present, the solution is
Choose...
blue-black
brownish yellow
clear
During the titration, the titrant, like thiosulfate, reduces the
Choose...
iodide ion into iodate polyatomic ion
neutral iodine into iodide ion
iodide ion into neutral iodine
When the iodine has completely reacted at the endpoint of the titration, the solution should become
Choose...
clear
blue-black
brownish yellow
Answer:
1. iodide ion into neutral iodine
2. blue-black
3. neutral iodine into iodide ion
4. clear
Explanation:
Hypochlorite oxidizes the iodide ion to iodine molecule according to the reaction equation;
ClO-(aq) + 2H+(aq) + 2I-(aq) ---------> 6 I2(l) + Cl- (aq)+ H2O(l)
When iodine is added, the colour of the starch solution immediately changes to blue-black.
A reduction reaction occurs when the titrant, thiosulfate is added as follows;
I2 + 2S2O32- → 2I- + S4O62-
The solution at end point is found to become clear again.
3 attempts left Be sure to answer all parts. Which indicators that would be suitable for each of the following titrations: (a) CH3NH2 with HBr thymol blue bromophenol blue methyl orange methyl red chlorophenol blue bromothymol blue cresol red phenolphthalein (b) HNO3 with NaOH thymol blue bromophenol blue methyl orange methyl red chlorophenol blue bromothymol blue cresol red phenolphthalein (c) HNO2 with KOH thymol blue bromophenol blue methyl orange methyl red chlorophenol blue bromothymol blue cresol red phenolphthalein
An indicator usually signals the endpoint of a neutralization reaction by undergoing a color change. They aid in discovering the point of equivalence of a titration.
The kind of indicator used depends on the nature of acid/base reacted.
In the case of CH3NH2 with HBr which strong acid and weak base titration, suitable indicators include; bromophenol blue, methyl orange, methyl red, and chlorophenol blue.
In the case of HNO3 with NaOH, this is a strong acid, strong base titration hence phenolphthalein, methyl red, chlorophenol, and bromothymol blue cresol red blue are suitable indicators.
In the case of HNO2 with KOH, this a weak acid, strong base titration and the suitable indicators are cresol red and phenolphthalein.
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Determine the number of water molecules in 0.2830g Na.
Answer:
7.38*10^21
Explanation:
2Na+2H20=2NaOH+H2
nNa=0.0123
number of water moles: 0.012*6*10^23=7.38*10^21
Ggggggggggggggggg666666666666666
LION
If 3.0L of helium at 20°C is allowed to expand to 4.4L, with pressure remain the same
Answer:
This question is asking to find the new temperature
The answer for the final temperature is 429.73K
Explanation:
Using Charles law equation as follows:
V1/T1 = V2/T2
Where;
V1 = initial volume (L)
V2 = final volume (L)
T1 = initial temperature (K)
T2 = final temperature (K)
According to this question;
V1 = 3.0L
V2 = 4.4L
T1 = 20°C = 20 +273 = 293K
T2 = ?
Using V1/T1 = V2/T2
3/293 = 4.4/T2
Cross multiply
293 × 4.4 = 3 × T2
1289.2 = 3T2
T2 = 1289.2 ÷ 3
T2 = 429.73K
Which of the following is used in EBRT?
O Silver tube
O Gold tube
O Copper tube
O Iron tube
Copper tube is used in EBRT.
What is meant by EBRT?External Beam Radiation. Therapy (EBRT) is a type of radiation therapy that directs a beam of radiation from outside the body, toward cancerous tissues inside the body.External beam radiation therapy (EBRT) is the most common type of radiation therapy. It directs high-energy radiation beams at the cancer.Copper tube is used in EBRT.
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Which of the following chemical reactions is reversible?
A. The neutralization of an acid.
B. The burning of wood.
C. The freezing of water into ice
D. The dehydration of copper sulfate (CuSO4).
The freezing of water into ice and the dehydration of copper sulfate are both reversible. The correct options are C and D.
What are reversible reactions?They are reactions in which the reverse can occur.
The freezing of water into ice can be undone. That is, the ice can be thawed back to water.
The dehydration of copper sulfate involves the removal of water molecules. As soon as water becomes available again, the reaction is reversed.
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write the chemistry of Epsom salt
Balance the following skeleton reaction and identify the oxidizing and reducing agents: Include the states of all reactants and products in your balanced equation. You do not need to include the states with the identities of the oxidizing and reducing agents.
NO_2(g) rightarrow NO_3^-(aq) +NO_2^- (aq) [basic]
The oxidizing agent is:______.
The reducing agent is:_______.
Answer:
a. 2NO₂ (g) + 2OH⁻ (aq) → NO₃⁻ (aq) + NO₂⁻ (aq) + H₂O (l)
b. i. NO₂⁻ is the oxidizing agent
ii. NO₃⁻ is the reducing agent.
Explanation:
a. Balance the following skeleton reaction
The reaction is
NO₂ (g) → NO₃⁻ (aq) + NO₂⁻ (aq)
The half reactions are
NO₂ (g) → NO₃⁻ (aq) (1) and
NO₂ (g) → NO₂⁻ (aq) (2)
We balance the number of oxygen atoms in equation(1) by adding one H₂O molecule to the left side.
So, NO₂ (g) + H₂O (l) → NO₃⁻ (aq)
We now add two hydrogen ions 2H⁺ on the right hand side to balance the number of hydrogen atoms
NO₂ (g) + H₂O (l) → NO₃⁻ (aq) + 2H⁺ (aq)
The charge on the left hand side is zero while the total charge on the right hand side is -1 + 2 = +1. To balance the charge on both sides, we add one electron to the right hand side.
So, NO₂ (g) + H₂O (l) → NO₃⁻ (aq) + 2H⁺ (aq) + e⁻ (4)
Since the number of atoms in equation two are balanced, we balance the charge since the charge on the left hand side is zero and that on the right hand side is -1. So, we add one electron to the left hand side.
So, NO₂ (g) + e⁻ → NO₂⁻ (aq) (5)
We now add equation (4) and (5)
So, NO₂ (g) + H₂O (l) → NO₃⁻ (aq) + 2H⁺ (aq) + e⁻ (4)
+ NO₂ (g) + e⁻ → NO₂⁻ (aq) (5)
2NO₂ (g) + H₂O (l) + e⁻ → NO₃⁻ (aq) + NO₂⁻ (aq) + 2H⁺ (aq) + e⁻ (4)
2NO₂ (g) + H₂O (l) → NO₃⁻ (aq) + NO₂⁻ (aq) + 2H⁺ (aq)
We now add two hydroxide ions to both sides of the equation.
So, 2NO₂ (g) + H₂O (l) + 2OH⁻ (aq) → NO₃⁻ (aq) + NO₂⁻ (aq) + 2H⁺ (aq) + 2OH⁻ (aq)
The hydrogen ion and the hydroxide ion become a water molecule
2NO₂ (g) + H₂O (l) + 2OH⁻ (aq) → NO₃⁻ (aq) + NO₂⁻ (aq) + 2H₂O (l)
2NO₂ (g) + 2OH⁻ (aq) → NO₃⁻ (aq) + NO₂⁻ (aq) + H₂O (l)
So, the required reaction is
2NO₂ (g) + 2OH⁻ (aq) → NO₃⁻ (aq) + NO₂⁻ (aq) + H₂O (l)
b. Identify the oxidizing agent and reducing agent
Since the oxidation number of oxygen in NO₂ is -2. Since the oxidation number of NO₂ is zero, we let x be the oxidation number of N.
So, x + 2 × (oxidation number of oxygen) = 0
x + 2(-2) = 0
x - 4 = 0
x = 4
Since the oxidation number of oxygen in NO₂⁻ is -1. Since the oxidation number of NO₂⁻ is -1, we let x be the oxidation number of N.
So, x + 2 × (oxidation number of oxygen) = 0
x + 2(-2) = -1
x - 4 = -1
x = 4 - 1
x = 3
Also, the oxidation number of oxygen in NO₃⁻ is -1. Since the oxidation number of NO₃⁻ is -1, we let x be the oxidation number of N.
So, x + 2 × (oxidation number of oxygen) = -1
x + 3(-2) = -1
x - 6 = -1
x = 6 - 1
x = 5
i. The oxidizing agent
The oxidation number of N changes from +4 in NO₂ to +3 in NO₂⁻. So, Nitrogen is reduced and thus NO₂⁻ is the oxidizing agent
ii. The reducing agent
The oxidation number of N changes from +4 in NO₂ to +5 in NO₃⁻. So, Nitrogen is oxidized and thus and NO₃⁻ is the reducing agent.