A solution is prepared by dissolving 55.55g of nh4cl into enough water to make 250.0 ml of solution. calculate the molarity.
Answers
The molarity of the solution prepared by dissolving 55.55 g of NH4Cl into enough water to make 250.0 ml of solution is 4.16 M.
To calculate the molarity of a solution, we need to divide the moles of solute by the volume of the solution in liters.
First, we need to calculate the number of moles of NH4Cl:
Molar mass of NH4Cl = 14.01 g/mol (atomic mass of nitrogen) + 4 * 1.01 g/mol (atomic mass of hydrogen) + 1 * 35.45 g/mol (atomic mass of chlorine)
= 14.01 + 4.04 + 35.45
= 53.50 g/mol
Number of moles = mass / molar mass
Number of moles of NH4Cl = 55.55 g / 53.50 g/mol
Number of moles of NH4Cl = 1.04 mol
Now, we can calculate the molarity (M):
Molarity = moles of solute / volume of solution (in liters)
Volume of solution = 250.0 ml = 250.0 ml * (1 L / 1000 ml) = 0.250 L
Molarity = 1.04 mol / 0.250 L
Molarity = 4.16 M
Therefore, the molarity of the solution prepared by dissolving 55.55 g of NH4Cl into enough water to make 250.0 ml of solution is 4.16 M.
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Find the volume of a cereal box that has the dimensions of 35.2 cm x 220 cm x 6.0 cm
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Answer:
Volume = 35.2×220×6.0 = 46464 centimeters³
Explanation:
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How can (r)-2-butanol be converted to (r)-2-methylbutanenitrile? select all that apply.
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What is the main function of the cell membrane in an animal cell?
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The rate constant k is dependent on the concentration of the reactant the nature of the reactants the temperature the order of the reaction none of these one of these two of these three of these all of these What would happen if the kinetic energy of the reactants was not enough to provide the needed activation energy? a. The products would be produced at a lower energy state. b. The rate of the reaction would tend to increase.
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What is missing from the solubility graph shown on the right? A graph with temperature in degrees Celsius ranging from 0 to 100 in units of 20 on the horizontal axis and unidentified tick marks from 0 to 300 in increments of 50 on the vertical axis. An orange line curves upward from 0, 90 through 50, 125 to 100, 245.
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Answer:
graph title, label for the y axis, and legend explaining the orange and blue colors
Explanation:
need help with the question
Answers
Answer:
for a i think it is choice 2
for b i think it is the first choice
Explanation:
Question 1 (8 points)
Name the following simple binary compounds.
1. NaBr =
2. KCl =
3. Mgo =
4. CaF2=
5. BaS =
6. Li20 =
7. NaF =
8. CS3N =
Answers
Answer:
Below
Explanation:
1. Sodium Bromide
2. Potassium Chloride
3. Magnesium Oxide
4. Calcium Fluoride
5. Barium Sulfide
6. Lithium Oxide
7. Sodium Fluoride
8. Cesium Nitride
A liquid is allowed to evaporate and leaves no residue. can you determine whether it was an element, a compound, or a mixture?
Answers
A liquid is allowed to evaporate and leaves no residue. It can not be determined whether it was an element, a compound, or a mixture.
A water vapor molecule stays roughly 10 days in the atmosphere after it has evaporated. Water vapor starts to cool back down as it ascends higher in the atmosphere. The water vapor condenses when it becomes cold enough, turning it back into liquid water. Eventually, individual water droplets will condense to create clouds and precipitation.
It is not possible to determined whether it would be an element, a compound and mixture because the size of the particle will be too less.
A method for separating homogenous mixtures with one or even more dissolved salts is called evaporation. The procedure separates the liquid from the solid components. Usually, the procedure entails heating the combination until there is no more liquid is present.
Therefore, it can not be possible to determined whether it was an element, a compound, or a mixture by evaporation.
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What is 2+2 plzzzzz help I’m only 83
Answers
Answer: 4
Explanation: if you move to spaces in a number line from 2 you would be at 4
Balance the chemical equation : CH₄ + O₂ → CO₂ + H₂O
Answers
Answer:
balanced equation: CH₄ + 2O₂ → CO₂ + 2H₂O
Explanation:
Given sample: CH₄ + O₂ → CO₂ + H₂O
The carbon is same, so no need to change. Hydrogen 2 less on left side so putted "2" before H₂O = "2H₂O"so now, there is total 4 oxygen on left side to balance put 2 before right side oxygen like this "2O₂"changes applied: CH₄ + 2O₂ → CO₂ + 2H₂O
practical work inside a laboratory and outside a laboratory (difference between)
Answers
The main difference between practical work inside a laboratory and outside a laboratory is the level of control that the experimenter has over the environment.
What differentiates work inside and outside a laboratory ?By conducting experiments in a controlled laboratory setting, the researcher has the ability to regulate various atmospheric conditions, such as temperature and humidity, thereby crucially contributing to the precision of the findings. It can be challenging to achieve precise outcomes when conducting experiments in non-laboratory settings due to the limited control over the surroundings.
The presence of specialized equipment sets apart practical work conducted within a laboratory from those performed outside. There is typically a vast selection of specialized equipment readily accessible in a laboratory, which can prove highly beneficial when conducting experimental procedures.
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Which describes the molecule shown below?
O A. Unsaturated alkyne
O B. Saturated alkane
O C. Saturated alkene
O D. Unsaturated alkane
Answers
The description of the molecules shown is it is unsaturated alkynes. The correct option is A.
What are unsaturated alkynes?Alkynes are unsaturated double bond containing compound that react with hydrogen when a catalyst is present.
The pie bond shows that the hydrogen atoms are lost from the alkanes.
Thus, the correct option is A, unsaturated alkynes.
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number 3 ...................
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Silicon is used for electronic devices because it is an element with very special properties. One of it's most important properties is that it is a semiconductor.
Draw the lewis structure for no2-. How many single bonds, double bonds, triple bonds, and lone pairs of electrons are on the central atom when considering a single contributing structure? ignore the averaging effects of resonance.
Answers
One unshared pair of electrons, one single bond, one double bond, and no triple bonds are present in nitrogen.
What is the number of bond pairs and lone pairs in NO2?There are two sigma bonds and one pair of unpaired electrons in nitrogen dioxide.
What percentage of NO2 is made up of single bonds?one bond only
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is Radon an elementary substance or a compound?
Answers
Answer:elementary substance
Explanation:
Which statement describes a gas condensing into a liquid?
A.
The atoms in the molecules get heavier so they move less.
B.
The molecules break into smaller pieces that take up less space.
C.
The molecules stick together to make bigger molecules.
D.
The molecules get closer together and move more slowly.
Answers
Answer: molecules get closer
Explanation:
Condensation is the process in which molecules of a gas slow down, come together, and form a liquid. When gas molecules transfer their energy to something cooler, they slow down and their attractions cause them to bond to become a liquid
hope this helps
pls mark me brainliest
Answer:
D. The molecules get closer together and move more slowly.
PLEASE HELP ME IM CONFUSED
Answers
Given -
T = 450K
∆H = 3000J
∆S = 4J/K
Using formula
∆G = ∆H - T∆S
∆G = 3000- 4*450
∆G = 1200 Joule
∆G is positive so the reaction is non spontaneous!
Just question 2 pleasee
Answers
The correct matching of the properties of the elements are;
Rubidium - catches fire as soon as it s exposed to air and burns fiercely
Lithium - hard to set alight but burns with a deep red flame
Potassium - Melts readily on heating with a Bunsen flame and catches fire easily. Burns with a mauve flame.
Sodium - Melts fairly readily with Bunsen flame then catches fire and burns with a deep yellow flame
What are the properties of the elements?We know that elements are classified into groups. The groups are a family of elements that have similar chemical and physical properties. Now we have the to match the properties of the elements as shown;
Rubidium - catches fire as soon as it s exposed to air and burns fiercely
Lithium - hard to set alight but burns with a deep red flame
Potassium - Melts readily on heating with a Bunsen flame and catches fire easily. Burns with a mauve flame.
Sodium - Melts fairly readily with Bunsen flame then catches fire and burns with a deep yellow flame
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At time t=0, an aluminum bar (thermal diffusivity k=0.86 ) of length Lcm with completely insulated lateral surfaces and constant thermal properties is removed from boiling water (u_B =100 degrees Celsius). Do the following i), ii), iii) for each of the scenarios, a-d, below i) Write down the initial-boundary value problem. That is, the PDE along with any initial and boundary conditions. ii) Without solving for u(x,t), describe the temperature distribution in the bar as t→[infinity] based on physical intuition. iii) Find the solution as t→[infinity] by solving the appropriate steady state equation. a) The two ends of the bar are immediately immersed in a medium with constant temperature 10 degrees Celsius. b) The end at x=0 is immersed in a medium with temperature 0 degrees Celsius and the end at x=L is completely insulated.
Answers
The initial-boundary value problem is 0. The temperature distribution in the bar as t ∞ for case b is u(x) = 100.
i) The initial-boundary value problem: Initial condition:
u(x, 0) = u0(x) = 100 °C
Boundary conditions:
Case a) u(0, t) = u(L, t)
= 10°C.
Case b) u(0, t) = 0°C,
uL(x) = ∂u/∂x|L
= 0.
ii) Temperature distribution: The temperature distribution in the bar as t→∞ for both cases will be linear and decreasing from 100°C to the imposed boundary conditions at either end of the bar. That is, a linear decrease of temperature from one end to the other.
iii) Solution as t→∞:
a) The appropriate steady-state equation to solve for case a is the ordinary differential equation:
d²u/dx² =0 with the boundary conditions:
u(0) = u(L) = 10°C.
The general solution of the ODE is u(x) = Ax+B.
Applying the boundary conditions gives u(x) = 10(L-x)/L
Thus, the temperature distribution in the bar as t→∞ for case a is u(x,∞ ) = 10(L-x)/L
b) The appropriate steady-state equation to solve for case b is the ordinary differential equation
d²u/dx²=0 with the boundary conditions:
u(0) = 0°C
The general solution of the ODE is u(x) = Ax + B
Applying the boundary conditions gives u(x) = x/l.
Thus, the temperature distribution in the bar as t→∞ for case b is u(x) = 100.
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As heat is added to a gas, the temperature will increase. As the temperature increases the pressure will change in which way?.
Answers
As heat is added to a gas, the temperature will increase. As the temperature increases, the pressure will change in the same way.
The average kinetic energy and the velocity of the gas particles striking the container walls both rise as the temperature rises. As the temperature rises, the pressure also rises, as well since pressure is the force exerted by particles per unit area exerted on the container.
The average kinetic energy of the gas's molecules directly relates to its temperature. Faster-moving particles will more frequently and violently hit the container walls. Because of this, the force acting on the container's walls increases, increasing the pressure.
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Mi One should know the properties of the components of the mixture to separate it. Explain with an example.
Answers
.. When water boils, you can see bubbles rising to the surface of
the water. Of what are these bubbles made?
a. air
b. hydrogen and oxygen gas
c. oxygen gas
d. water vapor
e. carbon dioxide gas
Answers
Answer:
Explanation:
It's water vapor. There is enough heat present to get the water to boil but not enough to break it into its chemical components (oxygen and hydrogen), so the answer is D.
What is the force that acts on an object and opposes its motion called?
Group of answer choices
Inertia
Friction
Thermal Energy
Answers
Answer:
Friction
Explanation:
Friction: A force that opposes the motion of objects that touch as they move past each other.
Hope this Helps!
Answer:
Friction
Explanation:
A force that opposes the motion of objects that touch as they move past each other. Friction acts at the surface where objects are in contact.
If there is a ball that is measured as 235 cm^3 and 530. 25g what is its density expressed in the correct number of significant figures?
Answers
PLZ DON"T JUST PICK AN ANSWER I REALLY NEED HELP
5. Sodium-24 has a half-life of approximately 15 hours. If only one-eighth of the sodium-24 remains, about how much time has passed?
A. 15 hours
B. 30 hours
C. 45 hours
D. 60 hours
Answers
Answer:
60 hours
Explanation:
18.0 __15__ 9 __15__ 4.5 __15__ 2.25 __15__ 1.125g
15(4)= 60
Sodium-24 has a half-life of approximately 15 hours. If only one-eighth of the sodium-24 remain , then it has been passed in 60 hours. So option D is correct.
What is half life and decay law ?The time required for half of the original population of radioactive atoms to decay is called the half-life. The relationship between the half-life, T₁ ÷ 2, and the decay constant is given by T₁/2 = 0.693/λ
According to radioactive decay law, the rate of decay of radioactive atoms at any instant is directly proportional to the number of atoms present at that instant.
dtd N ∝ N
dtd N = λ N
where, N is number of integrated nucleus present in the sample at any time t and λ is decay constant.
Thus ,option D 60 hours is correct.
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an enzyme-catalyzed reaction was carried out with the substrate concentration initially a thousand times greater than the km for that substrate. after 15 minutes, 1% of the substrate had been converted to product, and the amount of product formed in the reaction mixture was 25 mmol. if, in a separate experiment, one-fourth as much enzyme and twice as much substrate had been combined, how long would it take for the same amount (25 mmol) of product to be formed?
Answers
In the second experiment, it would take 60 minutes to produce the same amount of product (25 mmol).
We can start by using the Michaelis-Menten equation to determine the initial reaction rate in the first experiment, where the substrate concentration is initially a thousand times greater than the KM:
v₀ = Vmax × [S]/(KM + [S])
Since the substrate concentration is initially a thousand times greater than the KM, we can assume that [S] >> KM, and simplify the equation to:
v₀ = Vmax × [S] / KM
The amount of substrate converted to product after 15 minutes is 1% of the initial substrate concentration, so the remaining substrate concentration is 99% of the initial concentration. Let's define [S]0 as the initial substrate concentration, then:
[S] = 0.99 [S]₀
The amount of product formed in 15 minutes is 25 mmol, so we can calculate the initial reaction rate as:
v₀ = (25 mmol / 15 min) / (0.01 mol / mol) = 1666.7 nmol/min
where we have converted the units to nmol/min and used the molecular weight of the product to convert from moles to millimoles.
Now, let's consider the second experiment, where one-fourth as much enzyme and twice as much substrate are used.
Since the amount of product formed is the same in both experiments, we can write:
v₁ × t₁ = v₂ × t₂
where v₁ and t₁ are the initial reaction rate and reaction time in the first experiment, and v₂ and t₂ are the corresponding values in the second experiment.
Let's use the Michaelis-Menten equation to relate the initial reaction rate to the enzyme and substrate concentrations:
v₁ = Vmax × [S]₁ / (KM + [S]₁) × E₁
v₂ = Vmax × [S]₂ / (KM + [S]₂) × E₂
We can rearrange these equations to solve for [S]1 and [S]2 in terms of v1, v₂, E₁, E₂, and KM:
[S]₁ = (v₁ × KM) / (Vmax × E₁) - [S]₂
[S]₂ = (v₂ × KM) / (Vmax × E₂) - [S]₁
Substituting these expressions into the equation for the reaction time, we get:
t₂ = (v₁ × t₁ × Vmax × E₂) / (v₂ × Vmax × E₁ + v₁ × E₂ × KM / (Vmax × E₁) - v₁ × KM / (Vmax × E₁))
Substituting the values from the first experiment, we get:
t₂ = (1666.7 nmol/min × 15 min × 1/4 × 2) / (1666.7 nmol/min × 1/2 × 1/1000 mol/L + 1666.7 nmol/min × 2 × 0.01 mol/L / (1/1000 mol/L) - 1666.7 nmol/min × 0.01 mol/L / (1/1000 mol/L))
Simplifying, we get:
t₂ = 60 min
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Which of the graphs below shows a person walking at a constant rate with no change in speed or velocity?
All of these
Person b and person c
Person b only
Person b, c and d
Answers
\(▪▪▪▪▪▪▪▪▪▪▪▪▪ {\huge\mathfrak{Answer}}▪▪▪▪▪▪▪▪▪▪▪▪▪▪\)
The graph showing a person who is walking at the same rate with no change in speed or velocity is :
Person B
a 56.0 ml portion of a 1.30 m solution is diluted to a total volume of 248 ml. a 124 ml portion of that solution is diluted by adding 137 ml of water. what is the final concentration? assume the volumes are additive.
Answers
The final concentration after both dilutions is 0.104 M. There are two stages of dilution of 56.0ml portion of 1.30m solution to a total 248ml volume. First dilution: The concentration of the first dilution can be calculated as follows:
Initial concentration = 1.30 M
Initial volume = 56.0 mL
Final volume = 248 mL
Final concentration = (Initial concentration * Initial volume) / Final volume
Final concentration = (1.30 M * 56.0 mL) / 248 mL = 0.276 M
Second dilution: The concentration of the second dilution can be calculated as follows:
Initial concentration = 0.276 M
Initial volume = 124 mL
Final volume = 124 mL + 137 mL = 261 mL
Final concentration = (Initial concentration * Initial volume) / Final volume
Final concentration = (0.276 M * 124 mL) / 261 mL = 0.104 M
Therefore, the final concentration after both dilutions is 0.104 M.
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What is radius? When might you need to measure an object’s radius?
Answers
Answer: it’s halfway
into the circle, you would need it when trying to find the circumference
Explanation:
Radius is used to find the circumference of the circle or sphere.
What is the radius?A straight line extending from the centre of a circle or sphere to the circumference or surface: The radius of a circle is half the diameter.
To calculate the circumference of a circle, multiply the diameter of the circle with π (pi). The circumference can also be calculated by multiplying 2×radius by pi.
Hence, the radius is used to find the circumference of the circle or sphere.
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