How many milliliters of 0.111 M HClO4 solution are needed to neutralize 50.00 mL of 0.0789 M NaOH?How many milliliters of 0.111 M HClO4 solution are needed to neutralize 50.00 mL of 0.0789 M NaOH?
Answer: chalcogens.
Explanation: Strontium is an alkaline earth metal, it always exhibits a degree of oxidation in its compounds +2.
Chalcogens are a group of 6 chemical elements (oxygen O, sulfur S, selenium se, tellurium te, polonium Po) that have an oxidation state of -2 => Chalcogens will combine with strontium in a ratio of 1:1.
Answer: b. Fiona is correct because the diagram shows two individual simple machines.
Explanation:
A mechanical device using which we can change the direction or magnitude of force applied is known as simple machine.
For example, in the given diagram there are two individual simple machines.
The machine helps in changing the direction or magnitude of force applied by the man. As a result, it becomes easy for him to carry different things easily from one place to another.
Thus, we can conclude that the statement Fiona is correct because the diagram shows two individual simple machines, is correct.
Taking into accoun the ideal gas law, The volume of a container that contains 24.0 grams of N2 gas at 328K and 0.884 atm is 26.07 L.
An ideal gas is a theoretical gas that is considered to be composed of point particles that move randomly and do not interact with each other. Gases in general are ideal when they are at high temperatures and low pressures.
The pressure, P, the temperature, T, and the volume, V, of an ideal gas, are related by a simple formula called the ideal gas law:
P×V = n×R×T
where P is the gas pressure, V is the volume that occupies, T is its temperature, R is the ideal gas constant, and n is the number of moles of the gas. The universal constant of ideal gases R has the same value for all gaseous substances.
Explanation:
In this case, you know:
P= 0.884 atm
V= ?
n= 0.857 moles (where 28 g/mole is the molar mass of N₂, that is, the amount of mass that the substance contains in one mole.)
R=0.082
T= 328 K
Replacing in the ideal gas law:
0.884 atm×V= 0.857 moles× 0.082 ×328 K
Solving:
V= 26.07 L
The volume of a container that contains 24.0 grams of N2 gas at 328K and 0.884 atm is 26.07 L.
15 moles.Explanation:Hello,In this case, the undergoing chemical reaction is:Clearly, since carbon and oxygen are in a 1:1 molar ratio, 15 moles of carbon will completely react with 15 moles of oxygen, therefore 15 moles of oxygen remain as leftovers. In such a way, since carbon and carbon dioxide are also in a 1:1 molar ratio, the theoretical yield of carbon dioxide is 15 moles based on the stoichiometry:Best regards.
Answer: 25 g
Explanation: Given:
Original amount (N₀) = 100 g
Number of half-lives (n) = 11460/5730 = 2
Amount remaining (N) = ?
N = 1/2ⁿ × N₀
N = 1/2^2 × 100
N = 0.25 × 100
N = 25 g
Answer: 7.8125 g
Explanation: Given:
Original amount (N₀) = 500 g
Number of half-lives (n) = 9612/1602 = 6
Amount remaining (N) = ?
N = 1/2ⁿ × N₀
N = 1/2^6 × 500
N = 0.015625 × 500
N = 7.8125 g
Answer: B. carbon tetrachloride, CCI4
Explanation: The other options are incorrect. Let's write the correct formulas:
A. Diarsenic pentoxide - As2O5
C. Sodium dichromate - Na2Cr2O7
D. magnesium phosphide - Mg3P2
Answer: 306 L
Explanation: Using the ideal gas law,
PV = nRT
where R = 0.08206 L•atm/(mol•°K), solving for n gives
n = PV/(RT)
n = (845 mmHg) (270 L) / ((0.08206 L•atm/(mol•°K)) (24 °C))
Convert the given temperature to °K and the given pressure to atm:
24 °C = (273.15 + 24) °K ≈ 297.2 °K
(845 mmHg) × (1/760 atm/mmHg) ≈ 1.11 atm
Then the balloon contains
n = (1.11 atm) (270 L) / ((0.08206 L•atm/(mol•°K)) (297.2 °K))
n ≈ 12.3 mol
of He.
Solve the same equation for V :
V = nRT/P
Convert the target temperature to °K:
-50 °C = (273.15 - 50) °K = 223.15 °K
Then the volume under the new set of conditions is
V = (12.3 mol) (0.08206 L•atm/(mol•°K)) (223.15 °K) / (0.735 atm)
V ≈ 306 L
3
18 694
rating
0.857 moles (where 28 g/mole is the molar mass of N₂, that is, the amount of mass that the substance contains in one mole.)
×328 K
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