Answer : The volume of gas occupy at
is, 1.25 L
Explanation :
Charles' Law : It states that volume of the gas is directly proportional to the temperature of the gas at constant pressure and number of moles.
Mathematically,

where,
are the initial volume and temperature of the gas.
are the final volume and temperature of the gas.
We are given:

Putting values in above equation, we get:

Therefore, the volume of gas occupy at
is, 1.25 L
Answer:
This problem is providing a chemical equation between two hypothetical elements, X and Y and asks for the molesof X that are needed to
produce 21.00 moles of D in excess Y. After the following work, the answer turns out to be 15.75 mol X:Mole ratios:In chemistry, one the most crucial branches is stoichiometry, which allows us to perform calculations with grams, moles and particles (atoms, molecules and ions). It is based on the concept of mole ratios, whereby the moles of a specific substance can be converted to moles of another one, say product to reactant, reactant to reactant, reactant to product and product to product.
Calculations:In such a way, since 21.00 moles of D are given, we need the mole ratio of D to X in order to get the answer, which according to the reaction is 3:4 based on their coefficients in the reaction. Hence, we calculate the required as follows:
Explanation:
mark me brainliest!!
Answer:
–2.23 L
Explanation:
We'll begin by calculating the final volume. This can be obtained as follow:
Initial pressure (P₁) = 1.03 atm
Initial volume (V₁) = 3.62 L
Final pressure (P₂) = 2.68 atm
Final volume (V₂) =?
P₁V₁ = P₂V₂
1.03 × 3.62 = 2.68 × V₂
3.7286 = 2.68 × V₂
Divide both side by 2.68
V₂ = 3.7286 / 2.68
V₂ = 1.39 L
Finally, we shall determine the change in volume. This can be obtained as follow:
Initial volume (V₁) = 3.62 L
Final volume (V₂) = 1.39 L
Change in volume (ΔV) =?
ΔV = V₂ – V₁
ΔV = 1.39 – 3.62
ΔV = –2.23 L
Thus, the change in the volume of her lung is –2.23 L.
NOTE: The negative sign indicate that the volume of her lung reduced as she goes below the surface!
Answer:
11.39
Explanation:
Given that:


Given that:
Mass = 1.805 g
Molar mass = 82.0343 g/mol
The formula for the calculation of moles is shown below:

Thus,


Given Volume = 55 mL = 0.055 L ( 1 mL = 0.001 L)


Concentration = 0.4 M
Consider the ICE take for the dissociation of the base as:
B + H₂O ⇄ BH⁺ + OH⁻
At t=0 0.4 - -
At t =equilibrium (0.4-x) x x
The expression for dissociation constant is:
![K_{b}=\frac {\left [ BH^{+} \right ]\left [ {OH}^- \right ]}{[B]}](https://tex.z-dn.net/?f=K_%7Bb%7D%3D%5Cfrac%20%7B%5Cleft%20%5B%20BH%5E%7B%2B%7D%20%5Cright%20%5D%5Cleft%20%5B%20%7BOH%7D%5E-%20%5Cright%20%5D%7D%7B%5BB%5D%7D)

x is very small, so (0.4 - x) ≅ 0.4
Solving for x, we get:
x = 2.4606×10⁻³ M
pOH = -log[OH⁻] = -log(2.4606×10⁻³) = 2.61
<u>pH = 14 - pOH = 14 - 2.61 = 11.39</u>
Answer:
<u>ATGGCCTA</u>
Explanation:
For this we have to keep in mind that we have a <u>specific relationship between the nitrogen bases</u>:
-) <u>When we have a T (thymine) we will have a bond with A (adenine) and viceversa</u>.
-) <u>When we have C (Cytosine) we will have a bond with G (Guanine) and viceversa</u>.
Therefore if we have: TACCGGAT. We have to put the corresponding nitrogen base, so:
TACCGGAT
<u>ATGGCCTA</u>
<u></u>
I hope it helps!