Can someone plz help me???? will mark as brainliest!!!!!!!!!!!

As water molecules move through the hydrological cycle, water both gains and loses energy. in which case would the water be losi

zysi [14]

The correct answer is condensation

That is because energy is used to break bonds, so condensation where bonds are made actually loses energy because it turns into liquid.

6 0

3 years ago

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What is the specific heat of a substance if 300 J are required to raise the temperature of a 267-g sample by 12 degrees Celsius?

o-na [289]

The specific heat capacity of a substance if 300 j are required to raise the temperature of 207 g by 12 degrees is calculated using the formula below

Q=Mc deltaT
Q(heat required) = 300 j
M(mass) =207g
c(specific heat capacity) = ?
delta t(change in temperature)= 12 c

by making c the subject of the formula

C= Q/ M delta T

C= 300 j/ 267 g x 12 c = 0.0936 j/g/c

7 0

3 years ago

An ion with an atomic number of 34 and 36 electrons has a __________ charge

Solnce55 [7]

negative charge. The atomic number of 34 tells you immediately that the ion has 34 protons. Since it also has 36 electrons, it has 2 more electrons than needed to be neutral. So the ion will have a negative charge since electrons have a negative charge.

5 0

2 years ago

A buffer solution contains 0.306 M C6H5NH3Br and 0.418 M C6H5NH2 (aniline). Determine the pH change when 0.124 mol HCl is added

Ulleksa [173]

Answer: The pH change of the buffer is 0.30

Explanation:

To calculate the pH of basic buffer, we use the equation given by Henderson Hasselbalch:

$pOH=pK_b+\log(\frac{[\text{conjugate acid}]}{[\text{base}]})$

$pOH=pK_b+\log(\frac{[C_6H_5NH_3^+]}{[C_6H_5NH_2]})$ .....(1)

We are given:

$pK_b$ = negative logarithm of base dissociation constant of aniline = 9.13

$[C_6H_5NH_3^+]=0.306M$

$[C_6H_5NH_2]=0.418M$

pOH = ?

Putting values in equation 1, we get:

$pOH=9.13+\log(\frac{0.306}{0.418})\\\\pOH=8.99$

To calculate pH of the solution, we use the equation:

$pH+pOH=14\\pH_{initial}=14-8.99=5.01$

To calculate the molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

Moles hydrochloric acid solution = 0.124 mol

Volume of solution = 1 L

Putting values in above equation, we get:

$\text{Molarity of HCl}=\frac{0.124}{1L}\\\\\text{Molarity of HCl}=0.124M$

The chemical reaction for aniline and HCl follows the equation:

$C_6H_5NH_2+HCl\rightarrow C_6H_5NH_3^++Cl^-$

Initial: 0.418 0.124 0.306

Final: 0.294 - 0.430

Calculating the pOH by using using equation 1:

$pK_b$ = negative logarithm of base dissociation constant of aniline = 9.13

$[C_6H_5NH_3^+]=0.430M$

$[C_6H_5NH_2]=0.294M$

pOH = ?

Putting values in equation 1, we get:

$pOH=9.13+\log(\frac{0.430}{0.294})\\\\pOH=9.29$

To calculate pH of the solution, we use the equation:

$pH+pOH=14\\pH_{final}=14-9.29=4.71$

Calculating the pH change of the solution:

$\Delta pH=pH_{initial}-pH_{final}\\\\\Delta pH=5.01-4.71=0.30$

Hence, the pH change of the buffer is 0.30

8 0

3 years ago

At what temperature will 0.654 moles of helium gas occupy 12.30 liters at 1.95 atmospheres?

DedPeter [7]

Answer:

447,25k

Explanation:

According to the ideal gas law

$PV=nRT$

Where:

P: is the pressure of the gas in atmospheres.

V: is the volume of the gas in liters.

n: number of moles of the gas

R: ideal gas constant

T: absolute temperature of the gas in kelvin

now using:

$T=\frac{PV}{nR}\\ \\T=\frac{12,3L.1,95atm}{0,654mol.0,082(L.atm/K.mol)}\\ \\T=447,25K$

3 0

3 years ago