All categories

3 years ago

For the equation BaCl2 + Na2SO4 → BaSO4 + 2NaCl, how many chlorine atoms are on the reactants side and on the products side?

Chemistry

2 answers:

damaskus [11]3 years ago

8 0

Answer:

The balanced chemical reaction is written as follows:

bacl2 + na2so4 → baso4 + 2nacl

Therefore, the number of chlorine atoms in the reactants side and the products side would be 2 atoms in both sides which should be the case because the number of atoms in each side should have the same number.

Explanation:

Internet

lara [203]3 years ago

6 0

Answer:

fsadfdfsfds

Explanation:

You might be interested in

the atomic nubmer of an element is the total number of which particles in the nucleus?protons,neutrons,electrons,protons and ele

erma4kov [3.2K]

The atomic number is the total numbers of protons in an element

3 0

4 years ago

List two products derived from ethylene

laiz [17]

Ethylene is the starting material for the preparation of a number of two-carbon compounds including ethanol (industrial alcohol), ethylene oxide (converted to ethylene glycol for antifreeze and polyester fibres and films), acetaldehyde (converted to acetic acid), and vinyl chloride (converted to polyvinyl chloride).

8 0

2 years ago

A buffer with a pH of 4.31 contains 0.31 M of sodium benzoate and 0.24 M of benzoic acid. What is the concentration of [ H 3 O ]

Mnenie [13.5K]

Answer: The hydronium ion concentration in the solution is $1.29\times 10^{-4}M$

Explanation:

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 = 0.060 mol

Volume of solution = 1 L

Putting values in above equation, we get:

$\text{Molarity of HCl}=\frac{0.060}{1L}\\\\\text{Molarity of HCl}=0.060M$

The chemical reaction for aniline and HCl follows the equation:

$C_6H_5COO^-+HCl\rightarrow C_6H_5COOH+Cl^-$

Initial: 0.24 0.060 0.31

Final: 0.18 - 0.37

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

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

$pH=pK_a+\log(\frac{[C_6H_5COO^-]}{[C_6H_5COOH]})$

We are given:

$pK_a$ = negative logarithm of acid dissociation constant of benzoic acid = 4.2

$[C_6H_5COO^-]=0.18M$

$[C_6H_5COOH]=0.37M$

pH = ?

Putting values in equation 1, we get:

$pH=4.2+\log(\frac{0.18}{0.37})\\\\pH=3.89$

To calculate the hydronium ion concentration in the solution, we use the equation:

$pH=-\log[H_3O^+]$

pH = 3.89

Putting values in above equation, we get:

$3.89=-\log[H_3O^+]$

$[H_3O^+]=10^{-3.89}=1.29\times 10^{-4}M$

Hence, the hydronium ion concentration in the solution is $1.29\times 10^{-4}M$

3 0

3 years ago

As gaseous material condensed in the solar nebula, ice formed

Pachacha [2.7K]

Only far from the Sun material condensed in the solar nebula, ice formed

8 0

4 years ago

Calculate the specific heat capacity for a 22.7-g sample of lead that absorbs 237 J when its temperature increases from 29.8 °C

soldier1979 [14.2K]

Answer:

$\boxed {\boxed {\sf c\approx 0.159 \ J/ g \textdegree C}}$

Explanation:

We are asked to find the specific heat capacity of a sample of lead. The formula for calculating the specific heat capacity is:

$c= \frac{Q}{m \times \Delta T}$

The heat absorbed (Q) is 237 Joules. The mass of the lead sample (m) is 22.7 grams. The change in temperature (ΔT) is the difference between the final temperature and the initial temperature. The temperature increases from 29.8 °C to 95.6 °C.

ΔT = final temperature -inital temperature
ΔT= 95.6 °C - 29.8 °C = 65.8 °C

Now we know all three variables and can substitute them into the formula.

Q= 237 J
m= 22.7 g
ΔT = 65.8 °C

$c= \frac {237 \ J}{22.7 \ g \ \times \ 65.8 \textdegree C}$

Solve the denominator.

22.7 g * 65.8 °C = 1493.66 g °C

$c= \frac {237 \ J}{1493.66 \ g \textdegree C}$

Divide.

$c= 0.1586706479 J /g \textdegree C$

The original values of heat, temperature, and mass all have 3 significant figures, so our answer must have the same. For the number we found that is the thousandth place. The 6 in the ten-thousandth place tells us to round the 8 up to a 9.

$c \approx 0.159 \ J/g \textdegree C$

The specific heat capacity of lead is approximately 0.159 Joules per gram degree Celsius.

3 0

3 years ago