The reaction of propyl magnesium bromide with formaldehyde

The atmospheric pressure in Francisco on ascertain day was 97.6kpa what was the pressure on mmhg

vodka [1.7K]

Answer:

732.0601 mmHg

Explanation:

Given data:

Pressure = 97.6 KPa

Given pressure in mmHg = ?

Solution:

Kilo pascal and millimeter mercury both are units of pressure.

Kilo pascal is denoted as "KPa"

Millimeter mercury is denoted as " mmHg"

Kilo pascal is measure of force per unit area while also define as newton per meter square.

It is manometric unit of pressure. It is the pressure generated by column of mercury one millimeter high.

Conversation of kilopascal to mmHg:

97.6 × 7.501 = 732.0601 mmHg

3 0

3 years ago

Give me lil reasoning so I know your not lying for points

Natali5045456 [20]

Answer:

C. Tip had the lower pressure, it was 652.71mmHg

Explanation:

To answer this question you need to convert both values into a common unit so you can compare directly.

87kPa = 652.71 mmHg

4 0

3 years ago

A standard lanthanum solution is prepared by dissolving 0.1968 grams of lanthanum oxide (La2O3) in excess nitric acid and diluti

sweet-ann [11.9K]

Answer:

1.208x10⁻³M and 392.5ppm La(NO3)3

Explanation:

The reaction that occurs is:

La2O3 + 6HNO3 → 2La(NO3)3 + 3H2O

Molarity is defined as the moles of solute (In this case, LaO3) per liter of solution. And ppm, are mg of solute per liter of solution.

To solve this question we must find the moles of La(NO3)3 produced and its mass in milligrams to find molarity and ppm:

<em>Moles La2O3 -Molar mass: 325.81g/mol-</em>

0.1968g * (1mol / 325.81g) = 6.04x10⁻⁴ moles La2O3

<em>Moles La(NO3)3:</em>

6.04x10⁻⁴ moles La2O3 * (2mol La(NO3)3 / 1mol La2O3) = 1.208x10⁻³ moles La(NO3)3

<em>Molarity:</em>

1.208x10⁻³ moles La(NO3)3 / 1L =

<em>Mass La(NO3)3 -Molar mass: 324.92g/mol-</em>

1.208x10⁻³ moles La(NO3)3 * (324.92g / mol) = 0.392.5g La(NO3)3

In mg:

392.5mg La(NO3)3 / 1L =

392.5ppm La(NO3)3

7 0

2 years ago

What combination of substances will give a buffered solution that has a pH of 5.05? (Assume each pair of substances is dissolved

lilavasa [31]

Answer:

C

Explanation:

When the Kb is given, the Henderson-Hasselbalch equation can be used to calculate the pOH of a buffer solution:

pOH = pKb + log ([A⁻] / [HA]) = -log(Kb) + log ([BH+] / [B])

Here, moles can be used in place of the concentration since the pairs listed are both dissolved in 5L, which cancel due to the fraction in the logarithm.

a) pOH = -log(1.8 x 10⁻⁵) + log(1.5/1.0) = 4.92

pH = 14 - pOH = 14 - 4.92 = 9.08

b) pOH = -log(1.8 x 10⁻⁵) + log(1.0/1.5) = 4.57

pH = 14 - pOH = 14 - 4.57 = 9.43

c) pOH = -log(1.7 x 10⁻⁹) + log(1.5/1.0) = 8.95

pH = 14 - pOH = 14 - 8.95 = 5.05

d) pOH = -log(1.7 x 10⁻⁹) + log(1.0/1.5) = 8.59

pH = 14 - pOH = 14 - = 5.41

3 0

3 years ago

Determine the [H+] or [OH−] for each of a solutions at 25°C.

levacccp [35]

Answer:

Solution B: [OH-] = 1×10^–6 M

Solution C: [OH-] = 1×10^–6 M

Explanation:

Solution B:

Hydronium ion concentration, [H3O+] = 9.99×10^–9 M

Hydroxide ion concentration, [OH-] = ?

The Hydroxide ion concentration, [OH-] can be obtained as follow:

[H3O+] × [OH-] = 1×10^–14

9.99×10^–9 × [OH-] = 1×10^–14

Divide both side by 9.99×10^–9

[OH-] = 1×10^–14 / 9.99×10^–9

[OH-] = 1×10^–6 M

Therefore, the Hydroxide ion concentration, [OH-] is 1×10^–6 M

Solution C:

Hydronium ion concentration, [H3O+] = 0.000777 M

Hydroxide ion concentration, [OH-] = ?

The Hydroxide ion concentration, [OH-] can be obtained as follow:

[H3O+] × [OH-] = 1×10^–14

0.000777 × [OH-] = 1×10^–14

Divide both side by 0.000777

[OH-] = 1×10^–14 / 0.000777

[OH-] = 1.29×10^–11 M

Therefore, the Hydroxide ion concentration, [OH-] is 1.29×10^–11 M

6 0

3 years ago

The reaction of propyl magnesium bromide with formaldehyde​

The reaction of propyl magnesium bromide with formaldehyde