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Anna007 [38]
2 years ago
14

When a 0.764 g almond is burned, it heats 225 mL of water from 14.8oC to 18.4oC. Calculate the heat of combustion of the almond

in kcal/g
Chemistry
1 answer:
Soloha48 [4]2 years ago
8 0
It’s 1.071 BWIWNWOWLQQMKWBWJWNWKWNQKKQJWKWKKQKWKAMQNMWNWNWKWWKWKWNWKW
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HELP ME ON THIS
diamong [38]

Answer:

The attractive force between them decreases

Explanation:

This is because they become localised.

7 0
2 years ago
What is the pH of a 0.300 M NH₃ solution that has Kb = 1.8 × 10⁻⁵ ? The equation for the dissociation of NH₃ is: NH₃ (aq) + H₂O
nalin [4]

Answer:

11.4

Explanation:

Step 1: Given data

  • Concentration of the base (Cb): 0.300 M
  • Basic dissociation constant (Kb): 1.8 × 10⁻⁵

Step 2: Write the dissociation equation

NH₃(aq) + H₂O(l) ⇄ NH₄⁺(aq) + OH⁻(aq)

Step 3: Calculate the concentration of OH⁻

We will use the following expression.

[OH^{-} ]=\sqrt{Kb \times Cb } = \sqrt{1.8  \times 10^{-5} \times 0.300 } = 2.3 \times 10^{-3} M

Step 4: Calculate the pOH

We will use the following expression.

pOH =-log[OH^{-} ]= -log(2.3 \times 10^{-3} M) = 2.6

Step 5: Calculate the pH

We will use the following expression.

pH+pOH=14\\pH = 14-pOH = 14-2.6 = 11.4

8 0
3 years ago
What is the total of the amount of energy that is reflected back into space​
Tcecarenko [31]
Moon va half plus 22
7 0
3 years ago
why is the term "ionic" used when describing compounds formed from a metal element and a nonmetal element?
nordsb [41]
Ionic compounds form when metals transfer valence electrons to nonmetals. Ionic compounds exist as crystals rather than molecules.
8 0
2 years ago
An atom of argon has a radius rar = 88 pm and an average speed in the gas phase at 25°C of 172 m/s.
Rudik [331]

Answer:

1.2* 10³ rNe.

Explanation:

Given speed of neon=350 m/s

Un-certainity in speed= (0.01/100) *350 =0.035 m/s

As per heisenberg uncertainity principle

Δx*mΔv ≥\frac{h}{4\pi }

4π

h

..................(1)

mass of neon atom =\frac{20*10^{-3} }{6.22*10^{-23} } =3.35*10^{-26} kg

6.22∗10

−23

20∗10

−3

=3.35∗10

−26

kg

substituating the values in eq. (1)

Δx =4.49*10^{-8}10

−8

m

In terms of rNe i.e 38 pm= 38*10^{-12}10

−12

Δx=\frac{4.49*10^{-8} }{38*10^{-12} }

38∗10

−12

4.49∗10

−8

=0.118*10^{4}10

4

* (rNe)

=1.18*10³ rN

= 1.2* 10³ rNe.

Explanation:

This is the answer

7 0
2 years ago
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