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Mandarinka [93]
3 years ago
13

Block A is made of copper, which has a specific heat value of 0.385 J/g*C.

Chemistry
1 answer:
bulgar [2K]3 years ago
7 0

Answer:

im hungry

Explanation:

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Name the reactants and products of Na3PO4 + 3KOHàNaOH + K3PO4 <br> (Question 2 and 3)
Leona [35]

Answer:

Question 2: Na3PO4, KOH; Question 3: Na3PO4, KOH

Explanation:

Question 2

The reactants in a chemical equation are the species on the left side of the reaction arrow.

Thus the reactants are Na3PO4, KOH (sodium phosphate and potassium hydroxide).

Question 3.

The products in a chemical equation are the species on the right side of the reaction arrow.

Thus the products are NaOH, K3PO4 (sodium hydroxide and potassium phosphate).

6 0
3 years ago
Given that a for HCN is 6. 2×10^−10 at 25 °C. What is the value of b for cn− at 25 °C?
Kay [80]

If Ka for HCN is 6. 2×10^−10 at 25 °C, then the value of Kb for cn− at 25 °C is 1.6 × 10^(-5).

<h3>What is base dissociation constant? </h3><h3 />

The base dissociation constant (Kb) is defined as the measurement of the ions which base can dissociate or dissolve in the aqueous solution. The greater the value of base dissociation constant greater will be its basicity an strength.

The dissociation reaction of hydrogen cyanide can be given as

HCN --- (H+) + (CN-)

Given,

The value of Ka for HCN is 6.2× 10^(-10)

The correlation between base dissociation constant and acid dissociation constant is

Kw = Ka × Kb

Kw = 10^(-14)

Substituting values of Ka and Kw,

Kb = 10^(-14) /{6.2×10^(-10) }

= 1.6× 10^(-5)

Thus, the value of base dissociation constant at 25°C is 1.6 × 10^(-5).

learn more about base dissociation constant :

brainly.com/question/9234362

#SPJ4

4 0
2 years ago
How do the test variables (independent variables) and outcome variables (dependent variables) in an experiment compare? A. The o
SVETLANKA909090 [29]

Answer:

C.

Explanation:

8 0
3 years ago
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I NEED HELP RIGHT NOW
PolarNik [594]
There are a total of 4 elements
7 0
3 years ago
If I initially have a gas at a pressure of 12 kPa, a volume of 23 liters, and a temperature of 200 K, and then I raise the press
sertanlavr [38]

Answer:

The new volume of gas would be 30 L.

Explanation:

This is an example of a Combined Gas Laws problem.

5 0
3 years ago
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