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Tema [17]
3 years ago
6

The temperature of 170 g of a material decreases by 20.0⁰C when it loses 3050 J of heat. What is its specific heat

Chemistry
1 answer:
qaws [65]3 years ago
8 0

Answer:

0.897 J/g.⁰C

Explanation:

Given the following data:

m = 170 g (mass)

ΔT = 20.0⁰C (change in temperature)

q = 3050 J (amount of heat)

The amount of heat (q) is calculated as follows:

q = m x Cp x ΔT

Thus, we introduce the data in the mathematical expression to calculate the specific heat (Cp):

Cp = q/(m x ΔT) = 3050 J/(170 g x 20.0⁰C) = 0.897 J/g.⁰C

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polar beacuse there is no way its non polar.

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Water molecules stick to nearby water molecules due to
velikii [3]

Answer:

The answer will have to be A

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3 years ago
Give the word equation for the neutralization reaction of an acid and a base.
Grace [21]

Answer:Acid+Base--> salt+water

Explanation:the reaction between acids and bases is called neutralization and will always produce salt and water

7 0
3 years ago
2. Determine the molarity of the NaOH solution in each trial. a. Trial 1 Molarity: b. Trial 2 Molarity: 3. Calculate the average
butalik [34]

Answer:

This question is incomplete

Explanation:

This question is incomplete but...

1) You can calculate the molarity of the NaOH for each trial by following the steps below.

The formula for Molarity (M) is

M = number of moles (n) ÷ volume (V)

where the unit of volume must be in Litres or dm³

The unit of molarity is mol/dm³ or mol/L or molar conc (M)

The final answer must have the unit of molarity

If the number of moles is not provided, look out for the mass of NaOH used and then calculate your number of moles (n) as

n = mass of NaOH used ÷ molar mass of NaOH

Where the atomic mass of sodium (Na) is 23, oxygen (O) is 16 and hydrogen (H) is 1. Hence, molar mass for NaOH is 23 + 16 + 1 = 40 g/mol

n = mass of NaOH used ÷ 40

2) Average Molarity will be (Trial 1 Molarity +Trial 2 Molarity) ÷ 2

Answer must be in mol/dm³ or mol/L or M

3) Label the volumentric flask containing the NaOH solution with the answer gotten from (2) above

3 0
3 years ago
What is the concentration of a solution of HCl in which a 10.0 mL sample of acid required 50.0 mL of 0.150 M NaOH for neutraliza
puteri [66]

Answer:

The answer is 0.75M HCl

Explanation:

To calculate the concentration of 10 ml of HCl that would be required to neutralize 50.0 mL of 0.150 M NaOH, we use the formula:

To calculate the concentration of 10 ml of HCl that would be required to neutralize 50.0 mL of 0.150 M NaOH, we use the formula:

C1V1 = C2V2

C1 = concentration of acid

C2 = concentration of base

V1 = volume of acid

V2 = volume of base

From the information supplied in the question:

concentration of acid (HCl) is the unknown

volume of acid (HCl) = 10ml

concentration of base (NaOH) = 0.15M

volume of base (NaOH) = 50ml

C1 x 10ml = 0.15M x 50ml

C1 x 10 = 7.5

divide both side by 10

C1 = 0.75M

concentration of acid (HCl) is 0.75M

5 0
3 years ago
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