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Phoenix [80]
2 years ago
14

Does heating a cup of water allow it to dissolve more sugar? Temperature of the water is measured in degrees centigrade. Amount

of sugar is measured in grams
Chemistry
1 answer:
Alexeev081 [22]2 years ago
6 0

Answer:

Yes

Explanation:

Yes heating water allows it to dissolve more Sugars because the molecular distance increases and this distance can be covered by more sugar. In the given question, The independent variable would be the temperature of water.

Since to whatever temperature the water boils at the boiling temperature of  does not change remains hundred degree. Rest all the variables can vary the weight of the amount of sugar with the variable in the temperature of Boiling of water to remain constant.

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Under which conditions would the solubility of a gas be greatest? high pressure and high temperature high pressure and low tempe
kiruha [24]

Option B: high pressure and low temperature

A gas is more soluble under high pressure and low temperature conditions.

On increasing temperature of a gas, its kinetic energy increases. The increase in kinetic energy increases the motion of particles of gas this causes most of the gaseous particles to escape from the gas phase. Thus, less particles are available to dissolve in liquid and solubility decreases.

The effect of pressure on solubility of gas can be explained with the help of Henry's law. According to the law, at constant temperature, solubility of gas and partial pressure of gas are related to each other as follows:

p=k_{h}c

Here, p is the partial pressure of the gas, k_{h} is Henry's law constant, and

c is the concentrate of the gas.

According to above relation, concentration of gas decreases on decreasing partial pressure. Thus, on increasing pressure,  concentration of gas increases this increases the solubility of gas in liquid.

Therefore, solubility of gas is greatest at high pressure and low temperature.


4 0
3 years ago
Read 2 more answers
You are provided with a stock solution with a concentration of 1.0x10-5 M. You will be using this to make two standard solutions
artcher [175]

Answer:

1. V₁ = 2.0 mL

2. V₁ = 2.5 mL

Explanation:

<em>You are provided with a stock solution with a concentration of 1.0 × 10⁻⁵ M. You will be using this to make two standard solutions via serial dilution.</em>

To calculate the volume required (V₁) in each dilution we will use the dilution rule.

C₁ . V₁ = C₂ . V₂

where,

C are the concentrations

V are the volumes

1 refers to the initial state

2 refers to the final state

<em>1. Perform calculations to determine the volume of the 1.0 × 10⁻⁵ M stock solution needed to prepare 10.0 mL of a 2.0 × 10⁻⁶ M solution.</em>

C₁ . V₁ = C₂ . V₂

(1.0 × 10⁻⁵ M) . V₁ = (2.0 × 10⁻⁶ M) . 10.0 mL

V₁ = 2.0 mL

<em>2. Perform calculations to determine the volume of the 2.0 × 10⁻⁶ M solution needed to prepare 10.0 mL of a 5.0 × 10⁻⁷ M solution.</em>

C₁ . V₁ = C₂ . V₂

(2.0 × 10⁻⁶ M) . V₁ = (5.0 × 10⁻⁷ M) . 10.0 mL

V₁ = 2.5 mL

8 0
3 years ago
If the density of gold is 19.3 g/cm3 what would be the volume of 550 g of gold?​
RUDIKE [14]

Answer:

28.497 cm3

Explanation:

Formula

D=m/v

Given data:

density = 19.3g/cm3

mass = 550 g

Now we will put the values in formula:

V=m/d

V=550 g/ 19.3 g/cm3  = 28.497 cm3

So the volume of gold is 28.497 cm3.

7 0
3 years ago
How many grams of HNO3 are produced when 59.0 g of NO2 completely reacts?
Anettt [7]

Answer:

53.7 grams of HNO3 will be produced

Explanation:

Step 1: Data given

Mass of NO2 = 59.0 grams

Molar mass NO2 = 46.0 g/mol

Step 2: The balanced equation

3NO2 + H2O→ 2HNO3 + NO

Step 3: Calculate moles NO2

Moles NO2 = 59.0 grams / 46.0 g/mol

Moles NO2 = 1.28 moles

Step 4: Calculate moles HNO3

For 3 moles NO2 we need 1 mol H2O to produce 2 moles HNO3 and 1 mol NO

For 1.28 moles NO2 we'll have 2/3 * 1.28 =0.853 moles HNO3

Step 7: Calculate mass HNO3

Mass HNO3 = 0.853 moles * 63.01 g/mol

Mass HNO3 = 53.7 grams

53.7 grams of HNO3 will be produced

3 0
3 years ago
You need to produce a buffer solution that has a pH of 5.50. You already have a solution that contains 10 mmol (millimoles) of a
balandron [24]

Answer:

56.9 mmoles of acetate are required in this buffer

Explanation:

To solve this, we can think in the Henderson Hasselbach equation:

pH = pKa + log ([CH₃COO⁻] / [CH₃COOH])

To make the buffer we know:

CH₃COOH  +  H₂O  ⇄   CH₃COO⁻  +  H₃O⁺     Ka

We know that Ka from acetic acid is: 1.8×10⁻⁵

pKa = - log Ka

pKa = 4.74

We replace data:

5.5 = 4.74 + log ([acetate] / 10 mmol)

5.5 - 4.74 = log ([acetate] / 10 mmol)

0.755 = log ([acetate] / 10 mmol)

10⁰'⁷⁵⁵ = ([acetate] / 10 mmol)

5.69 = ([acetate] / 10 mmol)

5.69 . 10 = [acetate] → 56.9 mmoles

6 0
3 years ago
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