3 points Given 0.65 moles KNO3: (a) Convert to grams (b) Convert to particles. <br>

1. How many joules must be added to 10.0 g of water to raise its temperature from 10°C to<br> 15°C?

weqwewe [10]

Answer:

209.3 Joules require to raise the temperature from 10 °C to 15 °C.

Explanation:

Specific heat capacity:

It is the amount of heat required to raise the temperature of one gram of substance by one degree.

Formula:

Q = m × c × ΔT

Given data:

mass of water = 10 g

initial temperature T1= 10 °C

final temperature T2= 15 °C

temperature change =ΔT= T2-T1 = 15°C - 10°C = 5 °C

Energy or joules added to increase the temperature Q = ?

Solution:

We know that specific heat of water is 4.186 J/g .°C

Q = m × c × ΔT

Q = 10 g × 4.186 J/g .°C × 5 °C

Q = 209.3 J

6 0

3 years ago

An objects volume can be found by adding its mass by its?

Inessa05 [86]

Volume* density=MASS

4 0

3 years ago

Given that w for water is 2. 4×10−14 m^2 at 37 °C. Calculate the ph of a neutral aqueous solution at 37 °c, which is the normal

alexgriva [62]

The pH of a neutral aqueous solution at 37°C is 6.8.

Kw is defined as the dissociation, which is also known as self-ionization, constant of water. this is an equilibrium constant, and its expression is:

Kw = [OH⁻] . [H₃O⁺]

Neutral pH determines that the concentrations of OH⁻ and H₃O⁺ are equal.

<h3>Calculation</h3>

Let us suppose concentration of OH and H₃O⁺ is x, to calculate it:

Kw =[OH⁻] . [H₃O⁺] = x²

x² = 2.4 × 10⁻¹⁴ M²

x = 1.5919 × 10⁻⁷ M

Hence, the concentration of OH and H₃O⁺ (x) = [H₃O⁺] = [OH⁻] = 1.5919×10⁻⁷ M

pH = -log[H₃O⁺] = -log( 1.5919×10⁻⁷ M)

pH = 6.8

Thus, we find that the pH of a neutral aqueous solution at 37 °c (which is the normal human body temperature) is 6.8.

learn more about pH:

brainly.com/question/9529394

#SPJ4

3 0

1 year ago

Aspirin (C9H8O4) is synthesized by reacting salicylic acid (C7H6O3) with acetic anhydride (C4H6O3). The balanced equation isC7H6

prisoha [69]

<u>Answer:</u>

<u>For a:</u> The mass of acetic anhydride needed is 73.91 grams.

<u>For b:</u> The theoretical yield of aspirin is 130.43 grams.

<u>Explanation:</u>

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$ .....(1)

<u>For a:</u>

Given mass of salicylic acid = $1.00\times 10^2g=100g$

Molar mass of salicylic acid = 138.121 g/mol

Putting values in equation 1, we get:

$\text{Moles of salicylic acid}=\frac{100g}{138.121g/mol}=0.724mol$

For the given chemical reaction:

$C_7H_6O_3+C_4H_6O_3\rightarrow C_9H_8O_4+CH_3COOH$

By stoichiometry of the reaction:

1 mole of salicylic acid reacts with 1 mole of acetic anhydride.

So, 0.724 moles of salicylic acid will react with = $\frac{1}{1}\times 0.724=0.724mol$ of acetic anhydride.

Now, to calculate the mass of acetic anhydride, we use equation 1:

Moles of acetic anhydride = 0.724 moles

Molar mass of acetic anhydride = 102.09 g/mol

Putting values in equation 1, we get:

$0.724mol=\frac{\text{Mass of acetic anhydride}}{102.09g/mol}\\\\\text{Mass of acetic anhydride}=73.91g$

Hence, the mass of acetic anhydride needed is 73.91 grams.

<u>For b:</u>

By stoichiometry of the reaction:

1 mole of salicylic acid is producing 1 mole of aspirin.

So, 0.724 moles of salicylic acid will produce = $\frac{1}{1}\times 0.724=0.724mol$ of aspirin.

Now, to calculate the mass of aspirin, we use equation 1:

Moles of aspirin = 0.724 moles

Molar mass of aspirin = 180.158 g/mol

Putting values in equation 1, we get:

$0.724mol=\frac{\text{Mass of aspirin}}{180.158g/mol}\\\\\text{Mass of aspirin}=130.43g$

Hence, the theoretical yield of aspirin is 130.43 grams.

4 0

3 years ago

A sample of gas occupies a volume of 61.5 mL . As it expands, it does 130.1 J of work on its surroundings at a constant pressure

Lesechka [4]

Answer:

the final volume of the gas is $V_2$ = 1311.5 mL

Explanation:

Given that:

a sample gas has an initial volume of 61.5 mL

The workdone = 130.1 J

Pressure = 783 torr

The objective is to determine the final volume of the gas.

Since the process does 130.1 J of work on its surroundings at a constant pressure of 783 Torr. Then, the pressure is external.

Converting the external pressure to atm ; we have

External Pressure $P_{ext}$ :

$P_{ext} = 783 \ torr \times \dfrac{1 \ atm}{760 \ torr}$

$P_{ext} = 1.03 \ atm$

The workdone W = $P_{ext}$ V

The change in volume ΔV= $\dfrac{W}{P_{ext}}$

ΔV = $\dfrac{130.1 \ J \times \dfrac{1 \ L \ atm}{ 101.325 \ J} }{1.03 \ atm }$

ΔV = $\dfrac{1.28398717 }{1.03 }$

ΔV = 1.25 L

ΔV = 1250 mL

Recall that the initial volume = 61.5 mL

The change in volume V is $\Delta V = V_2 -V_1$

$- V_2= - \Delta V -V_1$

multiply through by (-), we have:

$V_2= \Delta V+V_1$

$V_2$ = 1250 mL + 61.5 mL

$V_2$ = 1311.5 mL

∴ the final volume of the gas is $V_2$ = 1311.5 mL

5 0

3 years ago

3 points Given 0.65 moles KNO3: (a) Convert to grams (b) Convert to particles. ​

3 points Given 0.65 moles KNO3: (a) Convert to grams (b) Convert to particles.