The ph of coffee is 5.6. The ph of grapefruit juice is 2.6. This means that the proton concentration in coffee is

Answers

Answer 1

Answer:

pH scale is used to measure the acidity or alkalinity on a scale of 0-14 where 0-6.9 is the acidic region , 7.1-14 is the basic region and 7 is for the neutral substance. We can calculate the concentration of proton from pH.

$pH=-log[H^+]$

where $[H^+]$ is the concentration of proton

As per the question ,the pH of coffee is 5.6 and we need to find the concentration of proton so putting the values in the above equation, we get

$5.6=-log[H^+]$

$-5.6=log[H^+]$

$antilog(-5.6)= [H^+]$

$[H^+]= 2.511*10^-^6 M$

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What volume will 12 g of oxygen gas (O2) occupy at 25 °C and a pressure of 53 kPa?

Answers

ANSWER

The volume of the oxygen gas is 17.5 L

EXPLANATION

Given that;

The mass of oxygen gas is 12 grams

The temperature of the gas is 25 degrees Celcius

The pressure of the gas is 53 kPa

To find the volume of the oxygen gas, follow the steps below

Step 1; Assume the gas behaves like an ideal gas

Therefore, apply the ideal gas equation to find the volume of the gas

$\text{ PV }=\text{ nRT}$

Where

P is the pressure of the gas

V is the volume of the gas

n is number of moles of the gas

R is the universal gas constant

T is the temperature of the gas

Step 2: Find the number of moles of the oxygen gas using the below formula

$\text{ mole }=\text{ }\frac{\text{ mass}}{\text{ molar mass}}$

Recall, that the molar mass of the oxygen gas is 32 g/mol

$\begin{gathered} \text{ mole }=\text{ }\frac{12}{\text{ 32}} \n \text{ mole }=\text{ 0.375 mol} \end{gathered}$

Step 3; Convert the temperature to degree Kelvin

$\begin{gathered} \text{ T }=\text{ t }+\text{ 273.15} \n \text{ t }=\text{ 25}\degree C \n \text{ T }=25\text{ }+\text{ 273.15} \n \text{ T }=\text{ 298.15K} \end{gathered}$

Step 4; Substitute the given data into the formula in step 1

Recall, that R is 8.314 L kPa K^-1 mol^-1

$\begin{gathered} \text{ 53 }*\text{ V }=\text{ 0.375}*\text{ 8.314}*\text{ 298.15} \n \text{ 53V }=\text{ 929.557} \n \text{ Divide both sides by 53} \n \text{ }\frac{\cancel{53}V}{\cancel{53}}\text{ }=\text{ }(929.557)/(53) \n \text{ V }=\text{ }(929.557)/(93) \n \text{ V }=\text{ 17.5 L} \end{gathered}$

Hence, the volume of the oxygen gas is 17.5 L

3.What do we call materials
that let heat pass through
them easily?
Thermal conductors
Thermal insulators
Transparent

4.
Which of these is a good
thermal conductor?
Plastic
Wood
Steel

5.
Which of these is a good
thermal insulator?
Steel
Iron
Polystyrene

6.
To save on heating bills, do
you think the roof of a
building should be lined with...
a thermal conductor
a thermal insulator
nothing

7.
How does heat travel?
From cold things to hotter things
From hot things to colder things
Between things of the same temperature

Answers

Answer:

thermal conductors

steel

polystyrene

thermal insulator

between things of the same temp?

sorry to ask but if its ok with you, may i get brainly, i need to rank up all i need is two more, if not its fine. thank you and yours truly golden

PLEASE HELP ME I WILL MARK AS BRAINLIEST ANSWER-100 points When we look at a plant, we notice that the leaves are green. This
is due to the (4). in the leave that allows the plant to absorb
sunlight. Plants use the light energy in a reaction called (5).
where they make food in the form of sugar and produce oxygen.
(6)___ which are only found in plant cells, is where the food
making reaction takes place. In this reaction: water, carbon dioxide,
and energy from the sun are needed for the reaction and are called
(7) The things that are formed in the reaction, or the
(8). are sugar and oxygen.

Answers

Answer: 4 chlorophyll 5 and photosynthesis 6 sugar glucose

Explanation:I'm not sure of seven or eight

Answer:

4:Chlorophyll. 6. Photosynthesis 8. Glucose

Explanation:

Express the equilibrium constant for the combustion of propane in the balanced chemical equation c3h8(g)+5o2(g)???3co2(g)+4h2o(g)

Answers

according to the balanced equation:
C3H8(g) + 5O2(g) ↔ 3CO2(g) + 4 H2O (g)
aA + bB ↔ cC + dD

according to K formula when:
K = concentration of the products / concentration of the reactants

K = [C]^c[D]^d / [A]^a[B]^b
when [A],[B],[C]and[D] is the concentrations
and a,b,c and d is the no of moles
∴ K = [CO2]^3[H2O]^4 / [C3H8] [ O2]^5

Answer:

combustion reaction

Explanation:

there is oxides in the equation c3h8(g)+5o2(g)3co2(g)+4h2o(g)

Be sure to answer all parts. The first step in HNO3 production is the catalyzed oxidation of NH3. Without a catalyst, a different reaction predominates: 4 NH3(g) +3 O2 (g) ⇌ 2 N2(g) + 6 H2O(g) When 0.0150 mol of NH3(g) and 0.0150 mol of O2(g) are placed in a 1.00−L container at a certain temperature, the N2 concentration at equilibrium is 1.96 × 10−3 M. Calculate Kc.

Answers

Answer: The value of $K_c$ for the reaction is $6.005* 10^(-6)$

Explanation:

We are given:

Initial moles of $NH_3=0.0150mol$

Initial moles of $O_2=0.0150mol$

Volume of the container = 1.00 L

Molarity of the solution = $\frac{\text{Number of moles}}{\text{Volume of container}}$

$[NH_3]_i=(0.0150)/(1.00)=0.0150M$

$[O_2]_i=(0.0150)/(1.00)=0.0150M$

The given chemical equation follows:

$4NH_3(g)+3O_2(g)\rightarrow 2N_2(g)+6H_2O(g)$

Initial: 0.0150 0.0150

At eqllm: 0.0150-4x 0.0150-3x 2x 6x

The expression of $K_c$ for above equation follows:

$K_c=([N_2]^2[H_2O]^6)/([NH_3]^4[O_2]^3)$ .......(1)

We are given:

Equilibrium concentration of $N_2=1.96* 10^(-3)$

Equating the equilibrium concentrations of nitrogen, we get:

$2x=1.96* 10^(-3)\n\nx=0.98* 10^(-3)M$

Calculating the equilibrium concentrations:

Concentration of $NH_3=(0.0150-4x)=0.0150-4(0.00098)=0.01108M$

Concentration of $O_2=(0.0150-3x)=0.0150-3(0.00098)=0.01206M$

Concentration of $N_2=2x=2(0.00098)=0.00196M$

Concentration of $H_2O=6x=6(0.00098)=0.00588M$

Putting values in expression 1, we get:

$K_c=((0.00196)^2* (0.00588)^6)/((0.01108)^4* (0.01206)^3)\n\nK_c=6.005* 10^(-6)$

Hence, the value of $K_c$ for the reaction is $6.005* 10^(-6)$

Final answer:

To calculate the equilibrium constant, Kc, for the reaction that produces HNO3 from NH3 and O2, you need to determine the equilibrium concentrations of NH3 and O2. The given information includes the initial moles and concentration of NH3 and O2, as well as the equilibrium concentration of N2. Using the stoichiometry of the reaction and the given data, you can calculate the equilibrium concentrations and substitute them into the Kc expression to determine the numerical value of Kc.

Explanation:

The question asks to calculate the equilibrium constant, Kc, for the reaction that produces HNO3 from NH3 and O2. The reaction equation is 4 NH3(g) + 3 O2(g) ⇌ 2 N2(g) + 6 H2O(g). The given information is that 0.0150 mol of NH3(g) and 0.0150 mol of O2(g) are placed in a 1.00-L container, and the N2 concentration at equilibrium is 1.96 × 10−3 M. To solve for Kc, we need to calculate the equilibrium concentrations of NH3 and O2.

Using the stoichiometry of the reaction, we can determine that the equilibrium concentration of NH3 is (0.0150 - 2*1.96 × 10−3) M and the equilibrium concentration of O2 is (0.0150 - 3*1.96 × 10−3) M. Substituting these values into the equilibrium expression for Kc, we can calculate the value of Kc.

In this case, the equilibrium constant, Kc, can be calculated as [N2]^2 / ([NH3]^4 * [O2]^3). Substitute the given equilibrium concentration of N2 and the calculated equilibrium concentrations of NH3 and O2 into the Kc expression to determine the numerical value of Kc.

Learn more about Calculating equilibrium constant here:

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