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2 years ago

7

As Danny was pouring cereal for his breakfast, he noticed that the cereal box says that the cereal contains 5 milligrams of iron

per serving. Danny decided to perform an experiment to test this claim. He found that the cereal contains 4.6 milligrams of iron per serving. Based on Danny's measurement, the value reported on the label is ( CORRECT OR INCCORECT) because A: (IT IS DIFFERENT FROM THE VALUE DANNY MEASURED) B: (IT IS REORTED TO THE NEAREST MILLIGRAM) C: (IT USES THE SAME UNITS THAT DANNY USED).

1 answer:

Vinil7 [7]2 years ago

6 0

Answer:

Correct, because B it is reported to the nearest miligram

Explanation:

4.6 rounded up is 5

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Heavy nuclides with too few neutrons to be in the band of stability are most likely to decay by what mode?

Margarita [4]

Answer:

Positron emission

Explanation:

Positron emission involves the conversion of a proton to a neutron. This process increases the mass number of the daughter nucleus by 1 while its atomic number remains the same. The new neutron increases the number of neutrons present in the daughter nucleus hence the process increases the N/P ratio.

A positron is usually ejected in the process together with an anti-neutrino to balance the spins.

6 0

2 years ago

Devon’s laboratory is out of material to make phosphate buffer. He is considering using sulfate to make a buffer instead. The pk

sesenic [268]

Answer:

Is not possible to make a buffer near of 7.

Optimal pH for sulfate‑based buffers is 2.

Explanation:

The dissociations of H₂SO₄ are:

H₂SO₄ ⇄ H⁺ + HSO₄⁻ pka₁ = -10

HSO₄⁻ ⇄ H⁺ + SO₄²⁻ pka₂ = 2.

The buffering capacity is pka±1. That means that for H₂SO₄ the buffering capacity is in pH's between <em>-11 and -9 and between 1 and 3</em>, having in mind that pH's<0 are not useful. For that reason, <em>is not possible to make a buffer near of 7.</em>

The optimal pH for sulfate‑based buffers is when pka=pH, that means that optimal pH is <em>2.</em>

<em />

I hope it helps!

4 0

2 years ago

In acidic solution, the breakdown of sucrose into glucose and fructose has this rate law: rate = k[H+][sucrose].

Karo-lina-s [1.5K]

Answer:

a)If concentration of [Sucrose] is changed to 2.5 M than rate will be increased by the factor of 2.5.

b)If concentration of [Sucrose] is changed to 0.5 M than rate will be increased by the factor of 0.5.

c)If concentration of $[H^+]$ is changed to 0.0001 M than rate will be increased by the factor of 0.01.

d) If concentration when [sucrose] and $[H^+]$ both are changed to 0.1 M than rate will be increased by the factor of 1.

Explanation:

Sucrose + $H^+\rightarrow$ fructose+ glucose

The rate law of the reaction is given as:

$R=k[H^+][sucrose]$

$[H^+]=0.01M$

[sucrose]= 1.0 M

$R=k[0.01M][1.0 M]$ ..[1]

a)

The rate of the reaction when [Sucrose] is changed to 2.5 M = R'

$R'=[0.01 M][2.5 M]$ ..[2]

[2] ÷ [1]

$\frac{R'}{R}=\frac{[0.01 M][2.5 M]}{k[0.01M][1.0 M]}$

$R'=2.5\times R$

If concentration of [Sucrose] is changed to 2.5 M than rate will be increased by the factor of 2.5.

b)

The rate of the reaction when [Sucrose] is changed to 0.5 M = R'

$R'=[0.01 M][0.5 M]$ ..[2]

[2] ÷ [1]

$\frac{R'}{R}=\frac{[0.01 M][0.5 M]}{k[0.01M][1.0 M]}$

$R'=2.5\times R$

If concentration of [Sucrose] is changed to 0.5 M than rate will be increased by the factor of 0.5.

c)

The rate of the reaction when $[H^+]$ is changed to 0.001 M = R'

$R'=[0.0001 M][1.0 M]$ ..[2]

[2] ÷ [1]

$\frac{R'}{R}=\frac{[0.0001 M][1.0M]}{k[0.01M][1.0 M]}$

$R'=0.01\times R$

If concentration of $[H^+]$ is changed to 0.0001 M than rate will be increased by the factor of 0.01.

d)

The rate of the reaction when [sucrose] and $[H^+]$ both are changed to 0.1 M = R'

$R'=[0.1M][0.1M]$ ..[2]

[2] ÷ [1]

$\frac{R'}{R}=\frac{[0.1M][0.1M]}{k[0.01M][1.0 M]}$

$R'=1\times R$

If concentration when [sucrose] and $[H^+]$ both are changed to 0.1 M than rate will be increased by the factor of 1.

5 0

2 years ago

The specific heat capacity of a pure substance can be found by dividing the heat needed to change the temperature of a sample of

mrs_skeptik [129]

Answer:

The answers to your questions are given below.

Explanation:

Data obtained from the question include:

Mass (M) = 420.0 g

Temperature change (ΔT) = 43.8 °C

Specific heat capacity (C) = 3.52 J/g °C

Heat needed (Q) =...?

The heat needed for the temperature change can be obtained by using the following formula:

Q = MCΔT

Where:

Q is the heat needed measured in joule (J).

M is the mass of substance measured in grams (g)

C is the specific heat capacity of the substance with unit J/g °C.

ΔT is the temperature change measured in degree celsius (°C).

Thus, we can calculate the heat needed to change the temperature as follow:

Q = MCΔT

Q = 420 x 3.52 x 43.8

Q = 64753.92 J

Therefore, the heat needed to cause the temperature change is 64753.92 J

4 0

2 years ago

A 100 mL reaction vessel initially contains 2.60×10^-2 moles of NO and 1.30×10^-2 moles of H2. At equilibrium the concentration

Sliva [168]

Answer:

<h2>The equilibrium constant Kc for this reaction is 19.4760</h2>

Explanation:

The volume of vessel used= $100$ ml

Initial moles of NO= $\frac{2.60}{10^2}$ moles

Initial moles of H2= $\frac{1.30}{10^2}$ moles

Concentration of NO at equilibrium= $0.161$ M

$Concentration(in M)=\frac{moles}{volume(in litre)}$

Moles of NO at equilibrium= $0.161(\frac{100}{1000})$

= $\frac{1.61}{10^2}$ moles

2H2 (g) + 2NO(g) <—> 2H2O (g) + N2 (g)

<u>Initial</u> :1.3*10^-2 2.6*10^-2 0 0 moles

<u>Equilibrium</u>:1.3*10^-2 - x 2.6*10^-2-x x x/2 moles

∴ $\frac{2.60}{10^2}-x=\frac{1.61}{10^2}$

⇒ $x=\frac{0.99}{10^2}$

$Kc=\frac{[H2O]^2[N2]}{[H2]^2[NO]^2} (volume of vesselin litre)$

<u>Equilibrium</u>:0.31*10^-2 1.61*10^-2 0.99*10^-2 0.495*10^-2 moles

⇒ $Kc=\frac{(0.0099)^2(0.00495)}{(0.0031)^2(0.0161)^2} (0.1)$

⇒ $Kc=19.4760$

3 0

2 years ago