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eimsori [14]
3 years ago
15

How are proteins carbohydrates and fats related to the discipline of chemistry

Chemistry
1 answer:
alexandr1967 [171]3 years ago
4 0

Answer:

Explanation:

The nutrition label on rice lists the amounts of proteins, carbohydrates, and fats in one serving. These substances are important for human nutrition. How are proteins, carbohydrates, and fats related to the discipline of chemistry? All are chemical compounds. All are found in living things. All are needed for nutrition. All are found in non-living things.

The answer for this question would be A) All are chemical compounds.

The answer is All are chemical compounds

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What is the average density of the Earth's inner core?
eduard
The answer is A hopefully this help you
5 0
3 years ago
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A 25 mL sample of 0.100 M HNO3 completely reacts with NaOH according to this equation:
Anastasy [175]
It is a good thing that you already have answered the first question. Now, moving on to the second question, there exist an equation for the neutralization of acid by a base that is shown below,
                                          M₁V₁ = M₂V₂
Now, all the variables in the equation are given except for our unknown which is the V₂. Substituting the known values from the given above,
                                       (0.1 M)(25 mL) = (0.05 M)(V₂)
The value of V₂ from the equation above is 50 mL. Therefore, 50 mL of 0.05 M NaOH solution will be needed to completely react with HNO3. 
6 0
4 years ago
Read 2 more answers
A volume of 120 mL of H2O is initially at room temperature (22.00 ∘C ). A chilled steel rod at 2.00 ∘C ∘C is placed in the water
aliina [53]

Answer:

\large \boxed{\text{28.5 g}}

Explanation:

There are two heat flows in this process and, since energy (heat) can neither be destroyed nor created, the energy change for the system must equal zero.

Data:

For Fe,    m₁ = ?;         C₁ = 0.452 J°C⁻¹g⁻¹; Ti =   2.00 °C; T_f = 21.50 °C

For H₂O, m₂ = 120 g; C₂ = 4.18    J°C⁻¹g⁻¹; Ti = 22.00 °C; T_f = 21.50 °C

Calculations:

1. Temperature changes

ΔT₁ = T_f - Ti = 21.50 °C -   2.00 °C = 19.50 °C

ΔT₂ = T_f - Ti = 21.50 °C - 22.00 °C = -0.50 °C

2. Mass of steel rod

\begin{array}{ccl}\text{Heat  gained by steel rod + heat lost by water} & = & 0\\m_{1}C_{1} \Delta T_{1} + m_{2}C_{2} \Delta T_{2}& = & 0\\m_{1} \times 0.452 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$}\times 19.50 \, ^{\circ}\text{C} + \text{120 g} \times 4.18\text{ J$^{\circ}$C$^{-1}$g$^{-1}$} \times (-0.50)\, ^{\circ}\text{C}& = & 0\\8.814m_{1}\text{ g}^{-1} - 250.8 & = &0\\\end{array}\\

\begin{array}{ccl}8.814m_{1}\text{ g}^{-1} & = &250.8\\m_{1} & = & \dfrac{250.8}{\text{8.814 g}^{-1}}\\\\& = & \textbf{28.5 g}\\\end{array}\\\text{The mass of the steel rod is $\large \boxed{\textbf{28.5 g}}$}

6 0
3 years ago
How many calories of heat are needed to raise the temperature of 33.0 g of diethyl ether from 14.0 ∘C to 25.0 ∘C ?
Xelga [282]

The amount of heat (in calories) needed to raise the temperature is 324.885 calories

<h3>How to determine the temperature change </h3>

We'll begin by obtaining the temperature change. This can be obtained as followed:

  • Initial temperature (T₁) = 14 °C
  • Final temperature (T₂) = 25 °C
  • Change in temperature (ΔT) = ?

ΔT = T₂ - T₁

ΔT = 25 - 14

ΔT = 11 °C

<h3>How to determine the heat (in Calories)</h3>

The amount of heat needed to raise the temperature can bee obtaimedals follow:

  • Mass (M) = 33 g
  • Change in temperature (ΔT) = 11 °C
  • Specific heat capacity (C) of diethyl ether = 0.895 cal/g°C
  • Heat (Q) =?

Q = MCΔT

Q = 33 × 0.895 × 11

Q = 324.885 calories

Thus, the amount of heat needed is 324.885 calories

Learn more about heat transfer:

brainly.com/question/10286596

#SPJ1

3 0
1 year ago
An inverted pyramid is being filled with water at a constant rate of 45 cubic centimeters per second. The pyramid, at the top, h
vaieri [72.5K]

Answer:

13.20 cm/s is the rate at which the water level is rising when the water level is 4 cm.

Explanation:

Length of the base = l

Width of the base  =  w

Height of the pyramid = h

Volume of the pyramid = V=\frac{1}{3}lwh

We have:

Rate at which water is filled in cube = \frac{dV}{dt}= 45 cm^3/s

Square based pyramid:

l = 6 cm, w = 6 cm, h = 13 cm

Volume of the square based pyramid = V

V=\frac{1}{3}\times l^2\times h

\frac{l}{h}=\frac{6}{13}

l=\frac{6h}{13}

V=\frac{1}{3}\times (\frac{6h}{13})^2\times h

V=\frac{12}{169}h^3

Differentiating V with respect to dt:

\frac{dV}{dt}=\frac{d(\frac{12}{169}h^3)}{dt}

\frac{dV}{dt}=3\times \frac{12}{169}h^2\times \frac{dh}{dt}

45 cm^3/s=3\times \frac{12}{169}h^2\times \frac{dh}{dt}

\frac{dh}{dt}=\frac{45 cm^3/s\times 169}{3\times 12\times h^2}

Putting, h = 4 cm

\frac{dh}{dt}=\frac{45 cm^3/s\times 169}{3\times 12\times (4 cm)^2}

=13.20 cm/s

13.20 cm/s is the rate at which the water level is rising when the water level is 4 cm.

3 0
3 years ago
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