False
Fact: Mammals and plants belong to the same domain, the Eukarya domain.
Evidence :All the organisms that possess a eukaryotic cell, plants, animals, protists, and fungi are in the Eukarya domain.
Explanation:
Given that,
Mass of metal cube, m = 20 g
Volume of cube, V =5 mL
We need to find the density of the cube. Mass per unit volume equals density.
So, the density of the cube is 4 g/ml.
We know that the density of Aluminium is 4 g/mL.
So, the cube is not Aluminium cube.
<h3>
Answer:</h3>
0.012 dekameters (dkm)
<h3>
Explanation:</h3>
<u>We are given;</u>
Required to identify the measurements that is not equivalent to 120 cm.
- Centimeters are units that are used to measure length together with other units such as kilometers(km), meters (m), millimeters (mm), dekameters (dkm), etc.
- These units can be inter-converted to one another using suitable conversion factors.
- To do this, we are going to have a table showing the suitable conversion factor from one unit to another.
Kilometer (km)
10
Decimeter (Dm)
10
Hectometer (Hm)\
10
Meter (m)
10
Dekameter (dkm)
10
Centimeter (cm)
10
Millimeter (mm)
Therefore;
To convert cm to km
Conversion factor is 10^5 cm/km
Thus;
120 cm = 120 cm ÷ 10^5 cm/km
= 0.0012 km
To convert cm to dkm
Conversion factor is 10 cm/dkm
Therefore,
120 cm = 120 cm ÷ 10 cm/dkm
= 12 dkm
To convert cm to m
The suitable conversion factor is 10^2 cm/m
Thus,
120 cm = 120 cm ÷ 10^2 cm/m
= 1.2 m
To convert cm to mm
Suitable conversion factor is 10 mm/cm
Therefore;
120 cm = 120 cm × 10 mm/cm
= 1200 mm
Therefore, the measurement that is not equal to 120 cm is 0.012 dkm
The answer for the following problem is described below.
<em><u> Therefore the standard enthalpy of combustion is -2800 kJ</u></em>
Explanation:
Given:
enthalpy of combustion of glucose(Δ
of 
) =-1275.0
enthalpy of combustion of oxygen(Δ of 
) = zero
enthalpy of combustion of carbon dioxide(Δ of 
) = -393.5
enthalpy of combustion of water(Δ of 
) = -285.8
To solve :
standard enthalpy of combustion
We know;
Δ = ∈Δ (products) - ∈Δ (reactants)
(s) +6 (g) → 6 (g)+ 6 (l)
Δ = [6 (-393.5) + 6(-285.8)] - [6 (0) + (-1275)]
Δ = [6 (-393.5) + 6(-285.8)] - [0 - 1275]
Δ = 6 (-393.5) + 6(-285.8) - 0 + 1275
Δ = -2361 - 1714 - 0 + 1275
Δ =-2800 kJ
<em><u> Therefore the standard enthalpy of combustion is -2800 kJ</u></em>
Isn't this a math problem?
If it is the the answer should be 102.
10 decimeters=1 meter
27x10=270
270-168=102
