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olga_2 [115]
3 years ago
12

An organism’s genotype is its ____.

Chemistry
1 answer:
never [62]3 years ago
4 0
B.Physical appearance
You might be interested in
Which type of nuclear radiation is being emitted here?
tekilochka [14]

Answer:

b.Beta

Explanation:

mass number remains constant while atomic number has been increased by 1 unit . beta is electron like element where its mass number is 0 and atomic number is -1.

7 0
3 years ago
What is the mass of volume= 5.4 mL  density= 2.5 g/mL​
ddd [48]

Answer:

13.5g

Explanation:

Mass is defined as the measure of the amount of matter in an object. Its unit is kg or g.

Mass can be calculated using the formula:

m= d × v

where,

d= density

m= mass

v= volume

m= 2.5×5.4

m= 13.5g

5 0
3 years ago
A hydrocarbon sample was burned in a bomb calorimeter. The temperature of the calorimeter and the 1.00 kg of water rose from 20.
fomenos

Answer:

The heat released by the combustion is 20,47 kJ

Explanation:

Bomb calorimeter is an instrument used to measure the heat of a reaction. The formula is:

Q = C×m×ΔT + Cc×ΔT

Where:

Q is the heat released

C is specific heat of water (4,186kJ/kg°C)

m is mass of water (1,00kg)

ΔT is temperature change (23,65°C - 20,45°C)

And Cc is heat capacity of the calorimeter (2,21kJ/°C)

Replacing these values the heat released by the combustion is:

<em>Q = 20,47 kJ</em>

6 0
3 years ago
A sample compound contains 5.723g Ag, 0.852g S and 1.695g O. Determine its empirical formula.
Lubov Fominskaja [6]

Answer:

Ag_2SO_4

Explanation:

Formula for the calculation of no. of Mol is as follows:

mol=\frac{mass\ (g)}{molecular\ mass}

Molecular mass of Ag = 107.87 g/mol

Amount of Ag = 5.723 g

mol\ of\ Ag=\frac{5.723\ g}{107.87\ g/mol} =0.05305\ mol

Molecular mass of S = 32 g/mol

Amount of S = 0.852 g

mol\ of\ S=\frac{0.852\ g}{32\ g/mol} =0.02657\ mol

Molecular mass of O = 16 g/mol

Amount of O = 1.695 g

mol\ of\ O=\frac{1.695\ g}{16\ g/mol} =0.10594\ mol

In order to get integer value, divide mol by smallest no.

Therefore, divide by 0.02657

Ag, \frac{0.05305}{0.02657} \approx 2

S, \frac{0.02657}{0.02657} \approx 1

O, \frac{0.10594}{0.02657} \approx 4

Therefore, empirical formula of the compound = Ag_2SO_4

7 0
3 years ago
Read 2 more answers
What STARTING element undergoes alpha decay to yield uranium-234 (U) as a PRODUCT?
nadezda [96]
Uranium-238 is the element that undergoes decay to yield uranium-234
3 0
3 years ago
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