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frutty [35]
3 years ago
15

PLEASE HELP 2) Common Factors: 2 cups = 1 pint 2 pints = 1 quart 4 quarts = 1 gallon

Chemistry
1 answer:
lyudmila [28]3 years ago
6 0
I think the answer might be 6 cups
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klio [65]
I think its cubic meters....
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3 years ago
Object a has a mass of 12 G and a volume of 8 cm3 object be has a mass of 20 G and a volume of 8 cm3 which object has a greater
solniwko [45]

Answer:

Option C = object B by 1 gram per cubic cm.

Explanation:

Given data:

Mass of object A = 12 g

Volume of object A = 8 cm³

Mass of object B = 20 g

Volume of object B = 8 cm³

Densities = ?

Solution:

Density:

Density is equal to the mass of substance divided by its volume.

Units:

SI unit of density is Kg/m3.

Other units are given below,

g/cm3, g/mL , kg/L

Formula:

D=m/v

D= density

m=mass

V=volume

Symbol:

The symbol used for density is called rho. It is represented by ρ. However letter D can also be used to represent the density.

Density of object A:

d = m/v

d = 12 g/ 8 cm³

d = 1.5 g/cm³

Density of object B:

d = m/v

d = 20 g/ 8 cm³

d = 2.5 g/cm³

object b has high density.

5 0
3 years ago
What does the MC1R gene code for
zlopas [31]
The MC1R gene provides instructions for making a protein called the melanocortin 1 receptor. This receptor plays an important role in normal pigmentation. The receptor is primarily located on the surface of melanocytes, which are specialized cells that produce a pigment called melanin.
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3 years ago
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I don’t understand what its means
docker41 [41]

Answer:

me neither as well

Explanation:

I don't know sorry

6 0
3 years ago
How many L of 3.0 M H2SO4 solution can be prepared by using 100.0 mL OF 18 M H2SO4?
max2010maxim [7]

Answer: A volume of 600 mL of 3.0 M H_{2}SO_{4} solution can be prepared by using 100.0 mL OF 18 M H_{2}SO_{4}.

Explanation:

Given: V_{1} = ?,        M_{1} = 3.0 M\\

V_{2} = 100.0 mL,       M_{2} = 18 M

Formula used to calculate the volume is as follows.

M_{1}V_{1} = M_{2}V_{2}

Substitute the values into above formula as follows.

M_{1}V_{1} = M_{2}V_{2}\\3.0 M \times V_{1} = 18 M \times 100.0 mL\\V_{1} = \frac{18 M \times 100.0 mL}{3.0M}\\= 600 mL

Thus, we can conclude that a volume of 600 mL of 3.0 M H_{2}SO_{4} solution can be prepared by using 100.0 mL OF 18 M H_{2}SO_{4}.

6 0
3 years ago
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