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KatRina [158]
3 years ago
13

A brownish fossil that can be found in rock and contains the remains of an organism is

Chemistry
1 answer:
My name is Ann [436]3 years ago
8 0
If it contains a fossil it must be a sedimentary rock.
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how much heat energy is needed to raise the temperature of 78.4g of aluminum from 19.4 degrees c to 98.6 degrees c.
Iteru [2.4K]

 The heat that is needed  to raise the   temperature  of 78.4 g of aluminium from 19.4 °c to 98.6°c  is    5600.77 j

 <u><em>calculation</em></u>

Heat(Q) = mass(M) x  specific heat capacity (C) x change in temperature(ΔT)


where;

Q=?

M = 78. 4 g

C=0.902 j/g/c

ΔT=98.6°c -19.4°c =79.2°c

Q is therefore = 78.4 g  x 0.902 j/g/c  x 79.2°c  =5600.77 j

7 0
3 years ago
Which of the following is NOT an example of a chemical change?
Katyanochek1 [597]

Answer:

hi there !!

B) Water evaporates into steam.

this is correct because only physical state changes, steam can be cooled down to get water again.

4 0
3 years ago
Read 2 more answers
(04.02 LC) Based on the cell theory, which of the following is true?
prohojiy [21]

Answer:

c

Explanation:

there are billions of cells in living things. We are made out of millions of cells

4 0
3 years ago
Atomic emission spectra are produced when the light emitted by an element passes through a prism. Which of the following stateme
xenn [34]
Option C: elements produce spectra with only few distinct lines.

The spectra are not continuos and are different for every element. This permits to identify elements.
3 0
3 years ago
Red #40 has an acute oral LD50 of roughly 5000 mg dye/1 kg body weight. This means if you had a mass of 1 kg, ingesting 5000 mg
FrozenT [24]

Answer:

350 g dye

0.705 mol

2.9 × 10⁴ L

Explanation:

The lethal dose 50 (LD50) for the dye is 5000 mg dye/ 1 kg body weight. The amount of dye that would be needed to reach the LD50 of a 70 kg person is:

70 kg body weight × (5000 mg dye/ 1 kg body weight) = 3.5 × 10⁵ mg dye = 350 g dye

The molar mass of the dye is 496.42 g/mol. The moles represented by 350 g are:

350 g × (1 mol / 496.42 g) = 0.705 mol

The concentration of Red #40 dye in a sports drink is around 12 mg/L. The volume of drink required to achieve this mass of the dye is:

3.5 × 10⁵ mg × (1 L / 12 mg) = 2.9 × 10⁴ L

8 0
3 years ago
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