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8090 [49]
3 years ago
7

Need help don’t understand this??!

Chemistry
1 answer:
soldier1979 [14.2K]3 years ago
6 0
The first one is remove heat

Second is add heat
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Determine the correct name for the compound MG3N2
SSSSS [86.1K]
Mg3N2 is Magnesium nitride
5 0
3 years ago
A gas sample is found to contain 39.10% carbon, 7.67% hydrogen, 26.11% oxygen, 16.82% phosphorus, and 10.30% fluorine. If the mo
Doss [256]

Answer:

C6H14O3F

Explanation:

The first step is to divide each compound by its molecular weight

Carbon

= 39.10/12

= 3.258

Hydrogen

= 7.67/1

= 7.67

Oxygen

= 26.11/16

= 1.63

Phosphorous

= 16.82/31

= 0.542

Flourine

= 10.30/19

= 0.542

The next step is to divide by the lowes value

3.258/0.542

= 6 mol of C

7.67/0.542

= 14 mol of H

1.63/0.542

= 3 mol of O

0.542/0.542

= 1 mol of P

0.542/0.542

= 1 mol of F

Hence the molecular formula is C6H14O3F

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3 years ago
Is cell division a form of reproduction? Explain. Why do you think this?
ivolga24 [154]
Cell division is the means of reproduction in organisms which reproduce asexually
8 0
3 years ago
Do ladybugs have different numbers of spots because of gender ?
STatiana [176]

Answer:

Determining whether a ladybug is male or female has nothing to do with the size or number of black spots on the insect's orange body. Both sexes have these spots. The male is generally slightly smaller than the female. The ladybug's reproductive organs share the abdomen, along with the digestive and respiratory organs.

Explanation:

5 0
3 years ago
Read 2 more answers
A 8.249 gram sample of copper is heated in the presence of excess fluorine. A metal fluoride is formed with a mass of 13.18 g. D
levacccp [35]

Answer:

CuF_2 the empirical formula of the metal fluoride.

Explanation:

Mass of copper heated = 8.249 g

Mass of copper fluoride formed = 13.18 g

Mass of fluorine gas in copper fluoride = x

13.18 g = 8.249 g + x\\x= 13.18 - 8.249 g = 4.931 g

Moles of copper :

= \frac{8.249 g}{63.546 g/mol}=0.1298 mol

Moles of fluorine:

= \frac{4.931 g}{18.998 g/mol}=0.2596 mol

For the empirical formula divide the smallest mole of an element with all the moles of elements present in the compound.

Copper= \frac{0.1298 mol}{0.1298 mol}=1\\Fluorine = \frac{0.2596 mol}{0.1298 mol}=2

The empirical formula of the copper fluoride = CuF_2

CuF_2 the empirical formula of the metal fluoride.

6 0
3 years ago
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