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2 years ago

The relative formula mass (M) of XCO, is 84

Chemistry

1 answer:

7nadin3 [17]2 years ago

5 0

Answer:

x = 56

relative atomic mass is 58.933

Explanation:

finding the relative molecular mass of x first...

x + (C ×1) + (O ×1) =84

x + (12 ×1) + ( 16 × 1) =84

x + 12 + 16 =84

collect like terms

x = 84-12+16

x= 56

now...

half of 56...

56÷2 =27

and 27 is the atomic number of cobalt...

so relative atomic mass of cobalt is 58.933.

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How many atoms of hydrogen combine with one carbon atom to form methane

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Explanation:

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2 years ago

When bases dissolve in water they release what type of negative ion?

Leto [7]

A base generally releases a hydroxide ion (OH-) when dissolved in water.

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3 years ago

A sample of propane(c3h8)has 3.84x10^24 H atoms.

deff fn [24]

Answer:

A) 14. 25 × 10²³ Carbon atoms

B) 34.72 grams

Explanation:

1 molecule of Propane has 3 atoms of Carbon and 8 atoms of Hydrogen.

The sample has 3.84 × 10²⁴ H atoms.

If 8 atoms of Hydrogrn are present in 1 molecule of propane.

3.84 × 10²⁴ H atoms are present in

$\mathfrak{ \frac{3.8 }{8} \times 10 ^{24}}$

= 4.75 × 10²³ molecules of Propane.

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

No. of Carbon atoms in 1 molecule of propane = 3

=> C atoms in 4.75× 10²³ molecules of Propane = 3 × 4.75 × 10²³

= 14.25 × 10²³ 

________________________________________

Gram Molecular Mass of Propane(C3H8)

= 3 × 12 + 8 × 1

= 36 + 8

= 44 g

1 mole of propane weighs 44g and has 6.02× 10²³ molecules of Propane.

=> 6.02 × 10²³ molecules of Propane weigh = 44 g

=> 4. 75 × 10²³ molecules of Propane weigh =

$\mathsf{ \frac{44 }{6.02 \times {10}^{23} } \times 4.75 \times {10}^{23} }$

$\mathsf{ = \frac{44 }{6.02 \times \cancel{{10}^{23} }} \times 4.75 \times \cancel{ {10}^{23}} }$

$\mathsf{ = \frac{44 }{6.02 } \times 4.75 }$

= 34.72 g

8 0

2 years ago

How much heat is needed warm 65.34 g of water from 18.43 to 21.75

BaLLatris [955]

Given that

Mass of water = 65.34 g

Amount of heat = mass of water * specific heat (temperature change

)

= 65.34 g * 4.184 J / g-C ( 21.75-18.43 )C

= 907.63 J

= 0.908 KJ

And

1 cal = 4.186798 J

907.63 J * 1 cal / 4.186798 J =216.78 cal

Or0.218 kcal

5 0

2 years ago