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daser333 [38]
2 years ago
12

Which contain covalent bonds?

Chemistry
1 answer:
Ilia_Sergeevich [38]2 years ago
3 0
The only compound that contains covalent bonds would be A. BCl4-.
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3. What is the volume of a salt crystal measuring 2.44 x 10-2 m by 1.4 x 10-3 m by 8.4 x 10-3 m?
AlexFokin [52]

Answer:

2.9× 10⁻⁷ m³

Explanation:

Given data:

Measurements of crystal = 2.44 × 10⁻² m by  1.4 × 10⁻³ m by 8.4  × 10⁻³ m

Volume = ?

Solution:

Volume is equal to the length by width by height.

<em>So from given measurement value:</em>

Volume =  2.44 × 10⁻² m by  1.4 × 10⁻³ m by 8.4  × 10⁻³ m

Volume = 2.9× 10⁻⁷ m³

So the volume from given values is 2.9× 10⁻⁷ m³.

8 0
2 years ago
During this reaction, p4 + 5o2 → p4o10, 0.800 moles of product was made in 15.0 seconds. what is the rate of reaction? 56.8 g/mi
Eduardwww [97]
The solution for this problem would be:
The mass of P4O10 is computed by: 0.800 mol x 284 g/mol = 227g t = 15.0 s ( 1 min / 60 s) = 0.25 min 
So solving for the rate will be mass over t = m/t = 227/0.25 = 908 g/min would be the answer for this problem.
3 0
2 years ago
In the chemical reaction NaHCO3 + CH3COOH → CH3COONa + H2O +CO2, 83 g of sodium bicarbonate reacts with 70 g of acetic acid. Whi
krok68 [10]
MNaHCO₃: 23+1+12+(48×3) = 84g
mCH₃COOH: 12+(1×3)+12+(16×2)+1 = 60g
.................
84g NaHCO₃ react with 60g CH₃COOH
83g NaHCO₃ react with...........

84g ----- 60g
83g ----- X
X = 59,29g CH₃COOH

We used 70g CH₃COOH, it' too much.
So, acetic acid is excess reagent, and sodium bicarbonate is limiting reagent.
_______________________________
B) Amount of CH3COOH is in excess.

:•)
3 0
2 years ago
What is the halflife of a radioisotope if a 20-g sample becomes 10g after 16 hours
ohaa [14]

Answer:

T½ = 16hours

Explanation:

Final mass (N) = 10g

Initial mass (No) = 20g

Time (t) = 16hours

T½ = ?

T½ = In2 / λ

But λ = ?

In(N/No) = -λt

In(10/20) = -(λ * 16)

In(0.5) = -16λ

-0.693 = -16λ

λ = 0.693 / 16

λ = 0.0433

Note : λ is known as the disintegration constant

T½ = In2 / λ

T½ = 0.693 / 0.0433

T½ = 16hours

The half-life of the sample is 16hours

5 0
2 years ago
If the lead can be extracted with 92.5% efficiency, what mass of ore is required to make a lead sphere with a 3.50 cm radius? le
ElenaW [278]

The volume of sphere can be calculated using the following formula:


V=\frac{4}{3}\pi r^{3}


Here, r is radius of the sphere which is 3.5 cm. Putting the value,


V=\frac{4}{3}\pi r^{3}=\frac{4}{3}(3.14)(3.5cm)^{3}=180 cm^{3}


This is equal to the volume of lead, density of lead is 11.34 g/cm^{3} thus, mass of lead can be calculated as follows:


m=d×V


Putting the values,


m=11.34 g/cm^{3}\times 180 cm^{3}=2041.2 g


Let the mass of ore be 1 g, 68.6% of galena is obtained by mass, thus, mass of galena obtained will be 0.686 g.


Now, 86.6% of lead is obtained from this gram of galena, thus, mass of lead will be:


m=0.686×0.866=0.5940 g


Therefore, 0.5940 g of lead is obtained from 1 g of ore for 100% efficiency, thus, for 92.5% efficiency

m=\frac{92.5}{100}\times 0.5940=0.54945 g

1 g of lead obtain from\frac{1}{0.54945}=1.82 grams of ore.


Thus, 2041.2 g of lead obtain from:


2041.2\times 1.82=3.715\times 10^{3}g or 3.715 kg


Therefore, mass of ore required to make lead sphere is 3.715 kg.


6 0
2 years ago
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