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ASHA 777 [7]
2 years ago
10

Please do question 9 and 10

Chemistry
1 answer:
DanielleElmas [232]2 years ago
8 0

Answer:

9) 60; 10) 0.2583

Explanation:

9) M(SiO₂)=28+16*2=60 (gr/mol);

10) if one mole is in 60gr., then for 15.5 gr.:

1/60=x/15.5; ⇒ x=15.5/60≈0.2583(mol).

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777 Joules of energy are applied to 25.0 grams of each of the following materials, which
4vir4ik [10]

Answer:

I think copper

Explanation:

Material IACS % Conductivity

Silver                105

Copper            100

Gold               70

Aluminum         61

Nickel              22

Zinc                          27

Brass                  28

Iron                        17

Tin                      15

Phosphor Bronze  15

Lead                      7

Nickel Aluminum Bronze 7

Steel                   3 to 15

the table might help- your indian brother

3 0
2 years ago
It took batman 5 minutes to catch robin in the bat-mobile. If he traveled 45 km before catching him, how fast was he going?
Nataly_w [17]

Answer:

he going zoom zoom

Explanation:

7 0
2 years ago
Suppose you have a spherical balloon filled with air at room temperature and 1.0 atm pressure; its radius is 17 cm. You take the
Sladkaya [172]
<span>Answer: 17.8 cm
</span>

<span>Explanation:
</span>

<span>1) Since temperature is constant, you use Boyle's law:
</span>

<span>PV = constant => P₁V₁ = P₂V₂


</span><span>=> V₁/V₂ = P₂/P₁</span>
<span>
2) Since the ballon is spherical:


</span><span>V = (4/3)π(r)³</span>
<span>
Therefore, V₁/V₂ = (r₁)³ / (r₂)³
</span>

<span>3) Replacing in the equation V₁/V₂ = P₂/P₁:


</span><span><span>(r₁)³ / (r₂)³ </span>= P₂/P₁</span>
<span>
And you can solve for r₂: (r₂)³ = (P₁/P₂) x (r₁)³


</span>(r₂)³ = (1.0 atm / 0.87 atm) x (17 cm)³ = 5,647.13 cm³
<span>
r₂ = 17.8 cm</span>

4 0
3 years ago
Read 2 more answers
Circle the solution with the highest concentration
choli [55]

Answer:

answer is option a..............

5 0
2 years ago
Read 2 more answers
<img src="https://tex.z-dn.net/?f=H_2PO_4%5E-%28aq%29%20%5Crightarrow%20H%5E%2B%28aq%29%20%2B%20HPO_4%5E%7B2-%7D%28aq%29" id="Te
klasskru [66]

Answer:

The pH of the buffer solution = 8.05

Explanation:

Using the Henderson - Hasselbalch equation;

pH = pKa₂ + log ( [HPO₄²-]/[H₂PO4⁻]

where pKa₂ = -log (Ka₂) = -log ( 6.1 * 10⁻⁸) = 7.21

Concentration of OH⁻ added = 0.069 M (i.e. 0.069 mol/L)

[H₂PO4⁻] after addition of OH⁻ = 0.165 - 0.069 = 0.096 M

[HPO₄²-] after addition of OH⁻ = 0.594 + 0.069 = 0.663 M

Therefore,

pH = 7.21 + log (0.663 / 0.096)

pH = 7.21 + 0.84

pH = 8.05

4 0
3 years ago
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