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murzikaleks [220]
2 years ago
13

How many molecules are in 15.0 g of oxygen gas (O2)?

Chemistry
1 answer:
frosja888 [35]2 years ago
3 0

Answer:

2.82*10^23

Explanation:

please see attached for work!

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If Log 4 (x) = 12, then log 2 (x / 4) is equal to
Alexus [3.1K]

The value of log₂(x/4) is 22. Using the properties of the logarithm, the required value is calculated.

<h3>What are the required properties of the logarithm?</h3>

The required logarithm properties are

logₐx = n ⇒ aⁿ = x; and logₐ(xⁿ) = n logₐ(x);

Where a is the base of the logarithm.

<h3>Calculation:</h3>

It is given that,

log₄(x) = 12;

On applying the property logₐx = n ⇒ aⁿ = x; here a = 4;

So,

log₄(x) = 12 ⇒ 4¹² = x

⇒ x = (2²)¹² = 2²⁴

Then, calculating log₂(x/4):

log₂(x/4) = log₂(2²⁴/4)

              = log₂(2²⁴/2²)

              = log₂(2²⁴ ⁻ ²)

              = log₂(2²²)

On applying the property logₐ(xⁿ) = n logₐ(x);

log₂(x/4) = 22 log₂2

We know that logₐa = 1;

So,

log₂(x/4) = 22(1)

∴ log₂(x/4) = 22.

Learn more about the properties of logarithm here:

brainly.com/question/12049968

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8 0
1 year ago
A sample of gas contains 0.1500 mol of HCl(g) and 7.500×10-2 mol of Br2(g) and occupies a volume of 9.63 L. The following reacti
Furkat [3]

Answer:

9.63 L.

Explanation:

Hello,

In this case, the undergoing chemical reaction is:

2HCl(g) + Br_2(g)\rightarrow 2HBr(g) + Cl_2(g)

So the consumed amounts of hydrochloric acid and bromine are the same to the beginning based on:

n_{Br_2}^{consumed}=0.1500molHCl*\frac{1molBr_2}{2molHCl}=0.075molBr_2

In such a way, the yielded moles of hydrobromic acid and chlorine are:

n_{HBr}=0.1500molHCl*\frac{2molHBr}{2molHCl}=0.1500molHBr \\n_{Cl_2}=0.1500molHCl*\frac{1molCl_2}{2molHCl}=0.075molCl_2

Thus, the volume of the sample, after the reaction is the same as no change in the total moles is evidenced, that is 9.63L.

Best regards.

7 0
3 years ago
The Great Pacific Garbage Patch is created by _____ currents. A. local B. surface C. deep ocean
Kamila [148]
The Great Pacific Garbage Patch is created by deep oceans. The Marine Debris is litter that ends up in oceans or other large bodies of water.

3 0
3 years ago
I need help can someone please do so?
larisa86 [58]

Answer:

0.296 J/g°C

Explanation:

Step 1:

Data obtained from the question.

Mass (M) =35g

Heat Absorbed (Q) = 1606 J

Initial temperature (T1) = 10°C

Final temperature (T2) = 165°C

Change in temperature (ΔT) = T2 – T1 = 165°C – 10°C = 155°C

Specific heat capacity (C) =..?

Step 2:

Determination of the specific heat capacity of iron.

Q = MCΔT

C = Q/MΔT

C = 1606 / (35 x 155)

C = 0.296 J/g°C

Therefore, the specific heat capacity of iron is 0.296 J/g°C

8 0
3 years ago
Round each number to four significant figures
ira [324]

Answer:

a.

84,791 » 8479

b.

256.75 » 256.8

c.

431,801 » 4318

d.

0.00078100 » 0.0007810

Explanation:

.

5 0
2 years ago
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