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KATRIN_1 [288]
2 years ago
5

A store sells 4 reams of paper for $23.08. Which proportion can be used to find the cost of 12 reams of paper?

Chemistry
1 answer:
sashaice [31]2 years ago
3 0
The answer is A. 12/23.08=x/4

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A light year measure the distance a lightbeam can travel in one year one light year is equivalent to 9500000000000 km why do sci
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Answer:

Light years are the only practical unit for measuring the distance of space

Explanation:

Practical... aka logical.

8 0
3 years ago
A mixture of gaseous CO and H2, called synthesis gas, is used commercially to prepare methanol (CH3OH), a compound considered an
mr_godi [17]

Answer : The value of equilibrium constant (K) is, 424.3

Explanation :  Given,

Concentration of H_2 at equilibrium = 0.067 mol

Concentration of CO at equilibrium = 0.021 mol

Concentration of CH_3OH at equilibrium = 0.040 mol

The given chemical reaction is:

CO+2H_2\rightarrow CH_3OH

The expression for equilibrium constant is:

K_c=\frac{[CH_3OH]}{[CO][H_2]^2}

Now put all the given values in this expression, we get:

K_c=\frac{(0.040)}{(0.021)\times (0.067)^2}

K_c=424.3

Thus, the value of equilibrium constant (K) is, 424.3

4 0
2 years ago
3. Did pushing (compression), pulling (tension), or sideways forces (shear) in the rock layers produce the folds and faults?
Neko [114]

Pushing(Compression) Causes the faults and folds


3 0
3 years ago
How many milliliters of 0.02 M HCl are needed to react completely with 100 mL of 0.01 M NaOH?
serious [3.7K]
The reaction of HCl and NaOH is HCl + NaOH = NaCl + H2O. So the mole number of HCl and NaOH is equal. So the volume of HCl =0.01*0.1/0.02=0.05 L =50 ml. So the answer is D).
3 0
3 years ago
Read 2 more answers
An atom of gold has a mass of 3.271 X 10-22 g. How many atoms of gold are in 5.00 g of gold? (Give your answer in scientific not
Kobotan [32]

Answer:

1.53 × 10²² atoms Ag

Explanation:

Step 1: Define conversions

3.271 × 10⁻²² g = 1 atom

Step 2: Use Dimensional Analysis

5.00 \hspace{3} g \hspace{3} Ag(\frac{1 \hspace{3} atom \hspace{3} Ag}{3.271(10)^{-22} \hspace{3} g \hspace{3} Ag} ) = 1.52858 × 10²² atoms Ag

Step 3: Simplify

We have 3 sig figs.

1.52858 × 10²² atoms Ag ≈ 1.53 × 10²² atoms Ag

5 0
2 years ago
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