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S_A_V [24]
3 years ago
14

49 grams of sulfuric acid, H2SO4, is dissolved in 1 liter of solution. Determine the molarity (M).

Chemistry
1 answer:
Igoryamba3 years ago
3 0

Answer: .5m

Explanation:

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In some areas of the Earth, the crust is squeezed and pushed upward. This is a _______ process in that it directly forms _______
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So..... I believe this is a Convergent boundary and mountains..
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3 years ago
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. In a titration, a 25.0 mL sample of 0.150 M HCl is neutralized with 44.45 mL of Ba(OH)2. a. Write the balanced molecular equat
choli [55]

Answer:

Equation of reaction:

a) 2HCl + Ba(OH)2 ==> CaCl2 + 2H2O

b) Molarity of base = 0.042 M.

Explanation:

Using titration equation

CAVA/CBVB = NA/NB

Where NA is the number of mole of acid = 2

NB is the number of mole of base = 1

CA is the molarity of acid =0.15M

CB is the molarity of base = to be calculated

VA is the volume of acid = 25 ml

VB is the volume of base = 44.45mL

Substituting

0.15×25/CB×44.45 = 2/1

Therefore CB =0.15×25×1/44.45×2

CB = 0.042 M.

5 0
3 years ago
Which of the following substances has
Dmitry [639]
D.  all of the above, although I do find it hard to believe something like flour would have a melting point, I looked up my answer to double check. 
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3 years ago
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A fuel tank holds 22.3 gallons of gasoline. If the density is 0.8206 g/mL, what is the mass in kilograms of gasoline in a full t
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Answer:

m=69.3kg

Explanation:

Hello!

In this case, since the density is computed by dividing the mass of the substance by its occupied volume (d=m/V), we first need to realize that 0.8206 g/mL is the same to 0.8206 kg/L, which means we first need to compute the volume in L:

V=22.3gal*\frac{3.78541L}{1gal}=84.415L

Then, solving for the mass in d=m/V, we get m=d*V and therefore the mass of gasoline in that full tank turns out:

m=0.8206g/L*84.415L\\\\m=69.3kg

Best regards!

8 0
3 years ago
What would A be? HELP ASAP
Minchanka [31]
Si has 4 available elections. Each Cl has 7.
7 x 4 = 28 + the 4 from your Si gives the total of 32 total electrons.
6 0
3 years ago
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