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3 years ago

How many grams of H3PO4 are in 234 mL of a 2.5 M solution of H3PO4?

Chemistry

1 answer:

motikmotik3 years ago

5 0

Answer:

57 grams of H3PO4

Explanation:

M= moles/ liters

convert mL to L

234 mL x 1L/1000mL = 0.234L

Rearrange the Molarity formula to solve for moles.

moles= MxL

moles= 2.5M x 0.234L

moles= 0.585 mol

Use the molar mass of H3PO4 to get to grams

0.585 mol x 97.994 grams/1 mol = 57.326 grams of H3PO4

round to two sig figs for 57 grams

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Infrared radiation if i’m correct

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3 years ago

Read 2 more answers

Determine the percent yield for the reaction

kkurt [141]

Answer:

The percent yield for Br₂ in the reaction = 96.15%

Explanation:

The balanced stoichiometric equation for the reaction is given as

2 NaBr + 1 Cl₂ → 2 NaCl + 1 Br₂

Percent yield

= 100% × (Actual yield)/(Theoretical yield)

To find the theoretical yield,

5.29 g of NaBr reacts with excess chlorine

gas; this means that NaBr is the limiting reagent because it is used up in the process of the reaction, hence, it determines the amount of products to be found.

So, we convert the 5.29 g of NaBr into number of moles.

Number of moles = (Mass)/(Molar mass)

Molar Mass of NaBr = 102.894 g/mol

Number of moles = (5.29/102.894) = 0.0514121329 = 0.05141 mole

From the stoichiometric balance of the reaction,

2 moles of NaBr give 1 mole of Br₂

0.05141 mole of NaBr will give (0.05141×1/2) mole of Br₂; that is, 0.0257 mole of Br₂

Theoretical yield = Mass of Br₂ expected from the reaction

= (Number of moles) × (Molar mass)

Molar mass of Br₂ = 159.808 g/mol

Theoretical yield of Br₂ = 0.0257 × 159.808 = 4.108 g

Percent yield

= 100% × (Actual yield)/(Theoretical yield)

Actual yield = 3.95 g

Theoretical yield = 4.108 g

Percent yield = 100% × (3.95/4.108) = 96.15%

Hope this Helps!!!

5 0

4 years ago

How much energy is required to melt a 500. gram block of iron? The heat of vaporization is 6090 J/g and the heat of fusion is 24

ki77a [65]

Answer:

Heat required = 1.23×10⁵J

Explanation:

Given:

Mass (m) = 500 gram

Specific heat = 6,090 J/g

heat of fusion = 247 J/g.

Find:

Heat required

Computation:

Heat required = 247 J/g× 500 g

Heat required = 1.23×10⁵J

4 0

4 years ago

A student is preparing to perform a series of calorimetry experiments. She first wishes to determine the calorimeter constant (C

Maru [420]

Explanation:

The given data is as follows.

$V_{1}$ = 50 ml, $T_{1}$ = 345 K

$T_{2}$ = 298 K, $T_{f}$ = 317 K,

$V_{2}$ = 50 ml

Now, we will calculate the heat energy as follows.

$Q_{hot} = m \times C_{p} \times (T_{1} - T_{f})$

= $50 g \times 4.184 \times (345 - 317)$

= 5857.6 J

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= $50 g \times 4.184 \times (317 - 298)$

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As, $Q_{hot} = -Q_{cold}$ so there will be loss of heat. And, some heat will go to the calorimeter.

Hence, $Q_{hot} = Q_{cold} + Q_{cal}$

$5857.6 = 3974.8 + Q_{cal}$

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dT = $T_{f} - T_{2}$

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= 19 K

Hence, we will calculate the specific heat as follows.

C = $\frac{Q}{dT}$

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= 99.1 J/K

Thus, we can conclude that the value of $C_{cal}$ for the calorimeter is 99.1 J/K.

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3 years ago

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4 years ago