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2 years ago

BaCl2(aq)+Na2SO4(aq)⟶BaSO4(s)+2NaCl(aq)

Chemistry

1 answer:

Bond [772]2 years ago

4 0

Answer:

1.46 moles NaCl

Explanation:

To find the moles of NaCl, you need to multiply the given value (0.731 moles BaCl₂) by the mole-to-mole ratio of the two relevant molecules. This ratio is made up of the coefficients in the balanced reaction. The numerator should contain moles NaCl to allow for the cancellation of moles BaCl₂. The final answer should have 3 sig figs to reflect the given value.

1 BaCl₂(aq) + Na₂SO₄(aq) ---> BaSO₄(s) + 2 NaCl(aq)

0.731 mole BaCl₂ 2 moles NaCl
--------------------------- x ------------------------- = 1.46 moles NaCl
1 mole BaCl₂

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A cylindrical grain has a base radius of 5m. the surface area is 1099m^2. What is the height of the bin to the nearest metre?

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The height of the bin is 29.9m.

The quantity of space enclosing a three-dimensional shape's exterior is its surface area.

The total surface area of the cylinder formula is simply the sum of the base surface area and the lateral surface area.

The radius of the cylinder, r = 5m

Surface area = 1099 m²

∴ A = 2πrh + 2πr²

$h = \frac{A- 2\pi r^2}{2\pi r}$

$h = \frac{1099- 2 (3.141) 5^2}{2 (3.141) 5}$

$h = \frac{1099- 157.05}{31.41}$

$h = \frac{941.95}{31.41}$

$h = 29.9 m$

Therefore, the height of the bin is 29.9m.

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2 years ago

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4 years ago

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If 0.60 L of a solution contains 6.6 g of NaBr, what is its molar concentration?

-BARSIC- [3]

Answer:

Molarity = moles ÷ liters

to get moles of NaBr divide grams of NaBr by its molar mass (mass of Na + mass of Bromine)

Na = 22.989769

Br = 79.904

molar mass of NaBr = 102.893769

6.6g ÷ 102.893769 = 0.064143826 moles of NaBr

0.064143826 moles ÷ 0.60 liters = 0.1069 molar concentration or 11 %

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3 years ago

2-bromo-3,4-dimethylpentane is combined with t-butoxide. The product of this reaction is.

Andrej [43]

2-bromo-3,4-dimethylpentane is combined with t-butoxide. The product of this reaction is 3,4 dimethyl - 1- pentene.

The reaction of 2-bromo-3,4-dimethylpentane is combined with t-butoxide forms 2 alkene in the elimination reaction due to steric hindrance. The least stable alkene 3,4 dimethyl - 1- pentene is easy to make. the t-butoxide is (CH₃)₃CO⁻. The reaction involves in this reaction is E2 elimination reaction. This reaction involves the one step reaction. The product will also form that is 3,4 dimethyl - 2 - pentene. so the reaction involve Elimination reaction and the product due to steric hindrance is 3,4 dimethyl - 1- pentene

Thus, 2-bromo-3,4-dimethylpentane is combined with t-butoxide. The product of this reaction is 3,4 dimethyl - 1- pentene.

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1 year ago

A student mixes 33.0 mL of 2.70 M Pb ( NO 3 ) 2 ( aq ) with 20.0 mL of 0.00157 M NaI ( aq ) . How many moles of PbI 2 ( s ) prec

GalinKa [24]

Answer: The moles of precipitate (lead (II) iodide) produced is $1.57\times 10^{-5}$ moles

Explanation:

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$ .....(1)

For lead (II) nitrate:

Molarity of lead (II) nitrate solution = 2.70 M

Volume of solution = 33.0 mL = 0.033 L (Conversion factor: 1 L = 1000 mL)

Putting values in equation 1, we get:

$2.70M=\frac{\text{Moles of lead (II) nitrate}}{0.033L}\\\\\text{Moles of lead (II) nitrate}=(2.70mol/L\times 0.0330L)=0.0891mol$

For NaI:

Molarity of NaI solution = 0.00157 M

Volume of solution = 20.0 mL = 0.020 L

Putting values in equation 1, we get:

$0.00157M=\frac{\text{Moles of NaI}}{0.020L}\\\\\text{Moles of NaI}=(0.00157mol/L\times 0.0200L)=3.14\times 10^{-5}mol$

For the given chemical reaction:

$Pb(NO_3)_2(aq.)+2NaI(aq.)\rightarrow PbI_2(s)+2NaNO_3(aq.)$

By Stoichiometry of the reaction:

2 moles of NaI reacts with 1 mole of lead (II) nitrate

So, $3.14\times 10^{-5}$ moles of NaI will react with = $\frac{1}{2}\times 3.14\times 10^{-5}=1.57\times 10^{-5}mol$ of lead (II) nitrate

As, given amount of lead (II) nitrate is more than the required amount. So, it is considered as an excess reagent.

Thus, NaI is considered as a limiting reagent because it limits the formation of product.

By Stoichiometry of the reaction:

2 moles of NaI produces 1 mole of lead (II) iodide

So, $3.14\times 10^{-5}$ moles of NaI will produce = $\frac{1}{2}\times 3.14\times 10^{-5}=1.57\times 10^{-5}moles$ of lead (II) iodide

Hence, the moles of precipitate (lead (II) iodide) produced is $1.57\times 10^{-5}$ moles

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3 years ago