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3 years ago

How many grams of sodium chromate, Na2CrO4, are needed to react completely with

Chemistry

1 answer:

JulsSmile [24]3 years ago

8 0

Answer:

We need 27.0 grams of Na2CrO4 to react with 56.7 grams of AgNO3

Explanation:

Step 1: data given

Mass of silver nitrate AgNO3 = 56.7 grams

Molar mass AgNO3 = 169.87 g/mol

Molar mass of Na2CrO4 = 161.97 g/mol

Step 2: The balanced equation

2AgNO3 + Na2CrO4 → Ag2CrO4 + 2NaNO3

Step 3: Calculate moles AgNO3

Moles AgNO3 = mass AgNO3 / molar mass AgNO3

Moles AgNO3 = 56.7 grams / 169.87 g/mol

Moles AgNO3 = 0.334 moles

Step 4: Calculate moles Na2CrO4 moles needed

For 2 moles AgNO3 we need 1 mol Na2CrO4 to produce 1 mol Ag2CrO4 and 2 moles NaNO3

For 0.334 moles AgNO3 we need 0.334 / 2 = 0.167 moles Na2CrO4

Step 5: Calculate mass Na2CrO4

Mass Na2CrO4 = 0.167 moles * 161.97 g/mol

Mass Na2CrO4 = 27.0 grams

We need 27.0 grams of Na2CrO4 to react with 56.7 grams of AgNO3

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Compare the electronegativity of H and O in the polar covalent bonds of a water

ExtremeBDS [4]

Answer:

The O atom will tend to attract the electrons.

Explanation:

The electronegativity of O (3.5) is much higher than H (2.1), which means it is more likely to attract electrons. The higher the electronegativity, the more attractive.

7 0

3 years ago

Water is a produced when 30.0 grams of hydrogen reacts with 80.0 grams of oxygen what is the limiting reagent

irakobra [83]

The balanced equation for the reaction is as follows
2H₂ + O₂ --> 2H₂O
stoichiometry of H₂ to O₂ is 2:1
number of H₂ moles - 30.0 g / 2 g/mol = 15 mol
number of O₂ moles - 80.0 g / 32 g/mol = 2.5 mol
limiting reactant is the reagent in which only a fraction is used up in the reaction
if H₂ is the limiting reactant
if 2 mol of H₂ requires 1 mol of O₂
then 15 mol of H₂ requires 1/2 x 15.0 = 7.5 mol of O₂
but only 2.5 mol of O₂ is required
this means that O₂ is the limiting reagentt and H₂ is in excess

7 0

3 years ago

Which term names the group of organisms able to interbreed and produce fertile offspring?

vladimir2022 [97]

Answer:

Species

Explanation:

Species is the group of organisms able to interbreed and produce fertile offspring.

Let's break down each word in the question:

"Organisms" means living thing. It can be a plant or animal like we usually think of, but it also includes the really small single-celled living things like some bacteria.

"Interbreed" means to mate with each other.

"Fertile" means that the living thing can also have babies.

"Offspring" means the children that are born.

"Fertile offspring" means that the children that are made must be able to have babies of their own. For example, if a frog and a bird could interbreed, they might produce offspring (children). But, if those frog-birds cannot also have children, then frog-bird is not a species.

7 0

3 years ago

Find the equation of the line passing through the points (2, 1) and (5, 10).

Vedmedyk [2.9K]

The equation : y=3x-5

<h3>Further explanation </h3>

Straight-line equations are mathematical equations that are described in the plane of cartesian coordinates

General formula

y-y1 = m(x-x1)

y = mx + c

Where

m = straight-line gradient which is the slope of the line

x1, y1 = the Cartesian coordinate that is crossed by the line

c = constant

The formula for a gradient (m) between 2 points in a line

m = Δy / Δx

Gradient

$\tt \dfrac{10-1}{5-2}=3$

Equation

$\tt y-1=3(x-2)\\\\y-1=3x-6\\\\y=3x-5$

3 0

3 years ago

Calculate the freezing point (in degrees C) of a solution made by dissolving 7.99 g of anthracene{C14H10} in 79 g of benzene. Th

mario62 [17]

<u>Answer:</u> The freezing point of solution is 2.6°C

<u>Explanation:</u>

To calculate the depression in freezing point, we use the equation:

$\Delta T_f=iK_fm$

Or,

$\Delta T_f=i\times K_f\times \frac{m_{solute}\times 1000}{M_{solute}\times W_{solvent}\text{ in grams}}$

where,

$\Delta T_f$ = $\text{Freezing point of pure solution}-\text{Freezing point of solution}$

Freezing point of pure solution = 5.5°C

i = Vant hoff factor = 1 (For non-electrolytes)

$K_f$ = molal freezing point depression constant = 5.12 K/m = 5.12 °C/m

$m_{solute}$ = Given mass of solute (anthracene) = 7.99 g

$M_{solute}$ = Molar mass of solute (anthracene) = 178.23 g/mol

$W_{solvent}$ = Mass of solvent (benzene) = 79 g

Putting values in above equation, we get:

$5.5-\text{Freezing point of solution}=1\times 5.12^oC/m\times \frac{7.99\times 1000}{178.23g/mol\times 79}\\\\\text{Freezing point of solution}=2.6^oC$

Hence, the freezing point of solution is 2.6°C

8 0

3 years ago