Ni4+ + _____e- LaTeX: \longrightarrow⟶ Nix+

Which of these changes would cause a decrease in the pressure of a contained gas?

Harman [31]

The equation is pv=nrt. So to decrease pressure, one would increase volume, decrease moles, or decrease temperature.

4 0

3 years ago

If 34.7 g of AgNO₃ react with 28.6 g of H₂SO₄ according to this UNBALANCED equation below, how many grams of Ag₂SO₄ could be for

Luba_88 [7]

The number of grams of Ag2SO4 that could be formed is 31.8 grams

<u><em> calculation</em></u>

Balanced equation is as below

2 AgNO3 (aq) + H2SO4(aq) → Ag2SO4 (s) +2 HNO3 (aq)

Find the moles of each reactant by use of mole= mass/molar mass formula

that is moles of AgNO3= 34.7 g / 169.87 g/mol= 0.204 moles

moles of H2SO4 = 28.6 g/98 g/mol =0.292 moles

use the mole ratio to determine the moles of Ag2SO4

that is;

the mole ratio of AgNo3 : Ag2SO4 is 2:1 therefore the moles of Ag2SO4= 0.204 x1/2=0.102 moles

The moles ratio of H2SO4 : Ag2SO4 is 1:1 therefore the moles of Ag2SO4 = 0.292 moles

AgNO3 is the limiting reagent therefore the moles of Ag2SO4 = 0.102 moles

<h3> finally find the mass of Ag2SO4 by use of mass=mole x molar mass formula</h3>

that is 0.102 moles x 311.8 g/mol= 31.8 grams

3 0

3 years ago

Is bread a molecule or a compound?

Morgarella [4.7K]

it's molecule.......

3 0

2 years ago

Read 2 more answers

The mass number of 3 isotopes of magnesium are 24,25,26.what is the mass number of the most abundant isotope of magnesium.

Anarel [89]

Answer:

24.4 amu or g/mole

Explanation:

24 x 0.790 = 19.0 amu

25 x 0.100 = 2.50 amu

26 x 0.110 = 2.86 amu

(Because of the 19.0, the sig figs go only to the 1/10 decimal place)

19.0 + 2.5 + 2.9 = 24.4 amu or g/mole

7 0

2 years ago

The concentration of a saturated BaCl2 solution is 1.75 M (mol/liter) and the concentration of a saturated Na2SO4 solution is 2.

Kitty [74]

Answer:

a) The theoretical yield is 408.45g of $BaSO_{4}$

b) Percent yield = $\frac{realyield}{408.45g}*100$

Explanation:

1. First determine the numer of moles of $BaCl_{2}$ and $Na_{2}SO_{4}$ .

Molarity is expressed as:

M= $\frac{molessolute}{Lsolution}$

- For the $BaCl_{2}$

M= $\frac{1.75molesBaCl_{2}}{1Lsolution}$

Therefore there are 1.75 moles of $BaCl_{2}$

- For the $Na_{2}SO_{4}$

M= $\frac{2.0moles[tex]Na_{2}SO_{4}$ }{1Lsolution}[/tex]

Therefore there are 2.0 moles of $Na_{2}SO_{4}$

2. Write the balanced chemical equation for the synthesis of the barium white pigment, $BaSO_{4}$ :

$BaCl_{2}+Na_{2}SO_{4}=BaSO_{4}+2NaCl$

3. Determine the limiting reagent.

To determine the limiting reagent divide the number of moles by the stoichiometric coefficient of each compound:

- For the $BaCl_{2}$ :

$\frac{1.75}{1}=1.75$

- For the $Na_{2}SO_{4}$ :

$\frac{2.0}{1}=2.0$

As the $BaCl_{2}$ is the smalles quantity, this is the limiting reagent.

4. Calculate the mass in grams of the barium white pigment produced from the limiting reagent.

$1.75molesBaCl_{2}*\frac{1molBaSO_{4}}{1molBaCl_{2}}*\frac{233.4gBaSO_{4}}{1molBaSO_{4}}=408.45gBaSO_{4}$

5. The percent yield for your synthesis of the barium white pigment will be calculated using the following equation:

Percent yield = $\frac{realyield}{theoreticalyield}*100$

Percent yield = $\frac{realyield}{408.45g}*100$

The real yield is the quantity of barium white pigment you obtained in the laboratory.

7 0

2 years ago