Please help!!

What is the SI unit used to measure the temperature of a substance?

navik [9.2K]

The SI unit for temperature is Kelvin.

7 0

3 years ago

For the following reaction, 38.3 grams of sulfuric acid are allowed to react with 33.5 grams of calcium hydroxide sulfuric acid(

Likurg_2 [28]

Answer:

What is the maximum amount of calcium sulfate that can be formed? 53.1 grams CaSO4

What is the FORMULA for the limiting reagent? H2SO4

What amount of the excess reagent remains after the reaction is complete? 4.59 grams of Ca(OH)2

Explanation:

Step 1: Data given

Mass of sulfuric acid = 38.3 grams

Molar mass of H2SO4 = 98.08 g/mol

Mass of calcium hydroxide = 33.5 grams

Molar mass of Ca(OH)2 = 74.09 g/mol

Step 2: The balanced equation

H2SO4 + Ca(OH)2 → CaSO4 + 2H2O

Step 3: Calculate moles of H2SO4

moles H2SO4 = mass H2SO4 / molar mass H2SO4

moles H2SO4 = 38.3 grams / 98.08 g/mol

moles H2SO4 = 0.390 moles

Step 4: Calculate moles of Ca(OH)2

moles Ca(OH)2 = 33.5 grams / 74.09 g/mol

moles Ca(OH)2 =0.452 moles

Step 5: Calculate limiting reactant

For 1 mol H2SO4, we need 1 mol of Ca(OH)2 to produce, 1 mol of CaSO4 and 2 mol of H2O

H2SO4 is the limiting reactant. It will completely be consumed (0.390 moles).

Ca(OH)2 is in excess. There will be consumed 0.390 moles

There will remain 0.452 - 0.390 = 0.062 moles

This is 0.062 * 74.09 g/mol = 4.59 grams

Step 6: Calculate moles of calcium sulfate

For 1 mol H2SO4, we need 1 mol of Ca(OH)2 to produce, 1 mol of CaSO4 and 2 mol of H2O

For 0.390 moles of H2SO4, there will be produced 0.390 moles of CaSO4

Step 7: Calculate mass of CaSO4

Mass CaSO4 = moles CaSO4 * molar mass CaSO4

Mass CaSO4 = 0.390 moles * 136.14 g/mol

Mass of CaSO4 = 53.1 grams

7 0

3 years ago

17 POINTS!!! JUST DO A B C D PLS!!!! PLS DONT COMMENT IF YOU DONT KNOW IT ITS DUE BY TOMORROW AND I CANT KEEP WASTING POINTS PLS

dimulka [17.4K]

Answer:

A B C D E

Explanation:

7 0

3 years ago

The solubility of silver(I)phosphate at a given temperature is 1.02 g/L. Calculate the Ksp at this temperature. After you get yo

Snezhnost [94]

<u>Answer:</u> The solubility product of silver (I) phosphate is $9.57\times 10^{-10}$

<u>Explanation:</u>

We are given:

Solubility of silver (I) phosphate = 1.02 g/L

To convert it into molar solubility, we divide the given solubility by the molar mass of silver (I) phosphate:

Molar mass of silver (I) phosphate = 418.6 g/mol

$\text{Molar solubility of silver (I) phosphate}=\frac{1.02g/L}{418.6g/mol}=2.44\times 10^{-3}mol/L$

Solubility product is defined as the product of concentration of ions present in a solution each raised to the power its stoichiometric ratio.

The chemical equation for the ionization of silver (I) phosphate follows:

$Ag_3PO_4(aq.)\rightleftharpoons 3Ag^{+}(aq.)+PO_4^{3-}(aq.)$

3s s

The expression of $K_{sp}$ for above equation follows:

$K_{sp}=(3s)^3\times s$

We are given:

$s=2.44\times 10^{-3}M$

Putting values in above expression, we get:

$K_{sp}=(3\times 2.44\times 10^{-3})^3\times (2.44\times 10^{-3})\\\\K_{sp}=9.57\times 10^{-10}$

Hence, the solubility product of silver (I) phosphate is $9.57\times 10^{-10}$

4 0

3 years ago

I need help with 17

Nezavi [6.7K]

Answer and explanation:

A chemical change occurs when new substances or materials are produced to have different chemical properties from the reactants.

Some examples of this are paper burning, fruit rotting, and metal rusting.

8 0

3 years ago