3. КОН + Н3РО4 1 K3PO4 + Н2O

Aq. sodium oxalate reacts with aq calcium chioride to form sd nd soum ce

STatiana [176]

Na2C2O4(aq) + CaCl2(aq) -----> 2NaCl(aq) + CaC2O4(s)

Here, CaC2O4(s) is a precipitate in the reaction as a result of precipitation reaction or double displacement reaction.

As we know that double displacement reaction two metal ions displaces each other from their salt solutions.

As we know that precipitation reaction is a reaction in which precipitate is formed.

4 0

3 years ago

A student sets up the following equation to convert a measurement. (The stands for a number the student is going to calculate.)

Crank

Answer:

0.090 J/(mmol·°C) × (1000 mmol/mole × 1 kJ/(1000 J)) = 0.090 kJ/mole

Explanation:

The unit of conversion from kilo-Joules to Joules is given as follows;

1000 Joules = 1 kilo-Joule

The unit of conversion from milimoles to moles is given as follows;

1000 milimoles = 1 Mole

Therefore, we have

The value of the given expression is 0.090 J/(mmol·°C) × 1000 mmol/mole × 1 kJ/(1000 J) = 0.090 kJ/mole

0.090 J/milimole = 0.09 kJ/mole.

3 0

3 years ago

What is the density (g/mL) of an object that has a mass of 0.03 kg and occupies a volume of 25 mL?

tatyana61 [14]

First convert the kg to g ----- 0.03kg = 30g
Then divide the mass by the volume ----- 30g ÷ 25mL = 1.2
The density is 1.2g/mL

5 0

3 years ago

Read 2 more answers

You are instructed to create 900. mL of a 0.29 M phosphate buffer with a pH of 7.8. You have phosphoric acid and the sodium salt

Aleksandr [31]

Answer:

B and C

Explanation:

When we have to do a buffer solution we always have to choose the reaction that has the pKa closer to the desired pH value. When we find the pKa values we will obtain:

$pKa_1=-Log[6.9x10^-^3]=2.16$

$pKa_2=-Log[6.2x10^-^8]=7.20$

$pKa_3=-Log[4.8x10^-^13]=12.31$

The closer value is pKa2 with a value of 7.2. Therefore we have to use the second reaction. In which $H_2PO_4^-^1$ is the acid and $HPO_4^-^2$ is the base. Therefore the answer for the first question is B and the answer for the second question is C.

8 0

3 years ago

Acetylene (C2H2), an important fuel in welding, is produced in the laboratory when calcium carbide (CaC2) reacts with water: CaC

Anastaziya [24]

Answer:

There are 1.287 grams of acetylene collected

Explanation:

Total gas pressure = 909 mmHg

Vapor pressure of water = 20.7 mmHg

Pressure of acetylene = 909 mmHg - 20.7 mmHg = 888.3 mmHg

1mmHg = 1 torr

22 ° C + 273.15 = 295.15 Kelvin

Ideal gas law ⇒ pV = nRT

⇒ with p = pressure of the gas in atm

⇒ with V = volume of the gas in L

⇒ with n = amount of substance of gas ( in moles)

⇒ with R = gas constant, equal to the product of the Boltzmann constant and the Avogadro constant (62.36 L * Torr *K^−1 *mol^−1)

⇒ with T = absolute temperature of the gas (in Kelvin)

888.3 torr * 1.024 L = n * 62.36 L * Torr *K^−1 *mol^−1 * 295.15 K

n = 0.04942 moles of C2H2

Mass of C2H2 = 0.04942 moles x 26.04 g/mole = 1.287 g

There are 1.287 grams of acetylene collected

6 0

3 years ago