All categories

3 years ago

5. Which type of substance yields hydrogen ions, H+, in

Chemistry

1 answer:

GaryK [48]3 years ago

7 0

Answer: An Arrhenius acid is a type of substance yields hydrogen ions, $H^{+}$ , in an aqueous solution.

Explanation:

Substances that give dissociate to give hydrogen ions or protons when dissolved in an aqueous solution are called Arrhenius acid.

For example, HCl is a strong acid and upon dissociation in water it gives the following ions.

$HCl \rightarrow H^{+} + Cl^{-}$

Thus, we can conclude that an Arrhenius acid is a type of substance yields hydrogen ions, $H^{+}$ , in an aqueous solution.

You might be interested in

If a 50 N force pulls on a 10 kg object,how much acceleration will occur?

meriva

Assuming that the force of friction does not equal to the force at which you are pulling at, hence Fnet is not zero, to solve for acceleration need to use the following equation:

Fnet = ma
a = Fnet/m
a = 50 N/10 kg
a = 5 N/kg or a = 5 m/s^2.

4 0

3 years ago

Read 2 more answers

When looking at the percentage of patients whose symptoms were relieved after taking a placebo, a scientist sees a 20% success r

9966 [12]

A. Evaluating test results is your answer. Let me know if this is correct!

6 0

3 years ago

A quantity of N2 gas originally held at 5.23 atm pressure in a 1.20 −L container at 26 ∘C is transferred to a 14.5 −L container

lara31 [8.8K]

Answer:

n (N₂) = 0.256 mol

n (O₂) = 1.0848 mol

n (Total) = 1.3408 mol

Pressure in new Container = 2.222 atm

Explanation:

Data Given:

For Nitrogen gas (N₂)

Pressure of N₂ gas = 5.23 atm

Volume of N₂ gas = 1.20 L

Temperature of N₂ gas = 26° C

Temperature of N₂ gas in Kelven (K) = 26° C +273

Temperature of N₂ gas in Kelven (K) = 299K

ideal gas Constant R = 0.08206 L atm K⁻¹ mol⁻¹

quantity of gas N₂ gas = ?

For Oxygen gas (O₂)

Pressure of O₂ gas = 5.21 atm

Volume of O₂ gas = 5.21 L

Temperature of O₂ gas = 26° C

Temperature of O₂ gas in Kelven (K) = 26° C +273

Temperature of O₂ gas in Kelven (K) = 299K

ideal gas Constant R = 0.08206 L atm K⁻¹ mol⁻¹

quantity of gas O₂ gas = ?

*we also have to find the total Pressure in the new container = ?

Formula Used

PV =nRT

n (N₂) = PV /RT . . . . . . . . . . . . . (1)

* Find the quantity of N₂

Put value in formula (1)

n (N₂) = 5.23 atm x 1.20 L / 0.08206 L atm K⁻¹ mol⁻¹ x 299K

n (N₂) = 6.276 atm .L / 24.52 L atm. mol⁻¹ x 299K

n (N₂) = 0.256 mol

* Find the quantity of O₂

Put value in formula (1)

n (O₂) = 5.21 atm x 5.10 L / 0.08206 L atm K⁻¹ mol⁻¹ x 299K

n (O₂) = 26.6 atm .L / 24.52 L atm. mol⁻¹

n (O₂) = 1.0848 mol

*Now to find the Total Quantity of both gases

n(Total) = n (N₂) + n (O₂)

n (Total) = 0.256 mol + 1.0848 mol

n (Total) = 1.3408 mol

**To find the Total Pressure in the new Container

Data to calculate Total Pressure in new container

Volume of gas = 14.5 L

Temperature of gases = 20° C

Temperature of gases in Kelven (K) = 20° C +273

Temperature of gases in Kelven (K) = 293K

ideal gas Constant R = 0.08206 L atm K⁻¹ mol⁻¹

Volume Pressure in new container = ?

Formula Used

PV =nRT

P = nRT / V . . . . . . . . . . . . . (2)

Put values in Equation (2)

P = 1.3408 mol x 0.08206 L atm K⁻¹ mol⁻¹ x 293 K / 14.5 L

P = 2.222 atm

8 0

4 years ago

What is the mass of 61.9 L of oxygen gas collected at STP?

Tcecarenko [31]

Answer:

D. 44.2 g O₂

General Formulas and Concepts:
Gas Laws

STP (Standard Conditions for Temperature and Pressure) = 22.4 L per mole at 1 atm, 273 K

Stoichiometry

Dimensional Analysis
Mole Ratio

Explanation:

Step 1: Define

Identify given.

61.9 L O₂ at STP

Step 2: Convert

We know that the oxygen gas is at STP. Therefore, we can set up and solve for how many moles of O₂ is present:

$\displaystyle 61.9 \ \text{L} \ \text{O}_2 \bigg( \frac{1 \ \text{mol} \ \text{O}_2}{22.4 \ \text{L} \ \text{O}_2} \bigg) = 2.76339 \ \text{mol} \ \text{O}_2$

Recall the Periodic Table (Refer to attachments). Oxygen's atomic mass is roughly 16.00 grams per mole (g/mol). We can use a mole ratio to convert from moles to grams:

$\displaystyle 2.76339 \ \text{mol} \ \text{O}_2 \bigg( \frac{16.00 \ \text{g} \ \text{O}_2}{1 \ \text{mol} \ \text{O}_2} \bigg) = 44.2143 \ \text{g} \ \text{O}_2$