The _____________ is measuring a volume of water in ________________

Zinc metal is added to a flask containing aqueous hydrochloric acid. The flask contains 0.400 mole of HCl. How much hydrogen gas

dusya [7]

Answer:

0.4 g of hydrogen gas would be produced.

1. 0.48 mole of HCl is needed to react completely with 15.5 g of zinc.

2. HCl is the limiting reactant.

3. 2.61 g of zinc is in excess

Explanation:

From the balanced equation of reaction:

$Zn(s)+ 2HCl(aq) --> H_2(g) + ZnCl_2(aq)$

1 mole of Zn requires 2 moles of HCl to produce 1 mole of hydrogen gas.

15.5 g of zinc = 15.5/65.3 = 0.24 moles of zinc.

0.24 moles Zn is supposed to require 0.24 x 2 moles HCl which is equivalent to 0.48 moles HCl.

But only 0.400 mole of HCl is present. Hence, HCl is the limiting reagent. For complete reaction with 15.5 g of Zinc, 0.48 mole HCl would be needed.

0.400 mole of HCl will require 0.2 mole of Zn for complete reaction. This thus means that 0.24 - 02 = 0.04 mole of Zn is in excess.

0.04 mole Zn = 0.04 x 65.3 = 2.61 g excess Zn.

Now, since HCl is the limiting reagent;

2 moles of HCl is required to produce 1 mole of H2 according to the equation.

0.400 mole HCl will therefore yield 0.400 x 1/2 = 0.2 mole H2

0.2 mole H2 = 2 x 0.2 = 0.4 g H2

Hence, 0.4 g of hydrogen gas would be produced.

6 0

3 years ago

The radioactive substance cesium-137 has a half-life of 30 years. The amount At (in grams) of a sample of cesium-137 remaining a

stealth61 [152]

Answer: The amount of sample left after 20 years is 288.522 g and after 50 years is 144.26 g

Explanation:

We are given a function that calculates the amount of sample remaining after 't' years, which is:

$A_t(t)=458\times (\frac{1}{2})^{\frac{t}{30}$

For t = 20 years

Putting values in above equation:

$A_t(t)=458\times (\frac{1}{2})^{\frac{20}{30}$

$A_t(t)=288.522g$

Hence, the amount of sample left after 20 years is 288.522 g

For t = 50 years

Putting values in above equation:

$A_t(t)=458\times (\frac{1}{2})^{\frac{50}{30}$

$A_t(t)=144.26g$

Hence, the amount of sample left after 50 years is 144.26 g

6 0

4 years ago

How many atoms are there in 7.80 moles of germanium

Orlov [11]

<h3>Answer:</h3>

4.70 × 10²⁴ atoms Ge

<h3>General Formulas and Concepts:</h3>

Math

Pre-Algebra

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right

Chemistry

Atomic Structure

Using Dimensional Analysis
Avogadro's Number - 6.022 × 10²³ atoms, molecules, formula units, etc.

<h3>Explanation:</h3>

Step 1: Define

7.80 mol Ge

Step 2: Identify Conversions

Avogadro's Number

Step 3: Convert

$\displaystyle 7.80 \ mol \ Ge(\frac{6.022 \cdot 10^{23} \ atoms \ Ge}{1 \ mol \ Ge} )$ = 4.69716 × 10²⁴ atoms Ge

Step 4: Check

We are given 3 sig figs. Follow sig fig rules and round.

4.69716 × 10²⁴ atoms Ge ≈ 4.70 × 10²⁴ atoms Ge

8 0

3 years ago

The pH of a solution in which the concentration of H+ is 0.010M will be:

shusha [124]

B: 2

If it's right plz give brainliest lol <3

7 0

3 years ago

Methane gas and chlorine gas react to form hydrogen chloride gas and carbon tetrachloride gas. What volume of hydrogen chloride

IrinaK [193]

The question is incomplete, here is the complete question:

Methane gas and chlorine gas react to form hydrogen chloride gas and carbon tetrachloride gas. What volume of hydrogen chloride would be produced by this reaction if 1.1 mL of methane were consumed? Be sure your answer has the correct number of significant digits.

Answer: The volume of hydrogen chloride produced in the reaction will be 4.4 mL

Explanation:

We are given:

Volume of methane gas = 1.1 mL

The chemical equation for the reaction of methane gas and chlorine gas follows:

$CH_4(g)+4Cl_2(g)\rightarrow 4HCl(g)+CCl_4(g)$

Moles of methane gas = 1 mole

Moles of hydrogen chloride gas = 4 moles

The relationship of number of moles and volume at constant temperature and pressure was given by Avogadro's law. This law states that volume is directly proportional to number of moles at constant temperature and pressure.

The equation used to calculate number of moles is given by:

$\frac{V_1}{n_1}=\frac{V_2}{n_2}$