Balance the following reactions CuO+S2

The addition of NaOH to water would result in which of the following?

EleoNora [17]

Answer:

A decrease in [H3O+] and an increase in pH (option a)

Explanation:

Equilibrium of water is shown in this equation

2H₂O ⇄ H₃O⁺ + OH⁻

When you add NaOH, you are modifying [OH⁻]

NaOH → Na⁺ + OH⁻

In equilibrium of water, the [OH⁻] increases

2H₂O ⇄ ↓ H₃O⁺ + OH⁻ ↑

As the [OH⁻] increases, by Le Chatellier, the equilibrium tends to decrease [H₃O⁺].

If the [OH⁻] is higher, pH is also high so the solution of water and sodium hydroxide would be totally basic.

3 0

4 years ago

Write a balanced net ionic equation for the following reaction. BaCI2(aq)+H2SO4(aq) -> BaSO4(s)+HCI (aq)

Sever21 [200]

Ba²⁺ + 2Cl⁻ + 2H⁺ + SO₄²⁻ = BaSO₄ (precipitate) + 2H⁺ + 2Cl⁻
Ba²⁺ + SO₄²⁻ = BaSO₄

4 0

3 years ago

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What is the mass of 8.23 x 10^23 atoms of Ag

Gnom [1K]

Answer:

$\boxed {\boxed {\sf Approximately \ 147 \ g\ Ag}}$

Explanation:

Convert Atoms to Moles

The first step is to convert atoms to moles. 1 mole of every substance has the same number of particles: 6.022 ×10²³ or Avogadro's Number. The type of particle can be different, in this case it is atoms of silver. Let's create a ratio using this information.

$\frac{6.022*10^{23} \ atoms \ Ag}{1 \ mol \ Ag}$

We are trying to find the mass of 8.23 ×10²³ silver atoms, so we multiply by that number.

$8.23 *10^{23} \ atoms \ Ag *\frac{6.022*10^{23} \ atoms \ Ag}{1 \ mol \ Ag}$

Flip the ratio so the atoms of silver cancel. The ratio is equivalent, but places the other value with units "atoms Ag" in the denominator.

$8.23 *10^{23} \ atoms \ Ag *\frac{1 \ mol \ Ag}{6.022*10^{23} \ atoms \ Ag}$

$8.23 *10^{23} *\frac{1 \ mol \ Ag}{6.022*10^{23} }$

Condense into one fraction.

$\frac{8.23 *10^{23} }{6.022*10^{23} } \ mol \ Ag$

$1.366655596 \ mol \ Ag$

Convert Moles to Grams

The next step is to convert the moles to grams. This uses the molar mass, which is equivalent to the atomic mass on the Periodic Table, but the units are grams per mole.

Ag: 107.868 g/mol

Let's make another ratio using this information.

$\frac {107.868 \ g \ Ag}{1 \ mol \ ag}$

Multiply by the number of moles we calculated.

$1.366655596 \ mol \ Ag*\frac {107.868 \ g \ Ag}{1 \ mol \ ag}$

The moles of silver cancel out.

$1.366655596 *\frac {107.868 \ g \ Ag}{1 }$

$1.366655596 * {107.868 \ g \ Ag}$

$147.4184058 \ g\ Ag$

Round

The original measurement of atoms has 3 significant figures, so our answer must have the same. For the number we calculated, that is the ones place.

147.4184058

The 4 in the tenths place tells us to leave the 7 in the ones place.

$147 \ g\ Ag$

8.23 ×10²³ silver atoms are equal to approximately 147 grams.

3 0

3 years ago

Describe the relationship between the light-dependent and the light-independent reactions.

Slav-nsk [51]

The products of the light-dependent reactions are used to help 'fuel' the light-independent reactions.

Example:
NADPH and ATP are produced during the light-dependent reaction for use in the light-independent reaction (the Calvin Cycle).

3 0

3 years ago

What is the initial temperature of a gas if the volume changed from 1.00 l to 1.10 l and the final temperature was determined to

Feliz [49]

Using charles law
v1/t1=v2/t2
v1=1l
v2=1.1l
t2=255+273=528
t1=?
1/t1=1.1/528
cross multiply
1.1t1=528 divide both sides by 1.1
t1=528/1.1
t1=480k or 207celcius

5 0

4 years ago

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