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9 days ago

One of the alkali metals reacts with oxygen to form a solid white substance. when this substance is dissolved in water, the solu

tion gives a positive test for hydrogen peroxide, h2o2. when the solution is tested in a burner flame, a lilac-purple flame is produced. what is the likely identity of the metal?

Chemistry

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If 4.168 kJ of heat is added to a calorimeter containing 75.40 g of water, the temperature of the water and the calorimeter incr

VMariaS [2998]

Answer:

The heat capacity of the calorimeter is $C_c$ = 54.4 $\frac{J}{c}$

Explanation:

Given the data

Heat supplied Q = 4.168 KJ = 4168 J

Mass of water $m_w$ = 75.40 gm

Change in temperature = ΔT = 35.82 - 24.58 = 11.24 °C

From the conditions provided

Q = $m_w C_w$ ΔT + $C_c$ ΔT

Plugging all values into the above equation yields

4168 = 75.70 × 4.18 × 11.24 + $C_c$ × 11.24

611.37 = $C_c$ × 11.24

$C_c$ = 54.4 $\frac{J}{c}$

This represents the heat capacity of the calorimeter.

7 0

2 months ago

Nicole has 2 glasses on the counter: one of water and one of sugar. She is baking a cake and needs to use the sugar-water for th

Alekssandra [3086]

Answer:Sugar-water is a mixture

Explanation:

If it consists of pure sugar, it's classified as neither; however, when mixed with water, it forms a homogeneous mixture.

8 0

2 months ago

Read 2 more answers

In an experiment a student mixes a 50.0 mL sample of 0.100 M AgNO₃(aq) with a 50.0 mL sample of 0.100 M NaCl(aq) at 20.0°C in a

KiRa [2933]

The enthalpy change associated with the precipitation reaction is 84 kJ/mole

Why?

The chemical equation for the reaction can be written as

AgNO₃(aq) + NaCl (aq) → AgCl(s) + NaNO₃(aq)

To determine the enthalpy change, the following equation applies

$\Delta H =\frac{Q}{n}$

To calculate the heat (Q):

$Q=m*C*\Delta T=(100g)*(4.2 J/g*^\circ C)*(21^\circ C-20^\circ C)\\\\Q=420J$

Next, we need to calculate the number of moles involved in the reaction (n):

$n=[AgNO_3]*v(L)=(0.1M)*(0.05L)=0.005moles$

With these two values, we can substitute them into the first equation:

$\Delta H= \frac{420J}{0.005moles}=84000J/mole=84kJ/mole$

Have a great day!

5 0

2 months ago

Sulfur and oxygen react to produce sulfur trioxide. In a particular experiment, 7.9 grams of SO3 are produced by the reaction of

VMariaS [2998]

Result:

94.7 %

Explanation:

The balanced reaction is:

2 S + 3 O₂ → 2 SO₃

The stoichiometric mole ratio is:

S: 2 moles

O₂: 3 moles

Moles are calculated as mass divided by molar mass:

n = w / m

where n = moles, w = mass, m = molar mass.

Given:

For sulfur: w = 6.0 g, molar mass = 32 g/mol, so n = 6 / 32 = 0.1871 mol

For oxygen: w = 5.0 g, molar mass = 32 g/mol, thus n = 5 / 32 = 0.15625 mol

Comparing to stoichiometric ratios, sulfur is in excess, so oxygen is the limiting reagent, controlling product formation.

Using proportions:

3 mol O₂ produce 2 mol SO₃, so 1 mol O₂ yields 2/3 mol SO₃.

Therefore, 0.15625 mol O₂ yields (2/3) × 0.15625 = 0.1042 mol SO₃.

Mass of SO₃ produced = n × molar mass = 0.1042 mol × 80 g/mol = 8.340 g

The percentage yield is actual yield divided by theoretical yield times 100:

Percent yield = (7.9 g / 8.340 g) × 100 = 94.7 %

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3 months ago

Which of the following substances would release the most amount of heat when they cool from 50oC to 25oC if you started with equ

lions [2927]

Answer:

pyvugdhgtcectbibycrc

Explanation:

c9yridutcltzojvifzlh ljf7gp please could you share the link for a moment 4

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2 months ago