Help please need to find the frequency of the wave

Which of the following is a homogeneous mixture? a. Na(s) b. H2O(l) c. Cl2(g) d. NaCl(aq)

iVinArrow [24]

Answer:

H²0 is answer is H20 ok please

7 0

3 years ago

1. An oxide of chromium is found to have the following % composition: 68.4% Cr

abruzzese [7]

Answer:

Empirical formula is Cr₂O₃.

Explanation:

Given data:

Percentage of Cr = 68.4%

Percentage of O = 31.6%

Empirical formula = ?

Solution:

Number of gram atoms of Cr = 68.4 / 52 = 1.3 2

Number of gram atoms of O = 31.6 / 16 = 1.98

Atomic ratio:

Cr : O

1.32/1.32 : 1.98/1.32

1 : 1.5

Cr : O = 1 : 1.5

Cr : O = 2(1 : 1.5)

Empirical formula is Cr₂O₃.

6 0

3 years ago

How many grams of CaCl2 are needed to make 277.8g of a solution that is 31.5% (m/m) in water? Note that mass is not technically

Sveta_85 [38]

<u>Answer:</u> The mass of calcium chloride present in given amount of solution is 87.5 g

<u>Explanation:</u>

We are given:

Mass of solution = 277.8 grams

Also, 31.5 % (m/m) of calcium chloride in water. This means that 31.5 g of calcium chloride is present in 100 g of solution.

To calculate the mass of calcium chloride in the given amount of solution, we use unitary method:

in 100 g of solution, the mass of calcium chloride present is 31.5 g

So, 277.8 g of solution, the mass of calcium chloride present is $\frac{31.5}{100}\times 277.8=87.5g$

Hence, the mass of calcium chloride present in given amount of solution is 87.5 g

8 0

3 years ago

Complete combustion of 8.10 g of a hydrocarbon produced 25.9 g of CO2 and 9.27 g of H2O. What is the empirical formula for the h

balu736 [363]

CxHy + O2 --> x CO2 + y/2 H2O

Find the moles of CO2 : 18.9g / 44 g/mol = .430 mol CO2 = .430 mol of C in compound

Find the moles of H2O: 5.79g / 18 g/mol = .322 mol H2O = .166 mol of H in compound

Find the mass of C and H in the compound:

.430mol x 12 = 5.16 g C

.166mol x 1g = .166g H

When you add these up they indicate a mass of 5.33 g for the compound, not 5.80g as you stated in the problem.

Therefore it is likely that either the mass of the CO2 or the mass of H20 produced is incorrect (most likely a typo).

In any event, to find the formula, you would take the moles of C and H and convert to a whole number ratio (this is usually done by dividing both of them by the smaller value).

8 0

3 years ago

Identify the Bronsted-Lowry acid, the Bronsted-Lowry base, the conjugate acid and the conjugate base for each of the following r

Vlad1618 [11]

Answer:

Acids → H₂CO₃ from equilibrium 1 and water, from equilibrium 2.

Bases → Water from equilibrium 1 and ammonia from equilibrium 2.

In 1st equilibrium, H₃O⁺ is the conjugate acid and HCO₃⁻ the conjugate base.

In 2nd equilibrium, NH₄⁺ is the conjugate acid, and OH⁻, the conjugate base.

Explanation:

By the Bronsted-Lowry you know that acids are the one that release protons and base are the ones that catch them.

For the first equilibrium:

H₂CO₃(aq) + H₂O(l) ⇄ H₃O⁺(aq) + HCO₃⁻(aq)

Carbonic acid is the acid → It donates the proton to water, so the water becomes the base. As H₂CO₃ is the acid, the bicarbonate is the conjugate base (it can accept the proton from water to become carbonic acid, again) and the hydronium is the conjugate acid (it would release the proton to become water).

For the second equilibrium:

NH₃(aq) + H₂O(l) ⇄ NH₄⁺ (aq) + OH⁻(aq)

This is the opposite situation → Water relase the proton to ammonia, that's why water is the acid and NH₃, the base (it accepted to become ammonium). The NH₄⁺ is the conjugate acid (it can release the H⁺ to become ammonia) and the OH⁻ is the conjugate base (It can accept the proton to become water, again).

5 0

3 years ago