True or false. an object's weight is constant in all circumstances

The reaction of an arrhenius acid with an arrhenius base produces water and

aleksandr82 [10.1K]

Answer: salt

Explanation:

Arrhenius acids are substances which dissociate in water to give $H^+$ ions.

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

Arrhenius bases are substances which dissociate in water to give $OH^-$ ions.

$NaOH\rightarrow Na^++OH^-$

Thus when they combine, they neutralize each other and produce salt and water.

$HCl+NaOH\rightarrow NaCl+H_2O$

6 0

3 years ago

Which BEST explains why the sun appears reddish at sunrise? A) Blue wavelengths of light have already been scattered before reac

romanna [79]

The answer is A. This is because sunlight is composed of various wavelengths in the electromagnetic spectrum. In the visible light spectrum, red light has a higher frequency (lower wavelength) than blue light that has a lower frequency (higher wavelength). Therefore red light has higher energy than blue light hence able to penetrate more into the atmosphere before scattering, compared to blue light. At sunrise and sunset, due to the relative angle of the sun to the observer, the sun rays have to penetrate a thicker layer to the atmosphere before reaching the observer

7 0

3 years ago

If an element has 4 naturally occuring isotopes, how many peaks will show up in the mass spectra?

BartSMP [9]

Base peak: The most intense (tallest) peak in a mass spectrum, due to the ion with the greatest relative abundance (relative intensity; height of peak along the spectrum's y-axis). Not to be confused with molecular ion: base peaks are not always molecular ions, and molecular ions are not always base peaks.

6 0

3 years ago

If 500.0 ml of 0.10 m ca2+ is mixed with 500.0 ml of 0.10 m so42−, what mass of calcium sulfate will precipitate? ksp for caso4

attashe74 [19]

Answer : The mass of calcium sulfate precipitate will be, 6.12 grams

Solution :

First we have to calculate the moles of $Ca^{2+}$ and $SO_4^{2-}$ .

$\text{Moles of }Ca^{2+}=\text{Molarity of }Ca^{2+}\times \text{Volume of }Ca^{2+}=0.10mole/L\times 0.5L=0.05\text{ moles}$

$\text{Moles of }SO_4^{2-}=\text{Molarity of }SO_4^{2-}\times \text{Volume of }SO_4^{2-}=0.10mole/L\times 0.5L=0.05\text{ moles}$

As, 0.05 moles of $Ca^{2+}$ is mixed with 0.05 moles of $SO_4^{2-}$ , it gives 0.05 moles of calcium sulfate.

Now we have to calculate the solubility of calcium sulfate.

The balanced equilibrium reaction will be,

$CaSO_4\rightleftharpoons Ca^{2+}+SO_4^{2-}$

The expression for solubility constant for this reaction will be,

$K_{sp}=(s)\times (s)$

$K_{sp}=(s)^2$

Now put the value of $K_{sp}$ in this expression, we get the solubility of calcium sulfate.

$2.40\times 10^{-5}=(s)^2$

$s=4.89\times 10^{-3}M$

Now we have to calculate the moles of dissolved calcium sulfate in one liter solution.

$\text{Moles of }CaSO_4=\text{Molarity of }CaSO_4\times \text{Volume of }CaSO_4=4.89\times 10^{-3}mole/L\times 1L=4.89\times 10^{-3}\text{ moles}$

Now we have to calculate the moles of calcium sulfate that precipitated.

$\text{Moles of }CaSO_4\text{ precipitated}=\text{Moles of }CaSO_4\text{ present}-\text{Moles of }CaSO_4\text{ dissolved}$

$\text{Moles of }CaSO_4\text{ precipitated}=0.05-4.89\times 10^{-3}=0.045\text{ moles}$

Now we have to calculate the mass of calcium sulfate that precipitated.

$\text{Mass of }CaSO_4\text{ precipitated}=\text{Moles of }CaSO_4\text{ precipitated}\times \text{Molar mass of }CaSO_4$

$\text{Mass of }CaSO_4\text{ precipitated}=0.045moles\times 136g/mole=6.12g$

Therefore, the mass of calcium sulfate precipitate will be, 6.12 grams

5 0

3 years ago

Read 2 more answers

Need help with chemistry question

luda_lava [24]

Answer:

See explanation

Explanation:

In this case, we have to check two variables:

1) The leaving group

2) The carbon bonded to the leaving group.

Let's check one by one:

2-chloro-3-methylbutane

In this molecule, the leaving group is "Cl", the carbon bonded to the leaving group has two neighbors. Therefore, we have a secondary substrate.

1-phenylpropan-1-ol

In this molecule, the leaving group is "OH", the carbon bonded to the hydroxyl group has two neighbors also. So, we have a secondary substrate.

(E)-pent-3-en-2-yl 4-methylbenzenesulfonate

In this case, the leaving group is "OTs" (Tosylate), the carbon bonded to the tosylate group has as a neighbor a double bond. Therefore, we have an allylic substrate.

3a-bromooctahydro-1H-indene

In this molecule, the leaving group is "Br", the carbon bonded to the bromine has three neighbors. So, we have a tertiary substrate.

1-iodo-3-methylbutane

In this molecule, the leaving group is "I", the carbon bonded to the iodide has only one neighbor. So, we have a primary substrate.

See figure 1

I hope it helps!

4 0

3 years ago

True or false. an object's weight is constant in all circumstances​

True or false. an object's weight is constant in all circumstances