Look at the diagram. What number needs to be added to balance the equation?

Please help ASAP

marta [7]

Answer:

0.1

Explanation:

We must first put down the equation of the reaction in order to guide our solution of the question.

2HNO3(aq) + Sr(OH)2(aq) -------> Sr(NO3)2(aq) + 2H2O(l)

Now from the question, the following were given;

Concentration of acid CA= ??????

Concentration of base CB= 0.299M

Volume of acid VA= 17.8ml

Volume of base VB= 24.7ml

Number of moles of acid NA= 2

Number of moles of base NB= 1

From;

CAVA/CBVB= NA/NB

CAVANB= CBVBNA

CA= CBVBNA/VANB

SUBSTITUTING VALUES;

CA= 0.299 × 24.7 ×2 / 17.8×1

CA= 0.8298 M

But;

pH= -log[H^+]

[H^+] = 0.8298 M

pH= -log[0.8298 M]

pH= 0.1

5 0

3 years ago

Which set of reagents will best convert 2,2-dimethylpropan-1-ol (neopentyl alcohol) to 4,4-dimethylpentan-2-ol

Sladkaya [172]

Answer:

SOCl2, Mg, CH3CHO, H3O^+

Explanation:

The reaction of the 2,2-dimethylpropan-1-ol with SOCl2 converts the alcohol to an alkyl halide. This now reacts with Mg metal to yield a Grignard reagent.

The Grignard reagent reacts with CH3CHO and acid to yield 4,4-dimethylpentan-2-ol as shown in the image attached to this answer.

6 0

2 years ago

A sample of chlorine gas occupies a volume of 707 ml at 2.90 atm. What is the new pressure in torr if the volume is compressed t

dolphi86 [110]

Answer:

The new pressure is 3850 torr.

Explanation:

The relation between volume and pressure is inverse as per Boyle's law. Its mathematical form is given by :

$P_1V_1=P_2V_2$

Here,

$P_1=2.9\ atm$

$V_1=707\ mL$

$V_2=0.533\ L$

Let $P_2$ is the new pressure. So using Boyle's law we get :

$P_2=\dfrac{P_1V_1}{V_2}\\\\P_2=\dfrac{2.9\times 707}{0.533}\\\\P_2=3846.71\ \text{torr}$

or

$P_2=3850\ \text{torr}$

So, the new pressure is 3850 torr.

3 0

3 years ago

An inference is a prediction based on prior knowledge or experiences.<br>O<br>True<br>Or<br>False

goldenfox [79]

Answer:

True

Explanation:

5 0

2 years ago

Read 2 more answers

Calculate the frequency of the n=2 line in the lyman series of hydrogen

Alona [7]

Answer:

Approximately $2.47\times 10^{15}\; \rm Hz$ .

Explanation:

The Lyman Series of a hydrogen atom are due to electron transitions from energy levels $n \ge 2$ to the ground state where $n = 1$ . In this case, the electron responsible for the line started at $n = 2$ and transitioned to

A hydrogen atom contains only one electron. As a result, Bohr Model provides a good estimate of that electron's energy at different levels.

In Bohr's Model, the equation for an electron at energy level $n$ (

$\displaystyle - \frac{k\, Z^2}{n^2}$ (note the negative sign in front of the fraction,)

where

$k = 2.179 \times 10^{-18}\; \rm J$ is a constant.

$Z$ is the atomic number of that atom. $Z = 1$ for hydrogen.
$n$ is the energy level of that electron.

The electron that produced the $n = 2$ line was initially at the

$\begin{aligned} &E_{n = 2} \cr &= -\frac{k\, Z^2}{n^2} \cr &= -\frac{2.179 \times 10^{-18} \times 1}{2^2} \cr & \approx -5.4475\times 10^{-19}\; \rm J\end{aligned}$ .

The electron would then transit to energy level $n = 1$ . Its energy would become:

$\begin{aligned} &E_{n = 1} \cr &= -\frac{k\, Z^2}{n^2} \cr &= -\frac{2.179 \times 10^{-18} \times 1}{1^2} \cr & \approx -2.179 \times 10^{-18} \; \rm J\end{aligned}$ .

The energy change would be equal to

$\begin{aligned}&\text{Initial Energy} - \text{Final Energy} \cr &= E_{n = 2} - E_{n = 1} \cr &= -5.4475 \times 10^{-19} - \left(-2.179 \times 10^{-18}\right) \cr & \approx 1.63425\times 10^{-18}\; \rm J \end{aligned}$ .

That would be the energy of a photon in that $n = 2$ spectrum line. Planck constant $h$ relates the frequency of a photon to its energy:

$E = h \cdot f$ , where

$E$ is the energy of the photon.
$h \approx 6.62607015\times 10^{-34}\; \rm J \cdot s$ is the Planck constant.
$f$ is the frequency of that photon.

In this case, $E \approx 1.63425 \times 10^{-18}\; \rm J$ . Hence,

$\begin{aligned} f &= \frac{E}{h} \cr &\approx \frac{1.63425\times 10^{-18}}{6.62607015\times 10^{-34}} \cr & \approx 2.47 \times 10^{15}\; \rm s^{-1}\end{aligned}$ .

Note that $1 \; \rm Hz = 1 \; \rm s^{-1}$ .

6 0

3 years ago