Calculate the average reaction rate. PLZ HELP TY!

When performing a gram stain, it is necessary to remove crystal violet stain from any gram negative bacteria on the slide. Micro

Snezhnost [94]

Answer:

10.11 mL

Explanation:

Given that :

Total volume of decolorizer = 500 mL

A common recipe is equal parts 95% ethanol and acetone.

So;

The volume of 95% ethanol in the decolorizer = 500mL/2 = 250 mL

Let represent V(ml) for the 99% of ethanol needed to make 95% of the 250 mL of ethanol;

Therefore:

V(ml) × 0.99 = 250 × 0.95

V(ml) × 0.99 = 237.5

V(ml) = 237.5/0.99

V(ml) = 239.89 mL

Hence; the amount of water to be added to 95% of ethanol = ( 250 - 239.89 )mL = 10.11 mL

5 0

3 years ago

BNCT relies on the initial targeting of tumor cells by an appropriate chemical compound tagged with 10 5B, which preferentially

zimovet [89]

Answer:

$^{10}B + ^{1}n_{th} \rightarrow ^{4}He + ^{7}Li + 2.79MeV$

Explanation:

BNCT therapy consists in the irradiation of the atoms of ¹⁰B that are contained in the tissue that has the tumor cells, with thermal neutrons to produce alpha particles and ⁷Li nucleus, according to the following nuclear reaction:

$^{10}_{5}B + ^{1}_{0}n_{th} \rightarrow ^{4}_{2}He + ^{7}_{3}Li + 2.79MeV$

<em>Since the tissues also contain H and N atoms, the neutron irradiation produces secondary reactions in which proton particles and gamma radiation are produced:</em>

$^{1}H + n_{th} \rightarrow ^{2}H + \gamma$

$^{14}N + n_{th} \rightarrow ^{14}C + p$

I hope it helps you!

4 0

4 years ago

If 2.5 grams of calcium bromide reacted with excess lithium oxide, how many grams of bromide product would be formed?

Ede4ka [16]

Answer: 2.17 g of bromide product would be formed

Explanation:

The reaction of calcium bromide with lithium oxide will be:

$CaBr_2+Li_2O\rightarrow 2LiBr+CaO$

To calculate the moles :

$\text{Moles of solute}=\frac{\text{given mass}}{\text{Molar Mass}}$

$\text{Moles of calcium bromide}=\frac{2.5g}{199.9g/mol}=0.0125moles$

As lithium oxide is in excess, calcium bromide is the limiting reagent.

According to stoichiometry :

1 mole of $CaBr_2$ produce = 2 moles of $LiBr$

Thus 0.0125 moles of $CaBr_2$ will require= $\frac{2}{1}\times 0.0125=0.025moles$ of $NH_3$

Mass of $LiBr=moles\times {\text {Molar mass}}=0.025moles\times 86.8g/mol=2.17g$

Thus 2.17 g of bromide product would be formed

3 0

3 years ago

How do the greenhouse gases in earths atmosphere affect our environment ?

Law Incorporation [45]

certin greenhouse gasses poke hole in our Ozone layer which can lead to higher temps and bigger storms and longer droughts

8 0

3 years ago

Read 2 more answers

Using 25 micromoles, pmoles, of CO2 consumed how many grams of glucose (MW = 180 g/mole) were produced?

Sladkaya [172]

Grams of glucose = 0.7488 mg=748.8 μg

<h3>Further explanation</h3>

A reaction coefficient is a number in the chemical formula of a substance involved in the reaction equation. The reaction coefficient is useful for equalizing reagents and products.

Reaction

6CO₂+6H₂O⇒C₆H₁₂O₆+6O₂

25 micromoles, μmoles, of CO₂ :

moles CO₂ : moles glucose - C₆H₁₂O₆ = 6 : 1, so :

$\tt moles~glucose=\dfrac{1}{6}\times 25\mu~moles=4.16\mu~moles=4.16\times 10^{-6}~moles$

mass of glucose

$\tt mass=mol\times MW\\\\mass=4.16\times 10^{-6}\times 180\\\\mass=7.488\times 10^{-4}g=0.7488~mg=748.8\mu g$

8 0

4 years ago