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3 years ago

How to calculate the sum of the ratio of the ions

1 answer:

6 0

If it is just ratio you must just put the mass number on the either side of the symbol ‘:’ and cancel but the same process is not used to find simplest ratio

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A 5.0-kg box is pushed across the floor with a force of 32 N. What is the frictional force on the box if the box's acceleration

Slav-nsk [51]

$From\ 2nd\ Newton\ Law\ resultuant\ force: \\\\\ F_R=m*a=5kg*3,6\frac{m}{s^2}=18N\\\\
F=32N\\\\
F_f- friction\ force\\\\
F_R=F-F_f\\\\
F_f=F-F_R=32N-18N=14N\\\\Friction\ force\ is\ equal\ to\ 14\ N.$

3 0

4 years ago

Read 2 more answers

How can I identify easliy Amphoteric oxide from basic oxide

never [62]

Answer:

A general guideline to determine if oxide is acidic, basic, or amphoteric is to use the periodic table. Typically, metals such as Ba form basic oxides (BaO), while nonmetals such as S form acidic oxides (SO3).

8 0

3 years ago

An unbalanced chemical equation is shown:

Sati [7]

Answer:

1. Two to decompose

Explanation:

It's going to be a decomposition so you can eliminate answers 3 and 4. It is a decomposition because it is splitting apart. Then you can figure out that it is going to be two instead of four by balancing the equation

H2O2 → 2H2O + O2

$2H_{2}O_{2} → 2H_{2}O + O_{2}$

6 0

3 years ago

Read 2 more answers

Be sure to answer all parts. A freshly isolated sample of 90Y was found to have an activity of 2.2 × 105 disintegrations per min

Leokris [45]

Answer:

The half-life is $t_h = 3.856*10^{3} minute$

Explanation:

From the question we are told that

The sample is 90 Y

The first activity is $A_1 = 2.2 *10^5$ per minute

The second activity is $A_2 = 5.8 *10^3$ per minute

The duration from 1:00 p.m. on December 3, 2006 to 2:15 p.m. on December 17, 2006 is

$t = 14 \ days \ 1 hr \ 15 min$

Converting to minutes we have

$t = (14 * 24 * 60) + (1* 60) + 15$

$t = 20235 \ minutes$

The first order rate constant for this disintegrations can be mathematically represented as

$ln \frac{A_2}{A_1} = - \lambda t$

Where $\lambda$ is the rate constant

Substituting values

$ln [\frac{5.8 * 10^{3}}{2.2 *10^{5}} ] = - \lambda * 20235$

$-3.6358 = - \lambda * 20235$

$\lambda = \frac{3.6358}{20235}$

$\lambda = 1.7968 *10^{-4} minute^{-1}$

The half life is mathematically represented as

$t_{h} = \frac{0.693}{\lambda }$

So $t_h = \frac{0.693}{1.7968 *10^{-4}}$

$t_h = 3.856*10^{3} minute$

5 0

3 years ago

What is the molecular formula for an oligosaccharide containing eight glucose monomers linked together by dehydration reactions?

melomori [17]

The answer is C24H42O21. Lack of hydration responses shapes covalent bonds between two monomers by the arrival of a water particle. Every monomer taking an interest in the bond discharges either a hydroxyl amass (- OH) or a hydrogen (H) for each bond shaped. Connecting four glucose monomers requires three covalent bonds. This outcome in the loss of three particles of water or six hydrogen molecules and three oxygen iotas.

7 0

4 years ago