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4 years ago

If an atom has 35 protons in the nucleus, how many will it have orbiting nucleus

1 answer:

8 0

It will have 35 ''electrons'' . Basically the number of protons in the nucleus of an atom is always equal to the number of electrons but its just that protons are positively charged and electrons are negatively charged. <span />

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A 1.0 g sample of a cashew was burned in a calorimeter containing 1000. g of water, and the temperature of the water changed fro

Savatey [412]

Answer:

The correct answer is option C.

Explanation:

1.0 g sample of a cashew :

Heat released on combustion of 1.0 gram of cashew = -Q

We have mass of water = m = 1000 g

Specific heat of water = c = 4.184 J/g°C

ΔT = 30°C - 25°C = 5°C

Heat absorbed by the water : Q

$Q=1000 g\times 4.184 J/g^oC\times 5^oC=20,920 J$

Heat released on combustion of 1.0 gram of cashew is -20,920 J.

3.0 g sample of a marshmallows :

Heat released on combustion of 3.0 g sample of a marshmallows = -Q'

We have mass of water = m = 2000 g

Specific heat of water = c = 4.184 J/g°C

ΔT = 30°C - 25°C = 5°C

Heat absorbed by the water : Q'

$Q'=2000 g\times 4.184 J/g^oC\times 5^oC=41,840 J$

Heat released on 3.0 g sample of a marshmallows= -Q' = -41,840 J

Heat released on 1.0 g sample of a marshmallows : q

$q =\frac{-Q'}{3} = \frac{-41,840 J}{3}=-13,946.67 J$

Heat released on combustion of 1.0 gram of marshmallows -13,946.67 J.

-20,920 J. > -13,946.67 J

The combustion of 1.0 g of cashew releases more energy than the combustion of 1.0 g of marshmallow.

5 0

3 years ago

For a reaction: aA → Products, [A]o -4.3 M, and the first two half-lives are 56 and 28 minutes, respectively. Calculate k (witho

Mkey [24]

Answer:

C. $3.8\times 10^{-2}$

Explanation:

We are given that

Initial concentration, $[A]_o=4.3 M$

First half life, $t_{\frac{1}{2}}=56$ minutes

Second half life, $t'_{\frac{1}{2}}=28$ minutes

We have to find K.

The given reaction is zero order reaction.

We know that for zero order reaction

$t_{\frac{1}{2}}=\frac{[A]_o}{2k}$

Using the formula

$56=\frac{4.3}{2k}$

$k=\frac{4.3}{2\times 56}$

$k=3.8\times 10^{-2}$

Hence, option C is correct.

4 0

3 years ago

A 42.0g sample of compound containing only C and H was analyzed. The results showed that the sample contained 36.0g of C and 6.0

user100 [1]

Answer:

What is the empirical formula of the compound?

Explanation:

When the relative masses of elements in a hydrocarbon are given, it is possible to use this information to obtain the empirical formula by dividing the given masses of each element by the relative atomic masses of the element. The lowest ratio is now used to divide through to obtain the empirical formula of the compound.

The empirical formula only shows that ratio of atoms of each element present in the compound. From the information provided, the empirical formula of the compound is CH2. Hence the answer.

5 0

3 years ago

To determine the molar mass of an unknown organic acid, HA, a 1.056 g sample is titrated with standardized NaOH. Calculate the m

meriva

Explanation:

Equation for the given reaction is as follows.

$HA(aq) + NaOH(aq) \rightarrow NaA(aq) + H_{2}O(l)$

Therefore, moles of NaOH and HA are calculated as follows.

Moles of NaOH = $0.256 M \times 0.03378 L$

= $8.64 \times 10^{-3}$

= 0.00864 mol

Moles of HA = 0.00864

Also, moles = $\frac{\text{weight}}{\text{molecular weight}}$

Molecular weight = $\frac{1.056 g}{0.00864 mol}$

= 122.22 g/mol

Thus, we can conclude that molar mass of given unknown organic acid is 122.22 g/mol.

4 0

3 years ago

At high temperatures, calcium carbonate (CaCO3) decomposes to form calcium oxide (CaO) and carbon dioxide (CO2) gas. The average

bulgar [2K]

Answer:

Option D is correct

Explanation:

The individual mass of reactants in the reaction are

Calcium Oxide (CaO) = $56.079$ amu

Carbon dioxide (CO2) $= 44.009$ amu

Since this reaction is a simple addition reaction in which no mass of the reactants is wasted. Also, no energy is released accompanied by conversion of mass into energy.

Hence, total mass of reactants would be equal to the mass of the product.

Thus, mass of calcium carbonate i.e CaCO3 is

$= 56.079 + 44.009 \\$

$= 100.08$ amu

Hence, option D is correct

4 0

4 years ago