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3 years ago

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Jeremy had a set of 500 toy blocks that could be connected by snapping them together. At first, he used all the blocks to build

a small house. Later, he rearranged all the blocks to build a small car. How did the mass of the house compare to the car?

2 answers:

valentina_108 [34]3 years ago

6 0

The masses were the same

Nat2105 [25]3 years ago

3 0

Answer:

The masses were the same.

Explanation:

study isalnd :)

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If the initial concentration of a is 0.00540 m, what will be the concentration after 795 s?

zavuch27 [327]

<span>Feb 19, 2014 - The units of k tell you that this is a second order reaction. So, to solve this, you need to use the integrated rate law for a 2nd order reaction: 1/[A] = kt + 1/[A]o 1/[A] = 0.540/Ms (835 s) + 1/0.00640 1/[A] = 607 [A] = 1.65X10^-3 M.</span><span>
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7 0

3 years ago

How many atoms are there in 4.13 moles of nickel (Ni)

PtichkaEL [24]

Answer:

24.87× 10²³ atoms of Ni

Explanation:

Given data:

Number of atoms of Ni= ?

Number of moles of Ni = 4.13 mol

Solution:

we will calculate the number of atoms of Ni by using Avogadro number.

It is the number of atoms , ions and molecules in one gram atom of element, one gram molecules of compound and one gram ions of a substance.

The number 6.022 × 10²³ is called Avogadro number.

1 mole = 6.022 × 10²³ atoms

4.13 mol × 6.022 × 10²³ atoms /1 mol

24.87× 10²³ atoms of Ni

6 0

3 years ago

A laboratory analysis of a sample finds it is composed of 38.8% carbon, 16.2% hydrogen, and 45.1% nitrogen. What is its empirica

Sladkaya [172]

Answer: The empirical formula for the given compound is $CH_5N$

Explanation : Given,

Percentage of C = 38.8 %

Percentage of H = 16.2 %

Percentage of N = 45.1 %

Let the mass of compound be 100 g. So, percentages given are taken as mass.

Mass of C = 38.8 g

Mass of H = 16.2 g

Mass of N = 45.4 g

To formulate the empirical formula, we need to follow some steps:

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{38.8g}{12g/mole}=3.23moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{16.2g}{1g/mole}=16.2moles$

Moles of Nitrogen = $\frac{\text{Given mass of nitrogen}}{\text{Molar mass of nitrogen}}=\frac{45.4g}{14g/mole}=3.24moles$

Step 2: Calculating the mole ratio of the given elements.

For the mole ratio, we divide each value of the moles by the smallest number of moles calculated which is 3.23 moles.

For Carbon = $\frac{3.23}{3.23}=1$

For Hydrogen = $\frac{16.2}{3.23}=5.01\approx 5$

For Oxygen = $\frac{3.24}{3.23}=1.00\approx 1$

Step 3: Taking the mole ratio as their subscripts.

The ratio of C : H : N = 1 : 5 : 1

Hence, the empirical formula for the given compound is $C_1H_5N_1=CH_5N$

3 0

3 years ago

Who was the first person to discover the existence of electrons

k0ka [10]

Electron was discovered by J. J. Thomson in 1897 when he was studying the properties of cathode ray.

4 0

3 years ago

Read 2 more answers

The volume of 0.05 M H2SO4 is needed to completely neutralise 15ml of 0.1 M NaOH solution is

Feliz [49]

V(NaOH)=15 mL =0.015 L
C(NaOH)=0.1 mol/L
C(H₂SO₄)=0.05 mol/L

2NaOH + H₂SO₄ = Na₂SO₄ + 2H₂O

n(NaOH)=V(NaOH)C(NaOH)=2n(H₂SO₄)
n(H₂SO₄)=V(H₂SO₄)C(H₂SO₄)

V(NaOH)C(NaOH)=2V(H₂SO₄)C(H₂SO₄)

V(H₂SO₄)=V(NaOH)C(NaOH)/{2C(H₂SO₄)}

V(H₂SO₄)=0.015*0.1/{2*0.05}=0.015 L = 15 mL

5 0

3 years ago