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2 years ago

Convert 9541mg to kg

Chemistry

2 answers:

Westkost [7]2 years ago

4 0

Answer: 9.541 kg

Explanation:

Alexus [3.1K]2 years ago

4 0

Answer:

i believe it is 0.009541

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a sample of gold required 2.1200j of heat to melt it from room temperature, 22.0 degrees celsius to its melting point, 1064.4 de

ElenaW [278]

Mass of Gold = 267.165 × 0.01552494829

⇒ 4.1477228099

The amount of heat(q) required to raise m grams of a substance-specific C from T1 to T2 is given by

q=m C (T2-T1) ........1

Given : q= 2.1200 J

the initial temperature of gold, T1 = 22.0Celcius

the final temperature of gold, T2 = 1064.4Celcius

specific heat of gold = 0.131

putting values in eq 1:

⇒ 2.1200 = m × 0.131 × (1064.4-22)

⇒ 2.1200 = m × 0.131 × 1042.4

⇒ 2.1200 / 136.5544

⇒ 0.01552494829

Since 1g= 0.01552494829 Pounds

Mass of Gold = 267.165 × 0.01552494829

⇒ 4.1477228099

Learn more about temperature here: brainly.com/question/11464844

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7 0

1 year ago

A 46.2 mL,0.568 M calcium nitrate solution is mixed with 80.5mL of 1.396M calcium nitrate solution.Calculate tge concentration o

Ne4ueva [31]

Answer:

1.09 M

Explanation:

Let's define the equation that will be used to calculate the final concentration of the resultant calcium nitrate solution. In order to calculate it, we need to find the total number of moles of calcium nitrate and divide by the total volume of the resultant solution:

$c=\frac{n}{V}$

This equation firstly helps us find the number of moles of calcium nitrate. Multiplying molarity by volume will yield the moles. Adding the moles from the first component to the second component will provide us with the total number of moles of calcium nitrate:

$n_{Ca(NO_3)_2}=46.2 mL\cdot0.568 M+80.5 mL\cdot1.396 M=138.62 mmol$

Now, the total volume of this solution can be found by adding the volume values of each component:

$V_total=46.2 mL+80.5 mL=126.7 mL$

Finally, dividing the moles found by the total volume will yield the final molarity:

$c_{final}=\frac{138.62 mmol}{126.7 mL}= 1.09 M$

6 0

3 years ago

in an experiment 3.425g of lead oxide was reduced to form 3.105g of lead the empirical formula of the lead oxide is?

Romashka-Z-Leto [24]

Answer:

Pb3O4

Explanation:

According to this question, 3.425g of lead oxide was reduced to form 3.105g of lead in an experiment. Since lead oxide contains both lead (Pb) and oxygen (O) element,

Mass of lead oxide = 3.425g

Mass of lead = 3.105g

Mass of oxygen = (3.425g - 3.105g) = 0.320g

Next, we convert each mass value to mole by dividing by respective molar mass

Pb = 3.105g ÷ 207.2 = 0.0149mol

O = 0.320g ÷ 16 = 0.02mol

Next, we divide each mole value by the smallest (0.0149)

Pb = 0.0149mol ÷ 0.0149mol = 1

O = 0.02mol ÷ 0.0149mol = 1.342

Multiply each ratio value by 3 to get:

Pb = 1 × 3 = 3

O = 1.342 × 3 = 4.026

The whole number ratio, approximately, of Pb and O is 3:4, hence, their empirical formula is Pb3O4.

4 0

2 years ago

A 25.0 mL sample of 0.100 M lactic acid (HC3H503, Ka = 1.4 E-4) is titrated with 0.100 M NaOH solution. Calculate the pH of the

aleksklad [387]

I’m a bit confused too

4 0

3 years ago

Balance the following reaction. A coefficient of "1" is understood. Choose option "blank" for the correct answer if the

Vera_Pavlovna [14]

2NH₂ + O₂ → N₂ + 2H₂O

<u>Explanation:</u>

Balancing the equation means, the number of atoms on both sides of the equation must be the same.

In the case of the given equation, we have to find out whether it is balanced or not.

2NH₂ + O₂ → N₂ + 2H₂O

Atoms Number of atoms before balancing after balancing

LHS RHS LHS RHS

N 1 2 2 2

H 2 2 4 4

O 2 1 2 2

To balance the N atoms, we have to put 2 in front of NH₂, and then to balance the H, O atoms, we have to put 2 in front of H₂O, so that each atom in left hand as well as right hand side of the equation was balanced.

8 0

3 years ago