All categories

3 years ago

What is a gift offered to a god by killing or burning something? Plz answer I’ll give brainlist and 50 points for free

Chemistry

2 answers:

wlad13 [49]3 years ago

5 0

Answer:

I believe it would be a sacrifice or a sacrificial ritual. hope this helped!! <3

umka2103 [35]3 years ago

4 0

Answer:

it is a sacrafice to god so the spirtit can go free

Explanation:

You might be interested in

A goldsmith melts 12.4 grams of gold to make a ring. The temperature of the gold rises from 26°C to 1064°C, and then the gold me

DiKsa [7]

Problem One

You will use both m * c * deltaT and H = m * heat of fusion.

Givens

m = 12.4 grams

c = 0.1291

t1 = 26oC

t2 = 1204

heat of fusion (H_f) = 63.5 J/grams.

Equation

H = m * c * deltaT + m * H_f

Solution

H = 12.4 * 0.1291 * (1063 - 26) + 12.4 * 63.5

H = 1660.1 + 787.4

H = 2447.5 or 2447.47 is the exact answer. I have to leave the rounding to you. I have no idea where to round it although I suspect 2450 would be right for 3 sig digs.

Problem Two

Formula and Givens

t1 = 14.5

t2 = 50.0

E = 5680

c = 4.186

m = ??

E = m c * deltaT

Solution

5680 = m * 4.186 * (50 - 14.5)

5680 = m * 4.186 * (35.5)

5680 = m * 148.603 * m

m = 5680 / 148.603

m = 38.22 grams That isn't very much. Be very sure you are working in joules. You'd leave that many grams in the kettle after drying it thoroughly.

m = 38.2 to 3 sig digs.

8 0

4 years ago

Read 2 more answers

What happens when magnesium reacts with dilute acid?

iogann1982 [59]

The magnesium dissolves to form magnesium chloride and hydrogen gas is released.

Mg + 2HCl -----> MgCl2 + H2(g).

5 0

3 years ago

Do this Q7 if someone do I will her or him brainliest

Simora [160]

The average speed :

1. 10.44 m/s

2. 10.42 m/s

3. 9.26 m/s

The distance 100 m have the greatest average speed

<h3>Further explanation </h3>

Given

Distance and time of runner

Required

Average speed

Solution

<em> Average speed : total distance : total time </em>

1. d = 100 m, t = 9.58 s

Average speed : 100 : 9.58 = 10.44 m/s

2. d=200 m, t=19.19 s

Average speed : 200 : 19.19 = 10.42 m/s

3. d=400 m, t = 43.18 s

Average speed : 400 : 43.18 = 9.26 m/s

The distance 100 m have the greatest average speed

7 0

3 years ago

A 7.0 g sample of a hydrocarbon (a molecule that has only hydrogen and carbon) is subject to combustion analysis. The mass of CO

Akimi4 [234]

Answer: The empirical formula for the given compound is $CH_2$

Explanation:

The chemical equation for the combustion of compound having carbon and hydrogen follows:

$C_xH_y+O_2\rightarrow CO_2+H_2O$

where, 'x' and 'y' are the subscripts of carbon and hydrogen respectively.

We are given:

Mass of $CO_2=22.0g$

We know that:

Molar mass of carbon dioxide = 44 g/mol

For calculating the mass of carbon:

In 44 g of carbon dioxide, 12 g of carbon is contained.

So, in 22.0 g of carbon dioxide, $\frac{12}{44}\times 22.0=6g$ of carbon will be contained.

For calculating the mass of hydrogen:

Mass of hydrogen = Mass of sample - Mass of carbon

Mass of hydrogen = 7.0 g - 6 g

Mass of hydrogen = 1.0 g

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

Step 1: Converting the given masses into moles.

Moles of Carbon = $\frac{\text{Given mass of Carbon}}{\text{Molar mass of Carbon}}=\frac{6g}{12g/mole}=0.5moles$

Moles of Hydrogen = $\frac{\text{Given mass of Hydrogen}}{\text{Molar mass of Hydrogen}}=\frac{1.0g}{1g/mole}=1.0moles$

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 0.5 moles.

For Carbon = $\frac{0.5}{0.5}=1$

For Hydrogen = $\frac{1.0}{0.5}=2$

Step 3: Taking the mole ratio as their subscripts.

The ratio of Fe : C : H = 1 : 2

Hence, the empirical formula for the given compound is $C_{1}H_{2}=CH_2$

4 0

4 years ago

A balloon is filled with 2.00 L of helium gas at sea level, 1.00 atm and 12.0ºC. The balloon is released and it rises to an alti

pshichka [43]

Answer:

THE VOLUME OF THE BALLOON IS 1.45 L

Explanation:

At sea level:

Volume = 2 L

Pressure = 1 atm

Temperature = 12 °C

At 30000 ft altitude:

Pressure = 0.30 atm

Temperature = -55°C

Volume = unknown

Using the general gas formula:

P1 V1 / T1 = P2 V2 / T2

Re-arranging the formula by making V2 the subject of the equation, we have;

V2 = P1 V1 T2 / T1 P2

V2 = 1 * 2 * 12 / 0.30 * 55

V2 = 24 / 16.5

V2 = 1.45 L

The volume of the balloon at the temperature of -55 C and 0.30 atm is 1.45 L

3 0

3 years ago