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3 years ago

How many calories of heat were added to 347.9 g of water to raise its temperature from 25oC to 55oC?

Chemistry

1 answer:

saw5 [17]3 years ago

5 0

Answer:

10437calories

Explanation:

The following data were obtained from the question given:

M = 347.9g

C = 4.2J/g°C

T1 = 25°C

T2 = 55°C

ΔT = 55 — 25 = 30°C

Q =?

Q = MCΔT

Q = 347.9 x 4.2 x 30

Q = 43835.4J

Converting this to calories, we obtained the following:

4.2J = 1 calorie

43835.4J = 43835.4/ 4.2 = 10437calories

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The nuclear fission of U-235 can occur by different pathways. If 1 neutron is absorbed by U-235 to produce 36 neutrons and Kr-94

motikmotik

Answer:

See explanation

Explanation:

What I have written in the image attached is called a nuclear equation. It differs from a chemical reaction equation in the sense that it involves transformations that occur in the nucleus of atoms.

The nuclear equation must be balanced. This means that the mass and charge on both sides of the reaction equation must be the same.

On the left hand side the U-235 interacts with a neutron. The total mass on the left hand side is 236 while the total charge is 92. If we sum up the masses and charges of Ba and Kr, we also get a total of 236 mass units and a charge of 92.

Hence the other nucleus is barium-141

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3 years ago

A chemical factory is making soda ash (NA2CO3) from sodium bicarbonate. The production manager calculates they will make 80 tons

grigory [225]

Answer:

$\boxed{\text{ 93 \%}}$

Explanation:

$\text{\% yield} = \dfrac{\text{actual yield}}{\text {theoretical yield}} \times\text{100 \%}$

Data:

$\begin{array}{rcr}\text{Actual yield} & = & \text{74.3 T}\\\text{Theoretical yield} & = & \text{80 T}\\\end{array}$

Calculation:

$\text{\% yield} = \dfrac{\text{74.3 T}}{\text {80 T}} \times\text{100 \%} = \textbf{93 \%}}\\\\\text{The percent yield was } \boxed{\textbf{93 \%}}$

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3 years ago

Use the information below to match each object with the metal it is made of.

USPshnik [31]

Answer:

Explanation:

Love the pfp

5 0

3 years ago

2. Na2SiO3(s) + HF(aq) !H2SiF6(aq) + NaF(aq) + H2O (l)

Alex

Answer:

a) Na2SiO3(s) + 8HF(aq) → H2SiF6(aq) + 2NaF(aq) + 3H2O (l)

b) 2.40 moles HF

c) 5.25 grams NaF

d)0.609 grams Na2SiO3

e) 0.89 grams Na2SiO3

Explanation:

Step 1: Data given

Step 2: The balanced equation

Na2SiO3(s) + 8HF(aq) → H2SiF6(aq) + 2NaF(aq) + 3H2O (l)

b) How many moles of HF are needed to react with 0.300 mol of Na2SiO3?

For 1 mol Na2SiO3 we need 8 moles HF to produce 1 mol H2SiF6, 2 moles NaF and 3 moles H2O

For 0.300 moles Na2SiO3 we'll need 8*0.300 = 2.40 moles HF

c. How many grams of NaF form when 0.500 mol of HF reacts with excess Na2SiO3?

For 1 mol Na2SiO3 we need 8 moles HF to produce 1 mol H2SiF6, 2 moles NaF and 3 moles H2O

For 0.500 moles we'll have 0.500 / 4 = 0.125 moles NaF

Mass NaF = 0.125 moles * 41.99 g/mol

Mass NaF = 5.25 grams NaF

d. How many grams of Na2SiO3 can react with 0.800 g of HF?

Moles HF = 0.800 grams / 20.01 g/mol

Moles HF = 0.0399 moles

For 1 mol Na2SiO3 we need 8 moles HF to produce 1 mol H2SiF6, 2 moles NaF and 3 moles H2O

For 8 moles we need 1 moles Na2SiO3 to react

For 0.0399 moles we need 0.0399/8 = 0.00499 moles Na2SiO3

Mass Na2SiO3 = 0.00499 moles * 122.06 g/mol

Mass Na2SiO3 = 0.609 grams Na2SiO3

Using your answer to part d-If you started with 1.5 g of Na2SiO3how many grams would be left after the reaction is complete?

Number of moles HF = 0.0399 moles

Number of moles Na2SiO3 = 1.5 grams / 122.06 g/mol = 0.0123 moles

For 1 mol Na2SiO3 we need 8 moles HF to produce 1 mol H2SiF6, 2 moles NaF and 3 moles H2O

For 8 moles we need 1 moles Na2SiO3 to react

For 0.0399 moles we need 0.0399/8 = 0.00499 moles Na2SiO3

There will remain 0.0123 - 0.00499 = 0.00731 moles Na2SiO3

Mass Na2SiO3 remaining = 0.00731 * 122.06 g/mol = 0.89 grams Na2SiO3

7 0

3 years ago

Alternative energy sources

sammy [17]

Answer:

Natural resources like crude oil or animal disposal(poop)

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