All categories

3 years ago

0.0278, how many significant figures

Chemistry

1 answer:

My name is Ann [436]3 years ago

3 0

Answer:

There are three significant figures

Explanation:

When counting sig figs you don't count the zeros unless it is between a number greater than zero. The two zeros aren't between the greater numbers so there are only 3.

You might be interested in

Determine the molecular formulas for compounds having the following empirical formula and molar mass:

boyakko [2]

Answer:

C6H24S3

Explanation:

To go from Empirical Formula to Molecular Formula, you have to divide the actual amount (given) by the molar mass of the EF:

192 / 2(12.01)+8(1.01)+(32.06)

= 2.99

= 3

Then you multiply every subscript in the EF by 3, which gives you C6H24S3

Hope this helps!

-Emma Victoria

5 0

3 years ago

The beaker shown below contains a 0.58 M solution of dye in water. How many moles of dye are there in the beaker?

seropon [69]

According to the equation of molarity:

Molarity= no.of moles / volume per liter of Solution

when we have the molarity=0.58 M and the beaker at 150mL so V (per liter) = 150mL/1000 = 0.150 L

by substitution:
∴ No.of moles = Molarity * Volume of solution (per liter)
= 0.58 * 0.150 = 0.087 Moles

7 0

3 years ago

Read 2 more answers

Nickel and carbon monoxide react to form nickel carbonyl, like this: (s)(g)(g) At a certain temperature, a chemist finds that a

horsena [70]

The question is incomplete, here is the complete question:

Nickel and carbon monoxide react to form nickel carbonyl, like this:

$Ni(s)+4CO(g)\rightarrow Ni(CO)_4(g)$

At a certain temperature, a chemist finds that a 2.6 L reaction vessel containing a mixture of nickel, carbon monoxide, and nickel carbonyl at equilibrium has the following composition:

Compound Amount

Ni 12.7 g

CO 1.98 g

$Ni(CO)_4$ 0.597 g

Calculate the value of the equilibrium constant.

<u>Answer:</u> The value of equilibrium constant for the reaction is 2448.1

<u>Explanation:</u>

We are given:

Mass of nickel = 12.7 g

Mass of CO = 1.98 g

Mass of $Ni(CO)_4$ = 0.597 g

Volume of container = 2.6 L

To calculate the number of moles for given molarity, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Given mass of solute}}{\text{Molar mass of solute}\times \text{Volume of solution (in L)}}$

$\text{Equilibrium concentration of nickel}=\frac{12.7}{58.7\times 2.6}=0.083M$

$\text{Equilibrium concentration of CO}=\frac{1.98}{28\times 2.6}=0.0272M$

$\text{Equilibrium concentration of }Ni(CO)_4=\frac{0.597}{170.73\times 2.6}=0.00134M$

For the given chemical reaction:

$Ni(s)+4CO(g)\rightarrow Ni(CO)_4(g)$

The expression of equilibrium constant for the reaction:

$K_{eq}=\frac{[Ni(CO)_4]}{[CO]^4}$

Concentrations of pure solids and pure liquids are taken as 1 in equilibrium constant expression.

Putting values in above expression, we get:

$K_{eq}=\frac{0.00134}{(0.0272)^4}\\\\K_{eq}=2448.1$

Hence, the value of equilibrium constant for the reaction is 2448.1

4 0

4 years ago

What is the total pressure in atmospheres in a 10.0L vessel containing 2.50 x 10-3 mol H2, 1.00 x 10-3

grigory [225]

Answer:

Total pressure = 16.42× 10⁻⁹atm

Explanation:

Given data:

Moles of H₂ = 2.50 × 10⁻³ mol

Moles of He = 1.00 × 10⁻³ mol

Mass of Ne = 3 × 10⁻⁴ mol

Volume = 10 L

Temperature = 35°C

Total pressure = ?

Solution:

Pressure of hydrogen:

P = nRT / V

P = 2.50 × 10⁻³ mol× 0.0821 atm. L.mol⁻¹ .k⁻¹ × 308 K / 10 L

p = 63.22× 10⁻³ atm. L /10 L

P = 6.3 × 10⁻³atm

Pressure of helium:

P = nRT / V

P = 1.00 × 10⁻³ mol× 0.0821 atm. L.mol⁻¹ .k⁻¹ × 308 K / 10 L

p = 25.29 × 10⁻³ atm. L /10 L

P = 2.53× 10⁻³ atm

Pressure of neon:

P = nRT / V

P = 3 × 10⁻⁴ mol× 0.0821 atm. L.mol⁻¹ .k⁻¹ × 308 K / 10 L

p = 75.86× 10⁻³ atm. L /10 L

P = 7.59× 10⁻³ atm

Total pressure of mixture:

P(mixture) = pressure of hydrogen + pressure of helium+ pressure of neon

P(mixture) = 6.3 × 10⁻³atm + 2.53× 10⁻³ atm + 7.59× 10⁻³ atm

P(mixture) = 16.42× 10⁻⁹atm

8 0

3 years ago

Please help..

Hoochie [10]

Answer:

Explanation:

Let's start with the hydrogen. If we have 4 grams of hydrogen, it would be enough for 4 * 9 = 36 grams of water. Well, that can't be possible ...

4 votes

8 0

2 years ago