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3 years ago

(34.6785 x 5.39) + 435.12

Chemistry

2 answers:

Alex73 [517]3 years ago

6 0

Hope I helped

622.037115

Shalnov [3]3 years ago

5 0

To solve this, we should follow order of operations. To start, we should multiply the values inside of the parentheses.

(34.6785*5.39)+435.12
186.917115+435.12

Now, we should add the 2 values we are left with together.

186.917115
+435.120000
 622.037115

Using the math above, we can see that this expression is equal to 622.037115.

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Write the correct name for each compound. a. N2O5 b. Si2F4 c. S3F d. OF2 e. H4P6 f. C2O4 g. HF3

PSYCHO15rus [73]

Answer:

a. Boron trifluoride

b. Propane

c. Dinitrogen pentoxide

d. Carbon Dioxide

e. Silicon Octafluroride?

Explanation:

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3 years ago

5. Identify each of the following as an element, a mixture, or a compound.

VashaNatasha [74]

Answer:

Aluminum foil ~ element

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4 0

3 years ago

Box 1

Alex73 [517]

10g

Explanation:

Box 1, Mass of A = 10g

Box 2, Mass of B = 5g

Box 3, = 1A + 1B

Unknown:

Mass of B that would combine with mass of 20g of A

Solution:

Mass ratio of A to B:

$\frac{mass of A}{mass of B}$ = mass ratio

$\frac{10}{5}$ = mass ratio

The mass ratio of A to B = 2: 1

Now, number of B that will combine with 20g of A;

$\frac{mass of A}{mass of B}$ = mass ratio

$\frac{20}{mass of B}$ = $\frac{2}{1}$

Mass of B = 10g

10g of B would combine with 20g of A

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7 0

3 years ago

Read 2 more answers

Element x reacts with oxygen to produce x2o3 in an experiment it is found that 1.0000 g of x produces 1.1xxx g of x2o3 what is t

8_murik_8 [283]

Since X is 1 g, therefore O must be 0.1 g. Therefore:

moles O = 0.1 g / (16 g / mol) = 0.00625 mol

We can see that for every 3 moles of O, there are 2 moles of X, therefore:

moles X = 0.00625 mol O (3 moles X / 2 moles O) = 0.009375 mol

Molar mass X = 1 g / 0.009375 mol

Molar mass X = 106.67 g/mol

4 0

3 years ago

The partial pressure of CO2 gas in a bottle of carbonated water is 4.60 atm at 25 ºC. How much CO2 gas (in g) will be released f

frutty [35]

If the partial pressure of CO₂ in a bottle of carbonated water decreases from 4.60 atm to 1.28 atm, the mass of CO₂ released is 0.265 g.

The partial pressure of CO₂ gas in a bottle of carbonated water is 4.60 atm at 25 ºC. We can calculate the concentration of CO₂ using Henry's law.

$C = k \times P = \frac{1.65 \times 10^{-3} M }{atm} \times 4.60 atm = 7.59 \times 10^{-3} M$

We can calculate the mass of CO₂ in 1.1 L considering its molar mass is 44.01 g/mol.

$\frac{7.59 \times 10^{-3} mol}{L} \times 1.1 L \times \frac{44.01 g}{mol} = 0.367 g$

Now, we will repeat the same procedure for a partial pressure of 1.28 atm.

$C = k \times P = \frac{1.65 \times 10^{-3} M }{atm} \times 1.28 atm = 2.11 \times 10^{-3} M$

$\frac{2.11 \times 10^{-3} mol}{L} \times 1.1 L \times \frac{44.01 g}{mol} = 0.102 g$

The mass of CO₂ released will be equal to the difference in the masses at the different pressures.

$m = 0.367 g - 0.102 g = 0.265 g$

If the partial pressure of CO₂ in a bottle of carbonated water decreases from 4.60 atm to 1.28 atm, the mass of CO₂ released is 0.265 g.

Learn more: brainly.com/question/18987224

The partial pressure of CO₂ gas in a bottle of carbonated water is 4.60 atm at 25 ºC. How much CO₂ gas (in g) will be released from 1.1 L of the carbonated water when the partial pressure of CO2 is lowered to 1.28 atm? At 25 ºC, the Henry’s law constant for CO₂ dissolved in water is 1.65 x 10⁻³ M/atm, and the density of water is 1.0 g/cm³.

5 0

2 years ago