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3 years ago

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The temperature of a sample of water in the liquid phase is raised 30 degrees Celsius by the addition of 3762 J. What is the mas

s of the water in grams?
A. 0.03
B. 0.30
C. 30
D. 300

1 answer:

Butoxors [25]3 years ago

3 0

The answer is A because it’s how you calculate the mass

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What are the side effects of Sulfas and Macrolides?

Step2247 [10]

Answer: Minor side effects of macrolides are nausea, vomiting, diarrhea, and ringing or buzzing in the ears. Serious side effects, are allergic reaction and cholestatic hepatitis

For Sulfas the side effects are nausea, vomiting, diarrhea, and loss of appetite may occur. Serious side effects are severe allergic reaction, skin rashes, kidney, liver, or blood disorders.

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3 years ago

Read 2 more answers

The wavelength of transmission is 9.7mm. What is the energy of this transition

Sonja [21]

Answer:

E = 2.05×10⁻²³ J

Explanation:

Given data:

Wavelength of transmission = 9.7 mm (9.7/1000 = 0.0097 m)

Energy of transition = ?

Solution:

9.7 mm (9.7/1000 = 0.0097 m)

9.7×10⁻³ m

Formula:

E = hc/λ

By putting values,

E = 6.63 ×10⁻³⁴ Js × 3×10⁸ m/s / 9.7×10⁻³ m

E = 19.89×10⁻²⁶ J.m / 9.7×10⁻³ m

E = 2.05×10⁻²³ J

3 0

3 years ago

when 10.00 g of phosphorus reacts with oxygen, it produces 17.77 g of a phosphorus oxide. This phosphorus oxide was found to hav

Alex_Xolod [135]

Answer:

Molecular formula = P₄O₆

Explanation:

P(s) + O₂(g)------------------------------------⇒ PₓOₙ (g)

10g (17.77-10)g 17.77g

10g 7.77g 17.77g (gramme ratio)

The molecular mass of Phosphorus (P) = 31g/mole

The molecular mass of Oxygen atom (O) = 16g/mole

Mole ratio is given by:

P : O

10/31 7.77/16

0.3226 : 0.4856 Mole ratio---------------------------- (1)

Divide (1) through by 0.3226

1 : 1.5-------------------------------------------- (2)

From (2), the empirical formula for Phosphorus oxide :

Empirical formula = P₁O₁.₅

= PO₁.₅

The molecular formula can be calculated from below:

Since the molecular formula is a multiple of the empirical formula we have

Molecular formula = (PO₁.₅)ₙ----------------------------------- (3)

Since we are given the molecular mass of the oxide formed, we have:

(PO₁.₅)ₙ = 220-----------------------------(4)

[31 + (16 x 1.5)] x n = 220

[31 + 24]n = 220

55n =220

n = 4

Substituting into (3), we have :

Molecular formula = (PO₁.₅)₄

= P₄O₆

8 0

3 years ago

What is the empirical formula of a compound that is 24.42 % calcium, 17.07 % nitrogen, and 58.5% oxygen?

REY [17]

Answer:

$CaN_{2} O_{6}$

Explanation:

When calculating an empirical formula from percentages, assume you have a 100g sample. This allows you to convert the percentages directly to grams, because X % of 100g is X grams.

So:

24.42 % = 24.42 g Ca, 17.07% = 17.07g N, 58.5% = 58.5g O

The next step is to divide each mass by their molar mass to convert your grams to moles.

24.42/40.08 = 0.6092 mol

17.07/14.01 = 1.218 mol

58.85/15.99 = 3.680 mol

Then you will divide all of your mol values by the SMALLEST number of moles. This gives you whole numbers that are the mole ratio (subcripts) of the empircal formula.

0.6092 mol/0.6092 mol = 1

1.218 mol/0.6092 mol = 2

3.680 mol/0.6092 mol = 6

So the empirical formula is $CaN_{2} O_{6}$

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3 years ago

I need an answer now please asap and you will me marked brainiest please it's missing I need it now.

BaLLatris [955]

I really hope this helps I’ve been there too and I’m just doing this so I get the 20 words and I have the picture

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3 years ago