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3 years ago

Which is a chemical change?

Chemistry

2 answers:

Veseljchak [2.6K]3 years ago

8 0

I believe the a is the correct answer is d

Nonamiya [84]3 years ago

3 0

Here is your answer:

Your answer will be A "ice melting"!

Reason: A chemical change is a change where a object changes form such as Ice melting because the ice is changing from a Solid to a liquid.

Another Example is iron rusting its changes form!

Your answer is A!

Hope this Helps!

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An unknown gas diffuses 0.25 times as fast as helium. What is it’s molar mass? What steps are to be done?

dmitriy555 [2]

Answer:

64.0 g/mol.

Explanation:

Thomas Graham found that, at a constant temperature and pressure the rates of effusion of various gases are inversely proportional to the square root of their masses.

∨ ∝ 1/√M.

where, ∨ is the rate of diffusion of the gas.

M is the molar mass of the gas.

∨₁/∨₂ = √(M₂/M₁)

∨₁ is the rate of effusion of the unknown gas.

∨₂ is the rate of effusion of He gas.

M₁ is the molar mass of the unknown gas.

M₂ is the molar mass of He gas (M₂ = 4.0 g/mol).

∨₁/∨₂ = 0.25.

∵ ∨₁/∨₂ = √(M₂/M₁)

∴ (0.25) =√(4.0 g/mol)/(M₁)

By squaring the both sides:

∴ (0.25)² = (4.0 g/mol)/(M₁)

∴ M₁ = (4.0 g/mol)/(0.25)² = 64.0 g/mol.

4 0

3 years ago

Iodine-131 is administered orally in the form of NaI(aq) as a treatment for thyroid cancer. The half-life of iodine-131 is 8.04

evablogger [386]

Answer:

16.6 mg

Explanation:

Step 1: Calculate the rate constant (k) for Iodine-131 decay

We know the half-life is t1/2 = 8.04 day. We can calculate the rate constant using the following expression.

k = ln2 / t1/2 = ln2 / 8.04 day = 0.0862 day⁻¹

Step 2: Calculate the mass of iodine after 8.52 days

Iodine-131 decays following first-order kinetics. Given the initial mass (I₀ = 34.7 mg) and the time elapsed (t = 8.52 day), we can calculate the mass of iodine-131 using the following expression.

ln I = ln I₀ - k × t

ln I = ln 34.7 - 0.0862 day⁻¹ × 8.52 day

I = 16.6 mg

8 0

3 years ago

Amino axit X có công thức H2NCxHy(COOH)2. Cho 0,1 mol X vào 0,2 lít dung dịch H2SO4 0,5M, thu được dung dịch Y. Cho Y phản ứng v

Taya2010 [7]

Answer:

nH2SO4 = 0,1 mol

Đặt nNaOH = a; nKOH = 3a (mol)

Quy đổi phản ứng thành: {X, H2SO4} + {NaOH, KOH} → Muối + H2O

Ta có: nH+ = nOH- → 2nX + 2nH2SO4 = nNaOH + nKOH

→ 2.0,1 + 2.0,1 = a + 3a → a = 0,1

→ nH2O = nH+ = nOH- = 0,4 mol

BTKL: mX + mH2SO4 + mNaOH + mKOH = m muối + mH2O

→ mX + 0,1.98 + 0,1.40 + 0,3.56 = 36,7 + 0,4.18 → mX = 13,3 gam

→ MX = 13,3/0,1 = 133

→ %mN = (14/133).100% ≈ 10,526%

7 0

2 years ago

Is PO4^-3 polar, non-polar, or ionic?

Gala2k [10]

The answer is PO4^-3 is non-polar.

8 0

3 years ago

Which type of reaction is the Haber process: N2(g) + 3 H2(g) → 2 NH3(g) + heat? *

Alik [6]

Answer:

exothermic, with a decrease in entropy

Explanation:

Whenever you produce heat as a product in a reaction, the reaction is exothermic. To determine entropy, we know we have 4 moles of gas on reactant (1 from N2 and 3 from H2) and in produce side we only have two moles (2 from NH3) thus since we are decreasing the number of gas molecules, there is going to be less disorder, hence decrease in entropy.

8 0

3 years ago